general chemistry ii lab manual - christopher kingchristopherking.name/genchemtwolab/general...

General Chemistry II Lab ManualTroy University Chemistry Faculty

Spring, 2017

Licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

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http://creativecommons.org/licenses/by-sa/4.0/



Table of ContentsSpontaneity and Reversal of Reactions..................................................................................................................3

Molecular Geometry....................................................................................................................................................... 9

Resonance Structures, Formal Charges, and Polarity...................................................................................15

Determination of Absolute Zero on the Celsius Scale...................................................................................22

Determining the Molar Mass of a Volatile Liquid by the Dumas Method............................................30

Molar Mass by Freezing Point Depression........................................................................................................37

Determining How Much Aspirin Is in Aspirin Tablets by Spectroscopy..............................................45

Value of the Equilibrium Constant for the Reaction of Iron(III) with Thiocyanate.........................54

Le Châtelier’s Principle...............................................................................................................................................64

Determining the Ka of an Acid from pH Measurements...............................................................................73

Change in pH on Adding Acid or Base to a Buffer...........................................................................................83

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Spontaneity and Reversal of ReactionsTroy University Chemistry Faculty


Objective1. Distinguish spontaneous and non-spontaneous reactions.2. Demonstrate that a non-spontaneous reaction can be made to take place if work is done

on the reaction system.

IntroductionA spontaneous reaction is one that takes place without any work having to be done. For example, manganese metal will spontaneously react with nickel(II) ion:

Mn(s) + Ni(NO3)2(aq) Mn(NO3)2(aq) + Ni(s)

Some spontaneous reactions can be made to go backwards by doing work. This lab uses a hand-held DC generator (Figure 1) to do the work. When the crank is turned, electricity is produced. Electricity can, of course, do work; this electrical work can cause a reaction to go in the opposite direction:

Work + Mn(NO3)2(aq) + Ni(s) Mn(s) + Ni(NO3)2(aq)

The chemical changes taking place are easier to see if these reactions are written as net ionic equations. (You are asked to write some of those for this lab.) Reminder: in a net ionic equation, all aqueous species are written with the ions separated; all phases are shown, and charges are given for all ions. The ionic equation for the spontaneous reaction above is:

Mn(s) + Ni2+(aq) + 2NO3–(aq) Mn2+(aq) + 2NO3

– (aq) + Ni(s)

Cancelling the spectator ions gives

Mn(s) + Ni2+(aq) Mn2+(aq) + Ni(s)

A net ionic equation for the non-spontaneous reaction can written as

Work + Mn2+(aq) + Ni(s) Mn(s) + Ni2+(aq)

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Figure 1. Genecon DC generator. The device is hand held. When the crank on is turned, the generator will produce a small electric current, creating an electrical potential of up to a 12 volts. (Source: NADA Scientific, Ltd)




Spontaneity and Reversal of Reactions

Procedure Suppose that two compounds, A and B, spontaneously react to form C and D.

A + B C + D

Now, consider what would happen if C and D were mixed. Because the reaction goes spontaneously from left to right; the reverse reaction will not spontaneously occur, so if C and D are mixed, nothing will happen, so they must be the products of the reaction.

The object of the next two experiments is to determine which pair of compounds is reactants, and which pair is products for a spontaneous reaction. Using that information, you can then write a net ionic equation for the reaction that is spontaneous.

Equipment CheckConnect the light bulb to the hand-cranked generator by connecting the

alligator clips from generator’s leads to the alligator clips on the light bulb. (It does not matter which color lead is connected to which.) Turn the generator crank, and confirm that the light bulb lights up. If it doesn’t, the instructor can provide a replacement bulb.

A Spontaneous ReactionAdding Cu(s) to ZnSO4(aq)Fill a test tube with ZnSO4(aq) solution, and place the tube in a test tube

rack. Bend a strip of Cu(s) metal into a hook shape and place it in the solution, as shown in Figure 2. After a short time—20 seconds is sufficient—remove the copper strip, examine it, and record your observations in Table 1. Be sure to note any color changes in the Cu(s) strip or in the solution. Does it appear that a chemical reaction has spontaneously occurred?

Adding Zn(s) to CuSO4(aq)Fill a test tube with CuSO4(aq) solution, and place the tube in a test tube

rack. Bend a strip of Zn(s) metal into a hook shape and place it in the solution, as shown in Figure 3. After a short time—20 seconds is sufficient—remove the zinc strip, examine it, and record your observations in Table 1. Be sure to note any color changes in the Cu(s) strip or in the solution. Does it appear that a chemical reaction has spontaneously occurred?

Write a balanced net ionic reaction for the spontaneous reaction involving the four substances used above.

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Figure 2. ZnSO4(aq) solution in test tube (left) and tube after inserting a Cu(s) strip (right).

Figure 3. CuSO4(aq) solution in test tube (left) and tube after inserting a Zn(s) strip.


A Non-spontaneous ReactionSensation of Doing WorkIn the following sections, we are going to sense the work we do with the generator and

make the non-spontaneous reaction take place by doing work on the system.

The ease of turning the generator crank depends on how much work the generator is doing. A qualitative comparison of the amount of work required to turn the crank will be made under the following conditions:

a. with nothing connected (load-free). b. with a 6 V light bulb connected to the leads.c. with the leads connected to power an electrochemical process (the reaction of

Zn(s) and Cu+2(aq)).

Generator under No LoadObtain a hand cranked DC generator. To test the frictional resistance of the generator,

separate the alligator clip ends of the wires that are plugged into the generator. Make sure the clips are not touching one another, and turn the generator crank in a clockwise direction. Note the resistance the generator offers to your turning the crank, and record your impression of this level of resistance under the “Task 1” row in the “Load Level” column of Table 2.

Generator Loaded with a Light BulbAttach the alligator clips on the

end of the leads from the generator to the leads attached to a flashlight light bulb. Turn the generator crank clockwise until the light bulb filament starts to glow. Continue turning the generator crank until you have a firmly established sensation of how hard it is to turn the crank. This assessment gives you an impression of the work needed to keep a small light bulb lighted. Record your impressions of the level of work required for this task under “Task 2” in Table 2.

Generator Loaded with Chemical ReactionPrepare a test tube as shown below. Use a large test tube, and partition it into two

chambers with a strip of cardstock paper cut to size. Fill the test tube to within 1 cm to 2 cm of the top of the tube with a solution of ZnSO4 (the previous ZnSO4 solution may be used here). Place the tube in a test tube rack. It should resemble the left portion of Figure 4.

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Figure 4. Test tube containing ZnSO4(aq) and two strips of Cu(s). A strip of cardstock divides the test tube into two separate chambers. Each chamber contains a Cu(s) strip.


The right portion of the figure shows the test tube after the two copper strips have been inserted into the two respective partitions that the cardstock creates in the test tube.

The cardstock serves to keep the two copper strips separated and prevents a short circuit that their contact would create. Make sure that the two Cu(s) strips are completely separated during the following process.

Attach the generator’s lead’s alligator clips to each of the Cu(s) strips in the test tube illustrated above. Then turn the crank of the generator in a clockwise direction. Note the force the new load has created for the power source (you!) driving the generator and record your observation concerning the load in “Task 3” of Table 2. Also record any noticeable color changes on either of the strips and in the solution. Continue turning the crank until you are certain that any chemical change that will occur does occur… this may take up to two minutes.

Disconnect the generator leads from the copper strips. Next, adjust the voltmeter to display voltage in volts. This is done by setting the maximum voltage the meter will experience; a setting of 20 DCV (direct current volts) will do the job. The voltmeter electrode tips are metallic; if they are not visible, remove their plastic covers. Press the two voltmeter electrodes down onto the two copper strips (this is easiest to do if the electrodes press each strip down onto the glass lip of the test tube). Record the voltage (it will fluctuate a bit) and your observations in table 2. Switch the voltmeter electrodes. What happens to the reading? What is another name for this apparatus? (Hint: it is portable, and is used in cell phones.)

Electrode CleanupThe Cu(s) strips need to be cleaned for use in other lab sections. To clean them, obtain

a large test tube, fill it to within 2-4 cm of the top with 1 M HCl(aq). Place the Cu(s) electrodes in the HCl. Note any visible reaction that takes place in “Task 4” of Table 2. This reaction may take a few moments to begin. When the reaction appears to be complete, pour the acid and the Cu(s) into a 150 ml beaker, remove the Cu(s) with tongs, and rinse the Cu(s) well under tap water. Return the Cu(s) to your desk area, and discard the acid in the spent acid disposal container provided. (All waste for General Chemistry II Lab goes in the hood behind the hood at the front of the room.)

Remove the black on the zinc by just wiping with a paper towel.

Turn in just the data sheets. Although you may work in pairs, each student should turn in a data sheet.

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Name: ______________________________ Lab Day & Time: _______________ Date: ______________

Data SheetA Spontaneous Reaction Table I. Results of test of reaction of metals in solutions

Metal Solution Observations Spontaneous?

Cu(s) ZnSO4(aq) Yes No

Zn(s) CuSO4(aq) Yes No

Write the net ionic equations for each spontaneous reaction observed:

A Non-spontaneous ReactionTable II. Assessing the level of work required to operate a DC Generator

Task # Task Description

Load Level

(low, medium, or high) Comments/Observations

1 Load Free – Clips not touching

2 Load to Light 6 V Flashlight Bulb

3 Load to Drive Reaction of Cu(s) in ZnSO4(aq)

Describe Color Change

Cu(s) #1

Cu(s) #2

Solution

Voltmeter Initial Reading:Reading after switching electrodes:

What this apparatus represents:

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Electrode CleanupObservations on Reactivity and Color of the Cu electrode:

What was the material on the copper strip?

What was the material on the zinc strip?

Post-Lab Questions:1. What is a “spontaneous” chemical reaction?

2. How is work related to whether a chemical reaction is spontaneous or nonspontaneous?

3. Give the net ionic equation for the reaction that takes place when the copper electrode is cleaned.

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Molecular GeometryTroy University Chemistry Faculty


Chemists can look at the formula of a simple compound and picture the compound in their minds. This ability is developed by making and examining models. This lab project will help you develop this ability. You will start by drawing a Lewis structure of a species, then make a model of it, and finally decide if the molecule is polar.

Read the material in chapter 8 of your chemistry textbook about bonding and geometry before doing this experiment. Bring your textbook to the lab for this experiment.

ProcedureFor each species in the following table, give the number of valence electrons in the species, draw the Lewis structure, and give the names of the electronic and molecular geometry of each species. Also, construct a model of each species and draw a sketch of the model.

Your sketches should have bonds in the plane of the paper drawn as plain lines, bonds coming out of the paper drawn as wedges, and bonds going behind the paper drawn as dashes. For example, CH4 could be drawn like this:

Your sketches may include or exclude double and triple bonds, as you choose. However, please omit lone electrons. Obtain your instructor’s signature on the lab pages before turning in the lab. The instructor will not sign the pages unless he or she has observed that the models were actually made.

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H

H

H

HC




Molecular Geometry

Molecular ShapeNumber of

Electron GroupsNumber ofLone Pairs Electronic Geometry Molecular Geometry

1 0 linear linear2 0 linear linear3 0 trigonal planar trigonal planar3 1 trigonal planar bent (angular)4 0 tetrahedral tetrahedral4 1 tetrahedral trigonal pyramidal4 2 tetrahedral Bent5 0 trigonal bipyramidal trigonal bipyramidal5 1 trigonal bipyramidal see-saw5 2 trigonal bipyramidal T-shaped5 3 trigonal bipyramidal linear6 0 octahedral octahedral6 1 octahedral square pyramidal6 2 octahedral square planar

From https://commons.wikimedia.org/wiki/File:VSEPR_geometries.PNG

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Molecular GeometryName: ____________________________________________________ Lab Day & Time: ______________ Date: ____________

Data SheetSpecies

Valence Electrons

Lewis Dot StructureElectronic Geometry

Molecular Geometry

Sketch

CH4

C 44H 4

8C

H

HH

H

tetrahedral tetrahedral CH

H

H

H

BeBr2

CO2

BF3

SiI4

NCl3

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Molecular Geometry

H2O

HClNo central atom, so skip this box.

AsF5

SF6

XeF2

XeF4

BrF5

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Molecular Geometry

CH2O (C is central atom)

BrF3

SiO32–

(Just 1 reson-ance form)

BrO4–

IF4–

IF2–

NH4+

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Molecular Geometry

H3O+

NO2+

O3

(Just 1 reson-ance form)

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Resonance Structures, Formal Charges, and PolarityTroy University Chemistry Faculty


IntroductionLewis structures are models representing covalent bonding between atoms. These

structures use dots around atoms to signify electrons and lines to signify bonds between atoms. Lewis structures that differ only in the arrangement of electrons are called resonance structures. The most likely Lewis structures is determined using formal charge, which is the charge of an atom in a molecule, assuming that electrons in a chemical bond are shared equally between atoms. Below are resonance structures of the carbonate anion, CO3

2-.

ProcedureFor each species in the worksheet:

1. Draw a Lewis structure. Since this lab is about resonance, nearly every structure will have a double bond.

2. Determine the formal charges on each atom of the Lewis structure. The formal charge are calculated using the formula:

FC=¿valence electrons−(¿ nonbondingelectrons+½¿bondingelectrons)3. Draw a structure with the most favorable formal charges. This may involve moving

the double bond around, or converting a single bond to a double bond, or converting something like X=M=X to X–MX or XM–X. The dominant structure has

a) the lowest formal charges, and

b) any negative formal charges on the most electronegative element.

Atoms in row 3 and beyond can have more than an octet. For such atoms, consider converting single bonds to double bonds to lower the formal change.

4. Draw all possible res1onance structures for this most likely structure.5. Include formal charges on all of your structures. Formal charges of 0 do not need to

be shown.6. Identify the electronic and molecular geometry. The tables on the following pages

may help.




Resonance Structures, Formal Charges, and Polarity

7. Construct a model of the probable structure using the model kit and show to the instructor.

8. Draw a sketch of the model. Your sketches should have bonds in the plane of the paper drawn as plain lines, bonds coming out of the paper drawn as wedges, and bonds going behind the paper drawn as dashes. (The attached table of VSEPR geometries has examples you can use.)

9. In the column labeled polarity, indicate if the molecule is polar or nonpolar. If everything attached to the central atom is the same, it is nonpolar; otherwise, it is polar. For example, CH4 is tetrahedral and non-polar; H2O has two lone pairs and two bonds, the things around it are not identical, so it is polar. Polar and nonpolar can even be assigned to ions.

10. Wait to leave until your instructor has finished checking your report to ensure that you understand everything, and that you get full credit on this lab report.

Electronegativity

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Molecular ShapeNumber of

Electron GroupsNumber ofLone Pairs

Electronic Geometry Molecular Geometry

1 0 linear linear2 0 linear linear3 0 trigonal planar trigonal planar3 1 trigonal planar bent (angular)4 0 tetrahedral tetrahedral4 1 tetrahedral trigonal pyramidal4 2 tetrahedral Bent5 0 trigonal bipyramidal trigonal bipyramidal5 1 trigonal bipyramidal see-saw5 2 trigonal bipyramidal T-shaped5 3 trigonal bipyramidal linear6 0 octahedral octahedral6 1 octahedral square pyramidal6 2 octahedral square planar

From https://commons.wikimedia.org/wiki/File:VSEPR_geometries.PNG

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Name: ____________________________________________________ Lab Day & Time: ______________ Date: ___________

Data SheetSpecies Resonance Structures Including Formal Charges

Electronic Geometry

MolecularGeometry

Sketch Polarity

NO3-

SO42-

Species Resonance Structures Including Formal ChargesElectronic Geometry

MolecularGeometry

Sketch Polarity


NO2-

SO2

N2O(N iscentral atom)


MolecularGeometry

Sketch Polarity

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SCN-

SO3

(S2CO)2-

(C is thecentralatom)


MolecularGeometry

Sketch Polarity

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IO3-

ClO4-

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Determination of Absolute Zero on the Celsius ScaleTroy University Chemistry Faculty


ObjectivesTo give a feel for how gases behave when heated.

Introduction

-300 -250 -200 -150 -100 -50 0 50 1000

0.2

0.4

0.6

0.8

1

1.2

1.4

(TH, VH)

(TL, VL)

Gas Volume vs. Temperature

Temperature/°C

Vol

ume/

L

The plot above shows how the value of absolute zero is determined experimentally. The volume and temperature of a fixed amount of gas at constant pressure are measured at two different temperatures, giving a pair of points: (Tlow, Vlow) and (Thigh, Vhigh). These two points of data are plotted. As temperature is lowered, the volume gets smaller. It looks like the volume would become zero if the temperature were made low enough. To determine at what temperature the volume would appear to go to zero, a line is drawn connecting the two data points, then the line is extended such that the volume gets less and less as the temperature gets lower and lower. Eventually, the line reaches a point where the volume would be zero. This occurs where the line touches the x-axis; the point is called the x-intercept (in contrast the y-intercept is where the line crosses the y-axis). This x-intercept corresponds to a particular temperature, which is called absolute zero. Getting colder than absolute zero makes no sense, because volume cannot be less than zero.

Extending the line in the plot beyond the range of the data points is called extrapolation. This is considered dangerous because a system may behave differently




Determination of Absolute Zero on the Celsius Scale

outside the range for which data exists. For example, gas is being treated as if it is an ideal gas, for which the gas particles have a volume of zero. Real gases have real volumes; as a real gas is cooled, it turns into a liquid, then a solid, both of which have non-zero volumes. In spite of that, this method of determining absolute zero works reasonably well, because the volume of the gas particles is small compared to the volume of the gas at higher temperatures.

Procedure:1. Prepare a boiling water bath. Add about 400 mL of tap water to a 500 mL beaker.

Heat the water to boiling on a hot plate. To heat the water quickly , set the hot plate temperature to high until the water boils, then turn it down to, say, 200 °C to keep it boiling.

2. Assemble and weigh the gas flask. Two one-hole stoppers on a glass tube with a short piece of tubing attached to the top should be provided. If not already present, attach a pinch clamp to the tubing, but don’t tightened the clamp, yet. This will be called the “stopper assembly”. Firmly attach the stopper assembly to an absolutely dry 125 mL Erlenmeyer flask; this will be the “flask assembly”. Weigh the flask assembly to at least the nearest 0.1 g. Record this data in the data sheet.

If the Erlenmeyer flask contains even a drop of water, the lab will not work. The volume of gas in the flask is to be measured in the boiling water bath and at room temperature. If the gas is water, instead of air, the gas will condense to liquid at room temperature, instead of staying gaseous.

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3. Heat the gas to the temperature of the boiling water bath. With the pinch clamp open, place the flask assembly in the boiling water bath such that the water level is even with the bottom of the bottom stopper. The flask should be at least 1 cm above the bottom of the beaker (if it touches the bottom, the gas may get hotter than the water bath) and does not touch the side of the beaker. If necessary, adjust the water level in the beaker by adding or removing water. Turn up the heat until the water boils again, and continue boiling for 5 min. Measure the temperature of the boiling water. This temperature is thigh. Record this data in the data sheet.

4. Prepare a cold water bath. Fill a plastic pan with around 4 inches of cold tap water.

5. Cool the gas to the temperature of the cold water bath. With the flask still in the boiling water bath, firmly close the pinch clamp. Using the clamp as the handle, remove the flask assembly from the beaker and completely submerge it in the pan. Make sure everything is kept below the water surface to prevent air from entering the flask.

6. Let water into the cool flask. With everything completely submerged, carefully open the pinch clamp on the plastic tubing so that some water will be sucked into the flask by the contraction of the air. (The pinch clamp stays open until step 8.) If no water is sucked in, then something was done wrong. Talk with the instructor to figure out what to do differently, and go back to the beginning.

7. Measure tlow. Keep the flask submerged for 10 min. with the clamp open (find something to weight it down, so you don’t have to hold it under water yourself). Measure the temperature of the tap water. This temperature is tlow. Record this in the data sheet. Leave the flask assembly submerged in the pan.

8. Seal the flask assembly at atmospheric pressure. Keeping the rubber tubing submerged, invert the flask so that the air is in the wide part of the flask. Lift the flask part way out of the water so that the top of the water in the flask is at the same level as the top of the water in the pan. Close the pinch clamp and remove the flask assembly from the pan. Remove the utility clamp.

In this lab only the volume and temperature are being varied; the pressure is kept constant. By having the water in the flask at the same level as the water outside the flask, the pressure of gas in the flask is the same as the pressure outside the flask, which is atmospheric pressure. When the flask was sealed in the hot water bath in step 5, it was also at atmospheric pressure, so the pressure has been kept constant.

9. Weigh the flask and water. Remove any water in the plastic tubing above the pinch clamp. This can be done by just “flicking” the tubing with a finger. Open the pinch clamp (but leave it on the tubing, so it gets weighed) so that any water in the tubing below the clamp runs into the flask. Thoroughly dry the outside of the flask assembly with paper towels. (Don’t remove the stopper assembly from the flask, yet.) Weigh the flask assembly containing water. Record this data in the data sheet.

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10. Determine the volume of water in the flask at Tlow. Subtract the mass of the flask assembly from the mass just measured to get the mass of water in the flask at the low temperature. Record this data in the data sheet. Convert this mass to volume using the density of water in the table below (you can estimate the density at the value of Tlow). Record this value in the data sheet.

Density of waterTemp / °C Density / g/mL

30 0.995725 0.997022 0.997820 0.998215 0.9991

The volume of the gas in the flask will be needed. That is the total volume of the flask assembly minus the volume of the water. That total volume will be determined in the next two steps.

11. Fill the flask assembly full of water. Place a piece of tape on the neck of the flask and mark the level of the bottom of the stopper. (That is the level of the gas when it completely fills the assembly.) Remove the Erlenmeyer flask from the stopper assembly. Fill the flask to the brim with tap water. Replace the stopper assembly and push it down to the level previously marked on the label. This should fill the glass and plastic tubing with water. If not, add water to do so. Close the pinch clamp again and remove any water in the tubing above the pinch clamp.

12. Determine Vhigh (the volume of gas in the flask assembly at Thigh). Remove the tape with the mark and thoroughly dry outside of the flask assembly. The volume of water now filling the apparatus is the same as the volume of air in the original sample. Weigh the water-filled assembly. Record this data in the data sheet. Calculate the mass of the water in the flask by subtraction; record this in the data sheet. Calculate the volume of water in the flask using density, as before. Since the water is at room temperature, Tlow is a good estimate of the temperature of the water. This volume of water is also the volume of the gas in the flask at the high temperature, so this volume is Vhigh. Record this on the data sheet.

13. Determine Vlow (the volume of gas in the flask assemble at Tlow). To determine Vlow, subtract the volume of water in the flask at Tlow from Vhigh. Record this in the data sheet.

14. Determine absolute zero graphically. Open the spreadsheet on the computer (look under “My Documents”; open the file, GC2 in Lab Calculations.xlsx). Find the “O K” tab (you may have to scroll the tabs at the bottom to find it; to scroll the tabs, click the arrows to the left of the tab names). Enter your values for Vhigh, Vlow, Thigh, and Tlow in the light yellow cells. The x-intercept of the line of your graph is the value of absolute zero on the Celsius scale. (If your value is close to the expected value, the line will also appear in the second graph,

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which is an expansion of the first graph.) Record your graphical estimate of the x-intercept on the data sheet.

15. Determine absolute zero algebraically. It works out that the volume of a gas is

directly proportional to the temperature of the gas, V high

V low=

T high

T low but only if the

absolute (Kelvin) temperature, T, is used. If temperature is measured using the

Celsius scale, t, then the relation becomes V high

V low=

thigh+Yt low+Y , where Y is the conversion

from °C to K. Solving this equation for Y gives Y=t low V high−t high V low

V low−V high. Substituting

your values into this will give a value of Y, which is the algebraic value of absolute zero on the Celsius scale, and is also the conversion constant for going between °C and K.

However, when the numbers are entered on a calculator, parenthesis should be used to ensure that the subtraction is done before the division, so you may

want to enter the equation like this: Y=(t low V high− thigh V low )

(V low−V high ).

The actual value of absolute zero on the Celsius scale is the negative of this value, so change the sign and record that value in the data sheet.

Here are the steps to solve V high

V low=

thigh+Yt low+Y for Y. First, multiply both sides by

(tlow + Y):

(t low+Y )V high

V low=

t high+Yt low+Y (T low+Y )

On cancelling, that gives (t low+Y )V high

V low=t high+Y

Expand the left-hand part:

t low(V high

V low)+Y (V high

V low)=t high+Y

Collect the Y terms alone on the right side:

t low(V high

V low)−thigh=Y−Y (V high

V low)

Combine the terms containing Y:

t low(V high

V low)−thigh=Y (1−

V high

V low)

Now there is just one Y in the equation. Solve for that Y:

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Y=(tlow (V high

V low )−t high)(1−

V high

V low)

That is one answer, but it can be greatly simplified by multiplying the top and bottom by Vlow.

Y=

V low( tlow (V high

V low )−t high)V low(1−

V high

V low)

That gives:

Y=t low V high−t high V low

V low−V high

16. Repeat the experiment with a second clean, dry flask.17. When finished, print out two copies (one for each of you) of the Excel data sheet

(click print without selecting anything to print what is needed).

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Name: ___________________________________________ Lab Day & Time: _______________ Date: _______

Data SheetRun

1 2 3

Flask assembly mass g g g

Boiling water temperature, Thigh °C °C °C

Room temperature water bath temperature, Tlow °C °C °C

Mass of flask assembly partially filled with water g g g

Mass of water in flask at Tlow g g g

Volume of water in flask at Tlow mL mL mL

Mass of flask assembly completely full of water g g g

Mass of just the water in the flask g g g

Volume of the flask filled with water, Vhigh mL mL mL

Volume of gas in the flask at Tlow: Vlow mL mL mL

Graphical estimate of absolute zero °C °C °C

Algebraic estimate of absolute zero °C °C °C

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Post Lab Questions1. What are the two main gases present in the gas samples studied in this lab?

2. This lab states that a drop of water in the 125 mL Erlenmeyer flask will spoil the experiment. Calculate the volume of a drop of water that has been converted to the gas phase by heating to 100 °C. First, use the following to get moles of water in a drop: A drop has a volume of around 0.05 mL; the density of water is 1.00 g/mL; the molar mass of water is 18 g/mol. Then, to find the volume of this as a gas, use the ideal gas law, PV = nRT, where P is the pressure, 1 atm (atmospheres), n is moles of water, R is the ideal gas law constant, 0.08206 (L·atm)/(mol·K), and T is the temperature in K (add 273.15 to the Celsius temperature).

3. Based on the answer to the previous question, explain why a drop of water in the flask would or would not have made much of a difference in this lab.

4. Suppose that VH was too large, due to, say, using a wet flask. What effect would this have on the value of Y, based on the plot on the first page of this lab?

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Determining the Molar Mass of a Volatile Liquid by the Dumas Method

Troy University Chemistry Faculty


Objectives1. Determine the density of the vapor of an unknown volatile liquid using the Dumas

method.

2. Calculate the molar mass of a volatile liquid using the ideal gas equation.

Introduction The object of this lab is to determine the molar mass of a volatile liquid—that is, a liquid

that vaporizes at a relatively low temperature. Molar mass, MM, is given by MM=massmoles ;

the mass will be measured with a balance, and the moles will be determined using the ideal

gas law: PV = nRT, so n=PVRT .

An Erlenmeyer flask will be filled with gas from the volatile liquid. To do this, a few milliliters of the volatile liquid are added to the flask, along with a boiling chip.

A liquid heated in a clean, smooth container can get superheated—its temperature can get higher than the normal boiling point. A superheated liquid tends to “bump”, which means it all boils at once, rather violently, sometimes sending liquid out of the container. Boiling chips (or stones) are used to “promote smooth boiling”. The jagged surfaces on these chips provide a place for boiling to start at.

The top of the flask is covered with aluminum foil, which has a tiny hole in it. The flask is placed in a water





bath, which causes the unknown liquid to evaporate. When it has completely evaporated, the flask will be filled with just the unknown vapor.

As the liquid evaporates, gas is formed on its surface. This gas has a larger density than air does, so it stays on the bottom, with the air floating on top of it. As more liquid evaporates, more gas is formed, pushing the air out the pin hole. (Apparently, the air is pushed out of the hole faster than the air mixes with the unknown gas in the flask.) When the air has gone, the excess gas leaves through the pin hole.

The flask is then cooled under running water, which causes the vapor to condense back into the liquid phase. The presence of this liquid causes the flask to weigh more now than when it was empty.

Example: Here is an example to show how the molar mass of an unknown liquid would be calculated. Suppose a 125 mL Erlenmeyer flask containing a boiling chip and covered with some aluminum foil is weighed and found to have a mass of 143.122 g. To this flask is added 5 mL of an unknown liquid. The aluminum foil cap (which has a pin hole in it) is replaced and the flask is placed in a beaker of boiling water. The temperature of the water bath is found to be 92 °C; that is also the temperature of the gas in the flask. The pressure in the flask is the same as the pressure in the room (because of the pinhole); that pressure is measured with a barometer and found to be 75.7 cm of Hg (it’s an old barometer). When the unknown liquid has completely vaporized, the flask is removed from the water bath and cooled under a stream of tap water.

It isn’t easy to tell when the liquid has completely evaporated. The usual way is to watch for a ring of liquid around the boiling chip. When that ring seems to not be changing any more (or, better yet, completely disappears), the evaporation is assumed to be complete.

Another way to do it is to look at the jet of gas coming out of the pinhole in the aluminum foil. The gas jet makes the image behind it shimmer. (This gas is hotter than the surrounding air, so it has a different index of refraction, so light passing through it bends a bit, and appears to shimmer.) When the shimmer goes away, no more gas is coming out of the flask.

As soon as the flask is removed from the boiling water, you’ll think, “I didn’t heat it enough, because liquid is still present,” but that is just the gas starting to condense as the flask cools. After cooling a couple of minutes, the flask is dried and found to weigh 145.233 g (that includes the boiling chip and aluminum foil). So, the mass of liquid that filled the flask when vaporized is 145.233 g – 143.122 g = 2.111 g.

This liquid contains the same number of moles as when it was a gas in the flask, which

is given by n=PVRT . The volume of the gas is the volume that the 125 mL Erlenmeyer flask

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holds, which is actually a bit more than 125 mL. (You can see that the “125 mL” mark on it is below the top of the flask.) The empty Erlenmeyer flask is filled with water, which is then transferred to a 100 mL graduated cylinder in two portions: 76 mL and 68 mL, giving a total of 144 mL, so the volume of the flask (and gas) is 144 mL.

Let us use a value of R of 0.08206 L∙atm/(mol∙K). Convert the volume to L:

? L=144 mL( 1×10−3L1 mL )=0.144 L Eq (1)

Convert the pressure to atm:

75.7 cm of Hg is 757 mm Hg

? atm=757 mm Hg ( 1atm760 mm Hg )=0.996 atm Eq (2)

Convert the temperature to kelvin: 92 °C is 365 K. (The temperature of the gas when it filled the flask was 92 °C, not room temperature.)

The number of moles of gas, then, is

n=PVRT

=(0.996 atm ) (0.144 L )

( 0.08206 L ∙atmmol ∙K )365 K

=0.00479 molEq (3)

(Can you see why the result has only 3 significant figures?)

Now, the molar mass can be calculated:

M M=massmoles

=2.111 g0.00479 mol

=441 g/mol Eq (4)

The density of a gas will also be calculated. Because the density of a gas is much less than the density of a liquid, the units used are g/L for the gas, rather than the more common g/mL. From the data above the density of the unknown liquid in the gaseous state is:

density=massvolume

=2.111 g0.14 4 L

=14.7 g/L Eq (5)

Chemical AlertVolatile unknowns are flammable, toxic, and irritant. Do not use near an open flame.

Prevent eye, skin, and clothing contact. Avoid inhaling vapor.

Procedure1. Prepare a hot-water bath by placing about 200 mL of tap water into a 400 mL

beaker. Put the beaker on a hot plate, and set the hot plate to around 250 °C. Since

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the unknown volatile liquid has a boiling point lower than 100 °C, the water does not have to boil.

2. Place a boiling stone into a dry 125 mL Erlenmeyer flask.

3. Use a ruler and scissors to obtain a piece of aluminum foil that is about a 5 cm square.

4. Place the foil over the mouth of the Erlenmeyer flask. Fold the sides down to make a tight seal. Later, when the flask is put in the water bath, the foil should not hang down into the water. To prevent that, squeeze the foil together so that it makes a little cap on the flask.

5. Use a small pin (in a container at the front of the lab) to make a small hole in the center of the foil.

6. Wipe the outside of the flask with a lab tissue to remove dust and fingerprints.

7. Determine the mass of the dry flask, boiling stone, and foil cap to the nearest 0.001 g (some of the balances only record to the nearest 0.01 g; don’t use those balances for this lab). Record this mass in the data sheet.

8. Carefully remove the foil cap, and add about 7 mL of an unknown liquid to the flask. Record the identification code of the unknown in the data sheet.

9. Replace the foil cap securely. Touch the flask and cap as little as possible.

10. Use a utility clamp to hold the flask submerged up to its neck in the water (as shown in the figure on the first page of this lab) . Suspend the flask at a slight angle so that both the liquid in the flask and the boiling stone are on one side of the flask bottom. (If necessary, tap or swirl the flask so that the boiling stone is submerged in the unknown liquid in the bottom of the flask.) Also, make certain that the water does not touch the foil cap. (On the other hand, you want as much of the flask in the water as possible, so that the gas in the flask will all be at the same temperature.)

11. Suspend a thermometer in the water near the flask (the thermometer reading would not be accurate if the probe tip is touching the bottom of the beaker). Adjust the hot plate to maintain the water bath at 85-90 °C. (To speed up heating, the temperature setting of the hot plate can be increased to, 300 °C.)

12. Allow the unknown liquid to vaporize completely. As the liquid vaporizes, its volume will decrease and bubbles form. Just before the liquid is totally vaporized, you will see a ring of liquid surrounding the boiling stone. The disappearance of the ring indicates that the vaporization is complete. Often, a tiny bit of water is present, and the ring won’t completely disappear. In that case, when the amount of liquid in the bottom of the flask stops changing, consider vaporization to be complete. It is easy to miss the complete vaporization of the unknown liquid so pay close attention to the inside of the flask. At the point of complete vaporization, record the water bath temperature in the data sheet.

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13. Immediately after the liquid completely vaporizes, remove the clamp-flask assembly from the water bath using the utility clamp as a holder. (As mentioned above, liquid will start to condense immediately; do not reheat.)

14. Cool the flask and its contents to room temperature by holding the flask by the clamp under cold running water. Be quite careful that water does not collect under the aluminum foil.

15. After the flask has cooled and the vapor condensed back into a liquid, remove the clamp from the flask.

16. Thoroughly dry the outside of the flask and foil cap using a paper towel. Wipe the outside of the flask carefully to remove fingerprints.

17. Using the same weighing scale as was used earlier, determine the mass of the flask, its contents, and the foil cap to the nearest 0.001 g. Record this mass in the data sheet.

18. Discard the liquid remaining in the flask into the waste bucket labeled “Unknowns”. Transfer the boiling stone into the container labeled “Discarded Boiling Stones”.

19. Thoroughly rinse the flask with tap water and fill it to the brim with tap water.

20. Measure the volume of the flask by measuring the volume of water that is used to fully fill the flask to the brim. Note that this volume is not equal to the volume printed in the flask. To do this, carefully pour the water in portions from the filled flask into a 100-mL graduated cylinder. Add the volumes of the portions to obtain the total volume of the flask. Record this total volume in data sheet.

21. Determine the pressure in the lab by reading the barometer in the lab. Record this pressure in the data sheet.

Calculations1. Calculate the mass of the unknown from the increase in mass of the flask.

2. Convert the water bath temperature from °C to K.

3. Convert the pressure from cm Hg to atm using equation Eq (2).

4. Convert the volume from mL to L using equation Eq (1).

5. Calculate the number of moles of unknown in the flask at the temperature of the hot water using Eq (3).

6. Calculate the molar mass using equation Eq (4).

7. Calculate the density in g/L using equation Eq (5).

8. Calculate the average molar mass.

9. Show the instructor your results. The instructor will then provide you with the chemical formula of the unknown. From the formula, calculate the accepted molar mass of the unknown.

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10. Calculate the percent error using %=(experimental value−accepted value )

accepted value×100.

22. Do a second determination by repeating steps 2-20. (Make sure the inside of the flask you use is completely dry!)

23. Do a third determination by repeating steps 2-20.

NoteThe instructor may deduct points if units are not included, and if the correct number of significant figures are not shown.

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Molar Mass of a Volatile Liquid by the Dumas Method


Data Sheet Unknown ID: ____________________________

Run

First Second ThirdMass of flask with liquid after cooling, including boiling chip and foil

Mass of empty flask, boiling chip, and foil

Mass of unknown

Water bath temperature when vaporization is complete / °C

Water bath temperature / K

Barometric pressure / cm Hg (Same) (Same)

Barometric pressure / atm

Flask volume / mL

Flask volume / L

Moles of unknown vapor / mol

Molar mass of unknown

Density of unknown vapor / (g/L)

Average molar mass of unknown

Formula of unknown (from instructor)

Accepted molar mass of unknown

Percent error

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Post Lab Questions1. The density of air is around 1.15 g/L at sea level and 35 °C. How does this relate to

why the unknown pushes air out of the flask on heating? (Hint: what density did you calculate for your unknown?)

2. For each of the following errors made in the procedure, determine the effect on the calculated molar mass, and circle larger, smaller or unchanged. (For example, since

molar mass is massmoles , if an error caused the mass to be too large or the moles too

small, the molar mass would be too large.) Explain your choice a. When the unknown liquid is added to the Erlenmeyer flask, a drop of water

(which isn’t very volatile) falls unnoticed into the flask.

Larger Smaller Unchanged

Reason:

b. The flask is heated so long that a lot of the unknown vapor escapes from the flask.


Reason:

c. The top of the flask is considerably above the water level in the water bath (i.e., the gas in the top of the flask is cooler than the water bath temperature).


Reason:

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Molar Mass by Freezing Point DepressionTroy University Chemistry Faculty


ObjectivesTo observe freezing point depression

To use freezing point depression to determine how many moles of solute are in a solution.

To use mass and moles of a substance to determine its molar mass.

IntroductionTo make home-made ice cream, the ice cream ingredients must get colder than 0 °C. This is achieved by placing the container of ingredients into ice that is sprinkled with lots of salt. The salt lowers the freezing point of the ice, so that the ice-salt water mixture is colder than 0 °C. This lowering of the freezing point of a solvent (water) when a solute (salt) is added is called freezing point depression, which is represented by ΔTf. This ΔTf, is the difference between the freezing point of the pure solvent and of the solution (for example, water and salt water). The amount by which the freezing point is lowered depends on the how many solute particles present, but not on their identity. So, a mole of NaCl would lower the freezing point by about the same amount as would a mole of KBr. The relation between number of particles and ΔTf is given by

TΔ f = kf m.In this equation kf is a constant called the “freezing point depression constant”, which only depends on the solvent; every solvent has its own value of kf. The m is a measure of

concentration, called the “molality” of the solution. Whereas molarity, M, is molesof soluteliters of solution

, molality, m, is moles of solute

kilograms of solvent . The two differences

between molarity and molality are both in the denominator: one has units of liters, the other kilograms; and one refers to the solution, the other to the solvent.

The freezing point will be determined by creating a plot called a cooling curve, Figure 2. In this plot, a solution is cooled down and the temperature is recorded as a function of time. When the solution reaches the freezing point, the temperature

Figure 1. Cooling curve, from Wikipedia, Courbe analyse thermique generique.svg





stays constant until all of the liquid has frozen, then the temperature drops again, as the solid cools. Figure 2 shows a liquid supercooling before freezing. In a very clean test tube, the solid has no place to start to form, and the solution can get colder than the freezing point, but then solid begins to form, and the temperature returns to the freezing point.

To determine freezing point depression, the freezing points of a pure solvent is determined, then a solute is added and the freezing point is determined again. The difference between these temperatures is the freezing point depression, Figure 3.

Freezing point depression can be used to determine the molality of a solution, from which the molar mass of a substance can be determined. The molar mass of a known quantity of a pure substance can be calculated by the following:

molar mass= massmoles

The mass is easily determined with a balance, but the number of moles is more difficult to determine. First, molality is determined from the freezing point depression. Simply solve eq(1) for molality:

m=∆ T f

k fEq (3)

The value of kf is given in the procedure, and ΔTf will be measured, allowing the molality to be determined. Then, using that molality, the number of moles of solute can be determined: since molality has units of mol/kg, just multiply molality by the mass of solvent in kg (which is measured using a balance:

moles of solute=m×massof solvent ∈kgThis may be clearer if m is replaced with its units:

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Figure 2. Plot of a liquid freezing showing supercooling. From Wikipedia: “Cooling curve showing supercooling.png”.

Figure 3. Freezing point depression. (Graph from logger pro). Two of the runs are for the pure solvent (freezing point about 24.8 °C. The other two are for the solvent to which an unknown has been added (freezing point about 18.2 °C). The difference between these is the freezing point depression, ΔTf.


moles of solute=(mol solutekg solvent )×massof solvent ∈kg

Finally, this number of moles of solute is substituted into Eq (2), allowing the molar mass to be calculated.

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Example Cyclohexanol is an organic solvent that has a value of kf of 39.3 °C kg/mol, which is a fairly large value. (A large value is convenient, because the larger the value of kf, the larger the freezing point depression, which makes it easier to measure.) Suppose that a test tube is filled one-third full with cyclohexanol, and is found to contain 35.5 g of that solvent. The freezing point of the pure solvent is determined by placing the test tube in a cold water bath and measuring the temperature at which the solvent appears to freeze, which is found to be 25.0 °C. Now the solvent is thawed and 4.1 g of a solute is added to the solvent. (The molar mass of the solute is to be determined.) The freezing point of the solution is measured and found to be 17.4 °C. The molality of the solution is found to be

m=∆T f

k f=25.0−17.4℃

39.3℃ kgmol

=0.193 mol /kgEq (6)

The number of moles of solute present is given by the following (the term in parentheses is the molality, with its units included).

moles of solute=(mol solutekg solvent )×massof solvent ∈kg Eq (7)

Or, equivalently,

moles of solute=m∈molkg

×mass of solvent∈kg Eq (8)

= 0.193 mol/kg × 0.0335 kg = 0.00685 mol

(Note that the mass of solvent had to be converted from grams to kilograms.) The molar mass of the solute is found as follows:

molar mass= mass solutemoles of solute

= 4.1g0.00685mol

=597 g/mol Eq (9)

ProcedureThis lab uses t-butyl alcohol, whose structure is shown on the

right. Alcohols have an –OH group attached to a carbon; “butyl” means something containing four carbons; the “t” stands for tertiary, because the central carbon is attached to three other carbons. This alcohol is very soluble in water, so if you get it on your hands just rinse it off with water. The melting point of this alcohol is rather high, so it is usually a solid at room temperature, and has to be melted before it can be used. The value of kf for t-butyl alcohol is 8.37 °C kg/mol.

Prepare a test tube containing a known mass of solventObtain a clean, dry, large test tube (the slightest bit of water in the tube will lower the

freezing point of the alcohol considerably). Record the mass of the empty test tube to the nearest hundredth of a gram (the balance that weighs to the thousandth of a gram could

41

C C H 3

O H

C H 3

CH 3


also be used). The contents of two bottles of t-butyl alcohol have been melted by leaving them in a 50 °C water bath overnight. Pour 25 mL of the alcohol into the large test tube. A graduated cylinder is inconvenient for this, because the liquid may solidify in the graduated cylinder, especially in a cold room. Instead, just pour the alcohol into the test tube to the same level as a tube containing 25 mL of water (this tube should be near the bottles of alcohol). Record the mass of the test tube with the alcohol in it. How to weigh a test tube with liquid in it? Tare a 100 mL beaker, place the test tube with solvent in that beaker to hold the tube upright, and record the mass; a 150 mL beaker weighs too much and will overload some of the balances).

Set up the Software1. Plug the USB end of the temperature probe into the computer.

2. Start the software by clicking on the “Logger Pro” icon, , on the desktop. If the

software is already running, click the “New” icon, , to reset the program (try this if the software doesn’t seem to respond).

3. Setup the software for data collection.

a. On the Main screen, click the “Data collection” icon, .b. Set the “Duration” to “360” seconds.c. Set the “seconds/sample” to “0.5” (note that this will automatically update the other field,

“sample/second”).d. Click “Done” to return to the Main screen.

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Part I. Determine the Freezing Temperature of Pure t-Butanol4. Fill a 600 mL beaker with about 400 mL of

water. Add some ice and mix until the temperature is about 10-15 °C. Place the stopper with mixing wire and temperature probe into the test tube containing the alcohol (the loop-end of the mixing wire goes into the liquid). Clamp the test tube assembly to the ring stand above the beaker (don’t lower the test tube into the water bath, yet). Warm the solvent in the test tube up to about 30 °C using the hair dryer.

5. Lower the test tube assembly into the water bath so that the liquid level in the test tube is slightly below the water level outside the test tube (so that all of the alcohol will be at the same temperature).

Note: if the tip of the temperature probe is touching the bottom or side of the test tube, the temperature won’t be that of the liquid.

Click to start data collection. (If the software is not responding, close the program and restart it.) Constantly mix the solution by moving the metal wire up and down until the alcohol is completely frozen, or until data collection stops (it will stop automatically after 6 min).

Mixing keeps the temperature the same throughout the solution, and considerably improves the appearance of the cooling curve.

6. The freezing temperature can be determined by finding the mean (i.e., average) temperature in the portion of the graph with nearly constant temperature.

Expand the graph by clicking on the “Autoscale Graph” icon, .With the mouse select (click and drag the mouse) the region with nearly constant

temperature. That region will then appear in gray. Click the “Statistics” icon, . Record the mean (average) temperature of this region as the freezing temperature of pure t-butanol.

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7. Do a second run. First, warm the test tube with a hair dryer to melt the t-butanol and raise its temperature to at least 30°C.

Second, click collect, , to start data collection. The software will ask if the last run should be kept; click yes, because both runs should be on the graph.

8. Have the instructor approve your work so far. Do not discard your t -butyl alcohol ; save that solvent for the next part. Also, do not close Logger Pro.Average the two freezing point measurements and record this mean freezing point on the data sheet.

Part II. Determine the Freezing Temperature of a Mixture of t-Butanol and an Unknown1. Weigh out about 2 grams of a solid unknown on weighing paper and record the mass.

Add it to the test tube of alcohol, above (warm the tube if the alcohol is still frozen). Make sure the unknown completely dissolves. The water bath should be about 10 °C, since the freezing point will be lowered. Determine the freezing point as previously described.

2. Estimate the freezing point of the mixture. Unlike pure t-butanol, the temperature of the mixture may gradually decrease during freezing. So, DO NOT use the average function. Instead, the freezing point is taken to be the temperature at which the mixture initially started to freeze. To estimate that temperature, click on the “Examine”

icon, . As the cursor is moved over the graph, the temperature and time values are displayed in a box on the graph. Locate the initial freezing temperature of the solution, as shown above in Figure 3, and record that as the freezing temperature in your data table.

3. Warm the mixture up about 10 °C above its current melting point and record the cooling curve one more time.

4. Print the graph., which should now have four cooling curves on the same plot.

5. Calculate the freezing point depression for each run. Use the mean freezing point for the pure solvent.

6. Calculate the molality of each solution; see equations (3) and (5). Use the value of kf for t-butyl alcohol, which is 8.37 °C kg/mol.

7. Calculate the number of moles of the unknown; see equations (4) and (6)

8. Calculate the molar mass of the unknown; see equations (2) and (7).

9. Ask the instructor for the actual molar mass. Calculate the percent error.

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Data SheetPart I, Freezing Point of Pure SolventMass of test tube with alcohol ________________

Mass of empty test tube ________________

Mass of alcohol ________________

DeterminationFirst Second

Freezing point of t-butyl alcohol

Mean freezing point

Part II, Freezing Point of the Mixture, and Molar Mass of UnknownMass of solid unknown ________________

DeterminationFirst Second

Freezing point of mixture

Freezing point depression

Molality

Moles of unknown

Molar mass of unknown

Mean molar mass ________________

Actual molar mass (from instructor) ________________

Percent error ________________

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Post Lab Questions1. When you add your unknown, if a small drop of water gets into your test tube, will

the ΔTf increase or decrease or stay the same? Why?

2. If black rubber particles dropped into the test tube during stirring, would the freezing point increase, decrease, or stay the same? Why?

46

Determining How Much Aspirin Is in Aspirin Tablets by Spectroscopy



Introduction The object of this lab is to determine how much aspirin is in aspirin tablets. The active

ingredient in aspirin is acetylsalicylic acid, abbreviated in the lab as ASA. Its structure is shown below. Originally, a compound found in bark from willow trees was found to reduce fevers. Since the willow family is called the salix family, the compound with analgesic properties was called salicylic acid. This substance is very hard on the stomach. It was found that adding an “acetyl group”, —COCH3, reduced the stomach problems. The substance with the acetyl group on it is called acetylsalicylic acid.

acetylsalicylic acid, ASA salicylic acid

The amount of ASA in aspirin tablets is less than 100%, because the tablets contain binders and, perhaps, fillers. The amount of ASA present will be determined by reacting the ASA with iron(III) ion, which makes the solution blue, and measuring the intensity of the blue.

The iron(III) ion reacts with the salicylate ion, but not with ASA, so the acetyl group has to be removed from the ASA. This is done by heating the ASA in solution with sodium hydroxide, as shown in equation 1, above. Next, Fe3+ is added, which combines with the salicylate ion to form a “complex” (a combination of a Lewis acid and Lewis base), equation 2. This complex is blue in solution.




Determining the Value of the Equilibrium Constant for the Reaction of Iron(III) with Thiocyanate

Eq (2)Blue complex

The intensity of the blue is measured using a spectrometer, Figure 1. The container holding the sample is called a cuvette. It usually has a square cross-section.

Figure 1. The light path in the spectrometer.

In the spectrometer light goes through the cuvette (the bluish square), the solution in which absorbs some of the light. The light then goes through a “diffraction grating”, which, like a prism, spreads white light out into its colors. The intensity of each color is measured with a detector (the “array detector” measures the intensity of all the colors at one time). For light having a wavelength of, say, 530 nm, the intensity of light entering the sample is called I0 (It is actually measured with the cuvette having just a blank sample in it). The intensity of light leaving the sample is called I. The amount of light transmitted by the

sample is defined as T= II 0

, where T is the transmittance. However, the quantity that is

directly proportional to concentration is something called the absorbance, A, which is given

by A = –log T¿−log ( II 0

), that is, the negative logarithm of T. Absorbance does not have any

units. The spectrometer used in the lab automatically calculates the absorbance.

This relation between absorbance and concentration is given by Beer’s Law:

A = bc,ϵ Eq (3)where c is the concentration in mol/L, and b is the path length of the light through the sample, which is the thickness of the cuvette; the path length is usually 1 cm. The ϵ is the molar absorptivity; it gives the relation between absorbance and concentration. The value of depends on the substance being analyzed and on the εwavelength. The units of are molarε –1 cm–1, which cancels out the units of b and c. According to Beer’s Law, a plot of absorbance vs. concentration is a straight line. The equation for a straight line is y = mx + b, where m is the slope and b is the y-intercept. Equation 3 doesn’t have any “+ b” part, which means that the y-intercept

48


is expected to be zero. Figure 2 shows several data points plotted on a graph. The best-fit straight line is also shown, along with the equation of the line. According to that equation, the slope of the line is 300, which, according to Beer’s Law, is εb, and, since b has a value of 1 cm, must be 300. So,ε from this plot, the molar absorptivity is found from the slope to be 300.

ProcedureCaution! This lab uses sodium hydroxide, which can cause severe eye damage. Also, hydrochloric acid is present in the iron chloride solution. Do not remove your safety goggles while in the lab.

Part 1, Prepare a Calibration PlotPrepare a Stock Solution of Salicylic AcidA stock solution is a concentrated solution from which dilute solutions are prepared.

The stock solution of salicylic acid is made from acetylsalicylic acid by heating the acetylsalicylic acid, ASA, in base (eq 1). To do this, weigh about 0.4 grams of ASA on weighing paper. Record the actual mass used on the data sheet to the nearest milligram, 0.001 g. Transfer this solid to a 125 mL Erlenmeyer flask. Using a graduated cylinder, transfer about 10 mL of 1.0 M NaOH into the Erlenmeyer flask. Heat the contents of this flask to boiling on a hot plate.

Transfer the contents of the Erlenmeyer flask to a 250 mL volumetric flask using a plastic funnel. Dilute to the mark with deionized water. Stopper the flask. Mix the contents of the flask by inverting the flask three times, swirling it each time it is inverted. This is the stock solution from which the other solutions are made. Label this solution “stock”.

Every time a volumetric flask is filled, it must be mixed by inverting and swirling as just described. Failure to mix the contents of a volumetric flask will dramatically decrease the accuracy of your results.

Calculate the concentration of this stock solution of salicylic acid in moles per liter. The molar mass of ASA is 180.2 g/mol; the number of moles of salicylic acid present after

49

Figure 2. Beer's Law plot.


heating with base is equal to the number of moles of ASA initially present. So, to get molarity, determine how many moles of ASA were weighed out, and divide by the final volume of the solution. Record this molarity on the data sheet.

Prepare Diluted Solutions of Salicylic AcidFive diluted solutions will be prepared from the stock solution. These solutions will

also contain iron(III), KCl, and HCl. The iron(III), of course, forms a complex with the salicylate, making these solutions blue. The more salicylate ion used to make the solution, the bluer the solution will be. The KCl is a source of chloride ion, which helps to keep the iron(III) in solution; the HCl also helps to keep the iron(III) in solution by reducing the concentration of OH– so that rust (Fe2O3) does not precipitate.

These five diluted solutions are prepared by diluting 5.00, 4.00, 3.00, 2.00, and 1.00 mL of the stock solution in a 100 mL volumetric flask. To start, obtain a 5 mL graduated pipet, then rinse it with about 1 mL of the stock solution to remove whatever was in there previously. (The instructor will show you the proper technique for using a pipet.) Also, obtain a 100 mL volumetric flask, and rinse it with a few mL of deionized water. Using the 5 mL pipet, transfer 5.00 mL of the stock solution to the clean 100 mL volumetric flask.

Fill the flask to the base of the neck of the flask using the solution that is labeled 0.02 M FeCl3–KCl–HCl. (This yellow solution is stored on the bench top in a large plastic container.) Then, fill a small beaker with that iron solution, and, using a disposable pipet, fill the flask up to the mark.

If you try to fill a volumetric flask to the mark using the spigot on the tank containing the iron solution, you are apt to overfill it. When filling volumetric flasks, it is a good practice to add the last few drops using a pipet or water bottle.

Stopper the flask and mix, as described previously. Store this solution by transferring the contents to an Erlenmeyer flask; label that flask “A” using a piece of labeling tape. Repeat this procedure using 4.00, 3.00, 2.00, and 1.00 mL of solution, labeling the Erlenmeyer flasks “B”, “C”, “D”, and “E”, respectively. Check that the color decreases as the amount of salicylate ion added decreases. If the color does not decrease, something is wrong with the solutions.

Calculate the concentrations of salicylate ion in these solutions. This is done using the dilution equation:

MconcentratedVconcentrated = MdiluteVdilute. Eq (4)In this equation the initial molarity is the concentration of the stock solution. Vconcentrated

is how much was taken, and Vdilute is what it was diluted to. Record these concentrations on the data sheet. (Note: these concentrations will be used in the next part, so you might as well calculate them now.)

50


Start the software (Logger Pro)1. Connect the SpectroVis Plus to a USB port on the computer.2. Double click the Logger Pro 3 icon on the desktop.

Calibrate SpectroVis PlusThe spectrometer is calibrated by having it measure how much light goes through the

solution in the cuvette when no substance to analyze is present. A solution containing everything but the substance being analyzed is called a blank. In this lab, the blank is the yellowish iron-KCl-HCl solution.

3. Choose Experiment Calibrate Spectrometer 1 from the Experiment menu. Wait until the system finishes warming up (90 seconds).

4. Meanwhile, fill a cuvette about ¾ full with the blank solution. Wipe the smooth surface of the cuvette with a tissue paper (finger prints block the light). Place the cuvette in the cuvette holder as shown, with the light going through the clear windows (the light goes through the cuvette from the white lamp symbol to the white arrowhead), and the “ribbed” windows are on the sides.

Figure 3. A cuvette being inserted into the spectrometer.

5. Click “Finish calibration” in the dialog box to complete the calibration Wait until it finishes, then click OK.

6. Discard the blank solution from the cuvette and rinse it with distilled water. Save the cuvette for the next steps.

Determine the Maximum Wavelength & Set the Data Collection Mode7. Rinse the blank cuvette twice with small amounts of a standard solution (the most

concentrated—darkest—ASA solution). Then, fill the cuvette ¾ full with that solution and place it in the spectrophotometer.

51

clearribbed


8. Set up the data collection mode by clicking the Configure Spectrometer button, . Choose “absorbance vs. concentration” as the Collection Mode. The wavelength of maximum absorbance, max, will be selected (it will be the wavelength with a checkmark next to it). It should be around 530 nm. Record this wavelength on the data sheet, where it says max. Click OK.

Collect Data (Beer’s Law Plot)9. At this point the cuvette should have been rinsed twice with the solution to be

tested, and should be about ¾ full of that solution. The cuvette should have been wiped with a tissue and should be in the cuvette holder of the SpectroVis Plus, oriented so the light isn’t going through the ribbed sides.

10. Click the button, wait a few seconds for the absorbance of the solution (shown in the box under the data) to somewhat stabilize (our spectrometers tend to have noise that never completely stops), then click “Keep”. (Don’t click Stop.) A box appears; enter the concentration of the sample (the format for entering numbers is “1.23e-4”, but use your own values) and click OK. A point will appear in the Absorbance vs Concentration plot. Record the absorbance of this sample in the data

sheet. Expand the graph by clicking on the “Autoscale Graph” icon, . If the absorbance is outside the range of 0.5 – 1.2, then the solutions have not been prepared properly.

11. Here is an easy check to make sure the instrument is properly set up: check that the blank still has an absorbance of nearly zero. To do this, fill the cuvette with the blank solution and put the cuvette in the spectrometer. Don’t click any buttons on the software; just observe what the absorbance is. If it isn’t very close to zero, then go back to step 3 and start over (you can click the “skip the warm up” button when starting over). If the blank reading is close to zero, then continue with the next step. (Checking the blank does not need to be done after each sample.)

12. Repeat the above steps for each of the remaining samples. When finished, click “Stop” to end data collection. (Don’t dispose of the samples, yet, in case something is wrong and they need to be re-measured.)

13. Click Analyze Linear Fit to see the best fit line equation for the standard solutions. Record the slope of this line on the data sheet. Recall that the equation for a straight line is y = mx + b, where m is the slope. If the y-intercept, b, of your data is a long ways from zero, then something was done wrong in the process. Record the slope, y-intercept, correlation coefficient (R2), and molar absorptivity in the data sheet (see the paragraph before the procedure for instructions on how to obtain this last value). Don’t close the software!

14. Ask the instructor to approve the data.

Part 2, Analyze Aspirin TabletsPrepare Aspirin Tablet Samples for AnalysisOnce a Beer’s Law plot is made you can use the plot to determine the concentration of

the unknown. (Don’t close the software!) First, though, the aspirin tablet has to be dissolved, and the ASA converted to salicylate ion with base. The tablet will not completely

52


dissolve, because it contains binders to hold it together, and the binder does not dissolve in water.

1. Obtain three aspirin tablets, all from the same bottle, which is on the bench at the front of the room. Record the brand name of these tablet on the data sheet. Obtain three 125 mL Erlenmeyer flasks. Label them “1”, “2”, and “3”. Place one of them on the balance, tare the balance, add one of the tablets, and record the mass on the data sheet. Do the same for the other two tablets. To each flask add 10 mL of 1.0 M NaOH. Heat the flasks to boiling, being careful not to lose any of the solution (we are trying to be quantitative here). After the solution boils a short time, it may be removed from the heat.

The following part is done with one sample at a time, because it uses a 250 mL volumetric flask.

a) Cool the sample by running tap water over the outside of the flask. Quantitatively transfer the contents of the flask to a 250 mL volumetric flask, as done previously. Also, as done previously, dilute to the mark with water, and thoroughly mix. The contents of the flask will probably be a bit cloudy, due to a binder being present. You may need to let the solids settle to the bottom before continuing.

b) Next, use a clean pipet to transfer 5 mL of the above solution from the 250 mL flask to a 100 mL volumetric flask. Dilute to the mark as before, using the 0.02 M FeCl3–KCl–HCl solution. Mix thoroughly.

c) Place this aspirin tablet sample in the cuvette holder. Choose Interpolation Calculator from the Analyze menu. A dialog box will appear, displaying the absorbance and concentration of the unknown. Record these on the data sheet. Click OK.

d) Discard the sample in the waste container under the hood.2. Repeat with the second sample and, if time permits, with the third sample.

Calculate Percent ASA in Aspirin TabletThe aspirin tablet was dissolved in 250 mL of solution. 5 mL of that solution was

diluted to 100 mL, and the concentration of that solution was measured. Use the dilution equation, equation (4), to determine the concentration of ASA in the 250 mL container. Mdilute is the concentration just recorded from the spectrometer. Calculate and record Mconc (“Concentration of aspirin in 250 mL flask”) on the data sheet.

The entire tablet is contained in the 250 mL flask, whose concentration was just calculated. Determine how many moles of ASA are in the flask (since molarity = moles/L, then moles = molarity × L; for L, use 0.250 L). Record this on the data sheet.

Convert the moles to grams using the molar mass of ASA, which is 180.2 g/mol. Record this on the data sheet.

The mass just calculated should be somewhat close to the mass the bottle says is in each tablet. Calculate the percent error by assuming that the bottle value is the true value

53


(we probably can trust the manufacturer, but who knows? Also, there may be a random variation in the amount of ASA in the product.). Percent error is given by

%=(experimental value−accepted value )

accepted value×100.

Calculate the percent ASA in the tablet (recall that percent is partwhole

×100).

Finally, calculate the mean percent ASA in the tablet (mean means average).

54



Data SheetPart 1, Calibration Plot

Mass of ASA ________________

Moles of ASA ________________

Concentration of salicylate ion in stock solution ________________

Solution

Concentration / M

Absorbance, A

A

B

C

D

E

Calibration curve: slope (m) __________ y-intercept (b) __________ correlation (R2) __________Wavelength of maximum absorbance, λmax ________________Slope of Beer’s Law plot ________________Molar absorptivity of the iron(III)—salicylate complex (M–1 cm–1) at the maximum wavelength ________________

Part 2, Analyzing Aspirin TabletsTablet brand: ____________________________

Determination

First

Second

Third

Mass of tablet

Absorbance of solution

55


Concentration of ASA in 100 mL flask / M

Concentration of ASA in 250 mL flask / M

Moles of ASA in tablet

Mass of ASA in tablet

Percent error

Percent ASA in tablet

Mean percent




Summary: Equilibrium constants are introduced. The equilibrium constant for the reaction of Fe(III) ion with thiocyanate ion is determined. The concentration of FeNCS2+ is determined from a Beer’s Law plot, constructed from solutions of known concentrations of FeNCS2+.

IntroductionChemical reactions are often treated as if they go to completion, which means that

reactants are completely converted to products (or, more properly, the limiting reactant is completely converted to product). However, all chemical reactions are actually equilibrium systems, so there is always some reactant still present at equilibrium. (For some reactions, the amount of reactant remaining at equilibrium is negligible, so the reaction can be treated as if it went to completion.)

Equilibrium is represented by a double arrow.

2HI(g) H⇌ 2(g) + I2(g)If mainly reactant is present at equilibrium, the system is said to lie close to the

reactants’ side of the reaction; likewise the system may lie close to the products’ side. The position of the equilibrium is characterized by an equilibrium constant, Kc, where the subscript “c” is for concentration. An equilibrium constant is the product of product

56





concentrations divided by the product of reactant concentrations, with each species’ concentration is raised to the power of the coefficient of that species. For equation (1) the equilibrium constant expression would be

K c=[ H2 ] [I2 ]

[HI ]2.

The square brackets, [ ], are used to indicate concentration in mol/liter, i.e ., molarity, M.

In this experiment the value of the equilibrium constant will be determined for the reaction between iron(III) ion and thiocyanate ion:

Fe3+(aq)+SCN−(aq) FeNCS⇌ 2+(aq)(Task: write an equilibrium expression for this chemical reaction and insert it in the

equilibrium constant expression row on page 65.)

Where the name thiocyanate comes from: Cyanide is CN–. It can be oxidized to form cyanate, OCN–. The oxygen can be replaced with sulfur, giving SCN–. The

presence of sulfur in a compound is indicated by “thio”, so SCN– is called the

thiocyanate ion. Its Lewis structure is :SCN:–. It has a lone pair on each end. When it combines with iron ion, it donates the lone pair on the nitrogen to the iron, which is why the iron compound is shown as FeNCS2+, instead of FeSCN2+.

When a solution containing Fe3+ ion (pale yellow) is mixed with a solution containing thiocyanate ion (colorless), deep red thiocyanatoiron(III) ion (FeNCS2+) forms. As FeNCS2+ forms, the concentrations of free Fe3+ and SCN− decrease, because every mole of FeNCS2+ that forms consumes one mole of Fe3+ and one mole of SCN−.

The value of Kc is constant (at a given temperature). So, mixtures can be prepared that have various concentrations of Fe3+ and SCN− (and FeNCS2+), but the ratio of these concentrations, as given by the equilibrium constant expression, should be a constant. In this experiment mixtures of various concentrations of Fe3+ and SCN– will be prepared, and the value of Kc will be determined for each mixture to see if it really does seem to be a constant. The equilibrium concentration of the red FeNCS2+ ion will be determined using a spectrometer. The concentrations of Fe3+ and SCN– at equilibrium can be determined from their initial concentrations, and by knowing how much their initial concentrations decrease due to the formation of FeNCS2+.

Example: Equilibrium concentrations are calculated using an ICE table, which stands for initial, change, and equilibrium. First, initial concentrations of everything are needed. Suppose a solution is made up by adding 10.0 mL of 0.500 M of Fe3+ and 3.00 mL of 0.200 M SCN– to a 100 mL volumetric flask, which is then filled to the mark with water (the numbers are just made up for this example). The initial

57


concentration of Fe3+ is calculated using the dilution equation, MdVd = McVc. The 10.0 mL was the volume concentrated; the 0.500 M was the molarity concentrated, the 100 mL is the volume dilute, and what is desired is the molarity dilute. Plugging those into the dilution equation gives 0.0500 M for the iron ion concentration in the volumetric flask. Likewise, the molarity of SCN– in the flask is 0.00600 M. No FeNCS2+ was present initially. These initial concentrations are entered on the first line of the ICE table. The concentration of FeNCS2+ at equilibrium is determined spectroscopically and is found to be 0.00100 M. That is entered on the third line. The change in concentration of FeNCS2+ is the same value (since no FeNCS2+ was present initially), which is entered on the “Change” row. Now, if [FeNCS2+] increase by 0.00100 M, then [Fe3+] must have decrease by the same amount, because the Fe3+ in FeNCS2+ had to come from somewhere. That is, the change in [Fe3+] is –0.00100 M. Likewise, the change for [SCN–] is –0.00100 M. The equilibrium concentrations of Fe3+ and SCN– are the sum of the initial values and the change. So the equilibrium concentrations of free Fe3+ and SCN– are 0.0490 M and 0.00500 M, respectively. Those values are entered in the last row of the table.

Table 1. An ICE Table

Fe3

+

(aq)

SCN−

(aq)

FeNCS2+

(aq)

Initial conc./ M: 0.0500 0.00600 0

Change in conc. at equilibrium/ M:

–0.00100 –0.00100 +0.00100

Equilibrium conc./ M: 0.0490 0.00500 0.00100

Finally, the value of the equilibrium constant is calculated from the last row of the table.

K c=¿¿.

The concentration of FeNCS2+ present in solutions will be determined using a spectrometer, which measures how much light is absorbed at particular wavelength. The amount of light absorbed is displayed as absorbance, A, which is related to the compound’s molar concentration, c, by Beer’s Law:

A = εbc In this equation b is the light path length (1 cm for the equipment in this lab) and is ε

the molar absorptivity. This indicates that absorbance is proportional to concentration. A plot of absorbance vs. concentration, which is called a Beer’s Law plot, should be linear. In this lab, five solutions of known concentrations of FeNCS2+ will be prepared. Solutions of

58


known concentrations are called standards. The absorbance of these standards will be measured, and a Beer’s Law plot will be prepared, and used to determine the concentration of FeNCS2+ in equilibrium mixtures.

To construct a Beer’s Law plot requires having solutions of known concentrations of FeNCS2+. This is tricky, because FeNCS2+ tends to dissociate into Fe3+ and SCN– ions. The trick is to have a high concentration of Fe3+; this causes essentially all of the SCN– to be converted to FeNCS2+.

To see this, look at the equilibrium constant expression you put on page 65. If [Fe3+] is made large, then, for the equilibrium constant expression to be a constant, [SCN–] would have to become very small. And, the way [SCN–] becomes low is by being converted into FeNCS2+. Nearly all of the SCN– is converted into FeNCS2+, so the equilibrium concentration of FeNCS2+ is essentially the same as the starting concentration of SCN–, but only because of the high concentration of Fe3+.

Iron3+(aq) tends to precipitate as iron oxide. The precipitation does not occur in acidic solutions. Therefore, all of the solutions used in this lab are also 0.1 M in the acid HNO3.

ProcedureHazard: As always, wear Safety glasses while performing this experiment. The nitric acid solutions can cause chemical burns.

Contamination Notes: If your flask is wet before you prepare your standard/sample solutions, it may be contaminated. To remove any contaminants, rinse the flask with a little of the 0.1 M HNO3.

Part I, Prepare Standard Solutions for the Beer’s Law PlotA standard solution is a solution of known concentration. Because these standard

solutions have high concentrations of iron, the concentration of FeNCS2+ in them is the same as the initial concentration of SCN-.

59

0 0.000002 0.000004 0.000006 0.000008 0.0000100

0.1

0.2

0.3

0.4

0.5

0.6

unknown concentration:[X VALUE] M

unknown absorbance: [Y

VALUE]

Absorbance vs. Concentration of FeNCS2+

FeNCS2+ Concentration / M

Abs

orba

nce


1. Label a 125 mL Erlenmeyer flask as “Blank”. Label five 50 mL volumetric flasks as “S1” through “S5”. (In industry, lab technicians have been fired for making up solutions, but not labeling the containers.)

A blank contains everything but the compound of interest (in this case FeSCN2+). Blanks are used to correct for interfering contaminants.

Several standard solutions will be prepared. The contents of the tubes are shown in the following table. The next three steps give directions for preparing these solutions.

Table 2. Components of Standard Solutions for Beer’s Law plot (As Prepared Below)S1 S2 S3 S4 S5

0.2 M Fe(NO3)3 / mL10.00

10.00

10.00

10.00

10.00

0.002 M NaSCN / mL1.00

2.00

3.00

4.00

5.00

0.10 M HNO3 / mLFill to the mark on the neck of the 50-mL volumetric flask

Total volume of solution / mL

50.00

50.00

50.00

50.00

50.00

2. Label a 150 mL beaker “Fe3+”. Dispense about 80 mL of 0.2 M Fe(NO3)3(aq) from a carboy into that beaker, so that the solution is conveniently available. Record the actual Fe(NO3)3 concentration, which is written on that carboy, on Data Sheet 1. Rinse the pipet before first using it, and before switching to a different solution. Pipet 10.00 mL of that solution into each of the six containers labeled in step 1. Note: if 0.002 M Fe(NO3)3 solution is used by mistake, the experiment will not work.

3. Label a 150 mL beaker “NaSCN”. Dispense about 50 mL of 0.002 M NaSCN (in 0.1 M HNO3) into that beaker from a carboy, so that the solution is conveniently available. Record the actual NaSCN solution concentrations, which is written on the carboy, on Data Sheet 1. Pipet 1.00, 2.00, 3.00, 4.00, and 5.00 mL of that solution into volumetric flasks “S1” through “S5”, respectively. If the solutions do not turn dark when the SCN– is added, then something is wrong. Record the volumes of NaSCN solution added to the volumetric flasks on Data Sheet 1. (These should be given with three significant figures, because pipettes can be quite accurate.)

4. Label a 400 mL beaker “0.1 M HNO3”. Place about 275 mL of 0.10 M HNO3 into that beaker. Using a graduated cylinder, transfer about 40 mL of that acid solution to the “Blank” flask (so the blank will have everything except the SCN– in it). Dilute the solutions in the volumetric flasks to 50 mL with the 0.10 M HNO3 by filling each to

60

Configure button Collect buttonLinear fit button


the mark with the acid solution. To prevent overfilling the flasks, transfer the last few drops of solution with a plastic dropper pipet (or, if it is available, a squirt bottle labeled “0.1 M HNO3”). Stopper the flask and mix it. The standard way to mix the contents of a volumetric flask is to invert the container, swirl it, turn it right side up and repeat for a total of three times to ensure that the flask is well-mixed. (Not mixing the solutions will cause the experiment to fail.)

5. Calculate the “initial SCN– concentration”, and the “equilibrium FeNCS2+ ion concentration” for each flask. These values are needed for the next step.

Initial SCN– concentration: use the dilution equation: MdVd = McVc, where the “c” and “d” refer to concentrated and dilute molarities and volumes (see the example calculation for help on this).

Equilibrium FeNCS2+ concentration: the concentration of iron is so large that all of the initial SCN- is converted to FeNCS2+ at equilibrium, so this value is the same as the initial SCN- concentration.

Part II, Prepare the Beer’s Law Plot1. Plug the USB cable from the SpectroVis Plus spectrometer into the computer. Start

the Logger Pro 3 software by double-clicking on the icon on the desktop. (If the software doesn’t work, try plugging the spectrometer in after the software is running.) Calibrate the spectrometer: the menu path is: Experiment/Calibrate/Spectrometer 1. The instrument will wait 90 seconds for the lamp to warm up. Obtain a cuvette. Fill it two-thirds full with your blank solution; this is the “blank cuvette”. Wipe the clear windows with a KimWipe. When the spectrometer has warmed up, the software will say “Place a blank cuvette in the device:”. Do so, ensuring that the clear windows are on the sides of the spectrometer with the white marks. Then click on the “Finish Calibration” button and, when enabled, click the “OK” button to exit that dialog box.

2. Configure the spectrometer using the solution from the flask containing the highest concentration of SCN–. Rinse the cuvette with that solution. Place the washings in a waste beaker on your table. (At the end of the lab, empty this into the “Discarded Equilibrium Mixtures” waste container.) Fill the cuvette (always about two-thirds full) with this sample solution. Place the cuvette in the spectrometer.

61


Click the “Configure” button. Change “Collection Mode” to “absorbance vs. concentration”. A mark should appear above the highest peak between 400 and 600 nm. (If it doesn’t, click on that maximum on the spectrum; the corresponding wavelength should now have a check mark in the displayed table.) The wavelength

of highest absorption is referred to as λmax (it should be around 447 nm). Click “OK”. The software asks “Do you want to store the latest run before switching Collection Mode?” “Yes” keeps the data, which looks nice, but isn’t essential.

3. Click the “Collect” button. Don’t press “Stop”! Instead, press the “Keep” button, which has appeared to the right of the “Stop” button. Enter the concentration of this sample (“S5”), and press “OK”. (The concentration can be entered as, e.g., 1.23e-4.) A point will appear in the absorbance vs. concentration plot. Don’t press “Stop”! Rinse the cuvette with solution “S4”, then fill the cuvette with that solution, wipe the cuvette (as always), insert the cuvette in the spectrometer, and wait for the reading to stabilize. Then press “Keep” again. (To spread the data out, press the “Autoscale

Graph” icon, .) Repeat for the remaining samples. When finished, click “Stop” to end data collection.

4. Look at the data: it should be on a straight line. If it isn’t, the concentrations may have been entered incorrectly. The concentrations can be edited to correct the values. The straight line should go through the origin, or close to it.

If the line does not pass close to the origin, the blank may have been done incorrectly. The blank can be reset by going back to the menu item, Experiment/Calibrate/Spectrometer 1; warm-up can be skipped.

Another possible cause of error is if the cuvette was inserted into the spectrometer with light going through the ribbed sides instead of the clear sides.

5. Click the “Linear Fit” button to see the best fit line equation for the standard solutions. On the Data Sheet, enter the parameters for this equation (that is, the slope, y-intercept, and correlation coefficient squared).

6. Have the instructor look at the graph before continuing. 7. Do not shut down the software, because the calibration plot just created will be

used to determine the concentration of FeNCS2+ in equilibrium mixtures in the following steps.

Part III, Prepare and Measure the Equilibrium MixturesSeveral equilibrium mixtures will be prepared. The contents of the tubes are shown in

the following table. The following steps give directions for preparing these solutions.

Table 2. Composition of the equilibrium mixturesE1

E2 E3 E4 E5

0.002 M Fe(NO3)3/mL 5.0

5.00

5.00

5.00

5.00

62


0

0.002 M NaSCN/mL1.00

2.00

3.00

4.00

5.00

0.10 M HNO3/mL4.00

3.00

2.00

1.00

0.00

Total volume of solution/mL

10.00

10.00

10.00

10.00

10.00

1. Obtain 5 clean, dry test tubes that can hold 10 mL of solution (18 mm × 150 mm tubes work). Label these tubes “E1” through “E5” (the “E” is for equilibrium).

2. Label a 50 mL beaker “0.002 M Fe3+”. Place about 40 mL of 0.002 M Fe(NO3)3(aq) into that beaker. Using a pipet, transfer 5.00 mL of the 0.002 M Fe(NO3)3 solution into each tube. Note: this is the dilute iron solution; if the concentrated solution is used the experiment will fail. In Data Sheet II, record the concentration of Fe(NO3)3 written on the carboys, and also the volume of Fe(NO3)3 added to each tube.

3. Pipet 1.00, 2.00, 3.00, 4.00, and 5.00 mL of the 0.002 M NaSCN solution into tubes “E1” through “E5”, respectively. In Data Sheet II, record the concentration of SCN– written on the carboys, and also the volume of SCN–added to each tube.

4. To make the total volume of each tube be 10.00 mL, 0.10 M HNO3 is added. Do this by pipetting 4.00, 3.00, 2.00, 1.00, and 0 mL of 0.10 M HNO3 solution into tubes “E1” through “E5”, respectively. Cover the tubes with Parafilm and invert them several times to mix. Check that these equilibrium solutions are lighter orange than the calibration solutions were. (If they are darker, then the wrong iron solution was used.)

5. Place sample “E1” in the spectrometer and, in the “Analyze” menu select “Interpolation Calculator”. A helper box will appear, displaying the absorbance and concentration of FeNCS2+ of the unknown. This is the equilibrium concentration of FeNCS2+ in this sample. Record that concentration in the ICE table for this sample in Data Sheet II. Click “OK”. Repeat for the other samples. Print this graph, and turn it in with the two data sheets (the previous graph does not need to be turned in).

6. Calculate the initial Fe3+ and initial SCN– concentrations using the dilution equation, as before. Record these in the ICE tables.

7. Calculate the change in FeNCS2+ by subtracting its initial concentration (zero) from its equilibrium concentration; enter this in the “Change” row of the ICE table; give the number a plus sign.

8. Enter the change in Fe3+ concentration in the ICE table. Since the FeSCN2+ concentration increased, the Fe3+ concentration must have decreased by the same

63


amount, since there is just one “Fe” in FeSCN2+. So, this value is the negative of the change of FeSCN2+.

9. Likewise, enter the change in concentration of SCN–.10. Calculate the equilibrium Fe3+ concentration by adding the initial concentration to

the change in concentration. Record this in the ICE table.11. Likewise, calculate the equilibrium SCN– concentration.12. Write the equilibrium expression for equation Fe3+(aq)+SCN−(aq) FeNCS2+(aq) ⇌

Eq (3) in the Data Sheet II (just before the ICE tables).13. Calculate the value of the equilibrium constant expression, Kc, using your

equilibrium concentrations.14. Calculate the average value of Kc.15. Discard all solutions into the “Discarded Equilibrium Mixtures” container.16. If this is the last lab section of the day, disconnect the spectrometer from the

computer (after closing the “Logger Pro” software).

64



Data SheetParts I & II, FeNCS2+ Concentrations of Standard Solution for Beer’s Law Plot

Concentration of Fe(NO3)3 in 0.10 M HNO3 stock solution ____________________ MConcentration of NaSCN in 0.10 M HNO3 stock solution ____________________ M

Flask IDS1

S2

S3

S4

S5

Volume of NaSCN in 0.10 M HNO3 added

mL

mL

mL

mL

mL

Initial SCN– concentration

M

M

M

M

M

Equilibrium FeNCS2+ concentration

M

M

M

M

M

Calibration curve: slope (m) __________ y-intercept (b) __________ correlation (R2) __________

Part III, Preparing and Measuring Equilibrium MixturesConcentration of Fe(NO3)3 in 0.10 M HNO3 stock solution ____________________ MConcentration of NaSCN in 0.10 M HNO3 stock solution ____________________ M

Test Tube ID

E1

E2

E3

E4

E5

Volume of Fe(NO3)3 in 0.10 M HNO3 added

mL

mL

mL

mL

mL

Volume of NaSCN in 0.10 M HNO3

added

mL

mL

mL

mL

mL

Equilibrium constant expression:

Kc = ___________________________

ICE Tables

Test Tube E1

65


Fe3+(aq) + SCN−(aq) ⇌FeNCS2+(aq)

Initial conc./ M: 0


Equilibrium conc./ M:

Value of Kc (no units) ___________________

66


Test Tube E2 Fe3+(aq) + SCN−(aq) ⇌FeNCS2+(aq)

Initial conc./ M: 0





Initial conc./ M: 0




Test Tube E4Fe3+(aq) + SCN−(aq) ⇌FeNCS2+(aq)

Initial conc./ M: 0





Initial conc./ M: 0




Mean Kc ________________

67


Post-Lab Questions1. Each of the five equilibrium solutions had a different color, due to different amounts

of FeNCS2+ being present. Were the values of the equilibrium constants reasonably “constant”?

68

Le Châtelier’s PrincipleTroy University Chemistry Faculty


IntroductionLe Châtelier’s principle applies to a chemical system that is initially at equilibrium. When a

change is made to such a system, the equilibrium position will shift so as to counteract the change. Suppose a solution contains the species chromate, CrO4

2-, and dichromate, Cr2O72-, in

equilibrium, with some concentration of reactants and some concentration of products present. 2CrO4

2–(aq) + 2H3O+(aq) ⇌ Cr2O72–(aq) + 3H2O(l)(aq)

yellow orangeTo this solution is added a few drops of a concentrated solution of Na2CrO4(aq). The task is to predict whether adding the Na2CrO4(aq) to the mixture will result in the reaction shifting to the left, to the right, or staying the same (that is, to predict how the equilibrium position will shift). To understand how Na2CrO4(aq) is related to this equilibrium requires recognizing that sodium salts dissociate: Na2CrO4(aq) 2Na+(aq) + CrO4

2-(aq). So, adding Na2CrO4(aq) increases the concentration of CrO42– in the

solution. According to Le Châtelier’s principle, the reaction will shift to counteract what was done to it, so the reaction will shift to the right, which will lower the concentration of CrO4

2–. It is not easy to predict the color change in this case. Adding yellow Na2CrO4(aq) to the beaker would, of course, make the solution more yellow. But, the shift in the reaction to the right would make the solution more orange.

Suppose that AgNO3(aq) was added to the original equilibrium mixture. Neither Ag+ nor NO3

– is directly involved in the equilibrium. However, in this case the silver ion will react with the chromate ion (CrO4

2–): 2Ag+(aq) + CrO42–(aq) Ag2CrO4(s). So, adding AgNO3(aq) to the

equilibrium mixture will precipitate the chromate ion, making its concentration in solution less. The reaction will shift to counteract that change, which corresponds to shifting to the left. Shifting to the left corresponds to a decrease in the concentration of Cr2O7

2–, so less orange would be present after adding AgNO3(aq).

Suppose that a few drops of concentrated phosphoric acid, H3PO4(aq), were added to the original equilibrium mixture. The problem is to figure out what H3PO4 has to do with the equilibrium being studied. There is no phosphate in that equilibrium. However, in solution H3PO4 dissociates: H3PO4 H+(aq) + H2PO4

–(aq). And, H+(aq) is the same thing as H3O+(aq), so adding H3PO4 to a solution increases the concentration of hydronium ion, H3O+. Consequently, the reaction would be expected to shift to the right, consuming some of the hydronium ion that




Le Châtelier’s Principle

was added; the solution would become less yellow, since CrO42– decreases on shifting to the

right, and more orange, since Cr2O72– increases.

Suppose a few drops of concentrated NaOH(aq) are added to the equilibrium. The equilibrium does not show OH– ion. However, hydroxide reacts with hydronium ion (that is, base neutralized acid): OH–(aq) + H3O+(aq) 2H2O(l). So, adding hydroxide ion will decrease the concentration of hydronium ion, H3O+. The reaction will shift to counteract what was done to it; that is, it will shift to the left, becoming more yellow and less orange.

For some equilibria, all of the components may be colorless, so a shift in the position of the reaction cannot be seen. However, if the reaction is an acid-base reaction, the shift may be made visible by adding an acid-base indicator. Suppose an indicator is used that is green in acid and yellow in base (this is not the indicator used in this lab). Also, suppose the following equilibrium is being examined with that indicator:

H3PO4(aq) + H2O(aq) ⇌ H3O+(aq) + H2PO4–(aq)

yellower greenerThis reaction produces hydronium ion, so if the reaction shifts to the right, the solution will become more acidic and greener. If it shifts to the left, some of the hydronium ion would be consumed, making the solution more basic, and so yellower.

Would this reaction mixture get yellower or greener if a few drops of concentrated H3PO4(aq) were added to it? Well, the H3PO4(aq) is on the reactants side, so the equilibrium would shift to the right to consume some of the added material. Shifting to the right produces more H3O+(aq), making the solution more acidic, so the indicator would become greener.

Le Châtelier’s principle also applies to chemical reactions involving heat. Heat can be added or removed from a reaction by heating or cooling the reaction mixture. If heat is added to the mixture by raising the temperature, the equilibrium will shift to produce less heat; if heat is removed from the mixture by lowering the temperature, the reaction will shift to produce more heat. This will be used to determine if a reaction is exothermic or endothermic. To do this, heat is included in the previous reaction. It is placed on one side of the reaction, then the evidence is examined to see if that is the correct side. In the following reaction, it is placed on the reactants side (that corresponds to the reaction being endothermic).

heat + H3PO4(aq) + H2O(aq) ⇌ H3O+(aq) + HPO4–(aq)

yellower greenerSuppose that on heating this system the indicator became more yellow. Adding more heat should have shifted the reaction to the right to consume some of the added heat, resulting in the solution becoming greener, not yellower. Therefore, the heat was placed on the wrong side: heat must be a product, not a reactant, and the reaction must be exothermic. The correct equation is the following:

H3PO4(aq) + H2O(aq) ⇌ H3O+(aq) + H2PO4–(aq) + heat

yellower greener

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ProcedureCaution: As always, wear safety glasses while performing this experiment. Concentrated acid, which causes burns, will be used.

PreparationHot and cold water baths:

Start heating a 150 mL beaker about half-full of tap water to near-boiling on a hot plate. (Do not boil the water so that it won’t splatter on you.)

Fill another 150 mL beaker with about half full with ice. Add water to make a slush (having water present improves cooling, because an object will have a greater contact area with a liquid than with ice).

Label four 15 × 125 mm test tubes as follows:

NaHSO4 Na2SO4 MgCl2 NaOH

Place the tubes in a test tube rack.

Equilibrium of a Weak Acid, HSO4–

HSO4–(aq) + H2O(l) H⇌ 3O+(aq) + SO4

2–(aq)

Eq (5) is the net ionic equation for the next equilibrium to be examined. The acid-base indicator used in this part is thymol blue, which becomes yellow-orange as a solution becomes more basic, and red as a solution becomes more acidic. Copy the above reaction to the data sheet. Below that reaction write the indicator color change expected after the reaction has shifted to the left and to the right, as in H3PO4(aq) + H2O(aq) H3O+(aq) + H2PO4–(aq) Eq (2⇌ ).

1. Prepare storage solutions. Transfer about 5 mL of the provided NaHSO4 solution to the test tube labeled “NaHSO4”. Transfer about 5 mL of the provided Na2SO4 solution to the test tube labeled “Na2SO4”. Place a disposable plastic transfer pipet into each tube.

2. Transfer NaHSO4 to the 24-well plate. Using the disposable pipet, transfer about 10 drops of the NaHSO4 solution into wells A1, A2, and A3, and 20 drops to wells A4 and A5.

3. Add indicator. Add 1 drop of the thymol blue indicator from the dropper bottle to each of the five wells. Mix the solution with a

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Figure 5. 24-well plate. Notice the letters along the side and numbers along the top, which allow each well to be identified. For example, the bottom right well is D6. Picture by Joseph Elsbernd, on Flickr; creative commons SA-BY.


toothpick (some people mix by swirling the 24-well plate on the desk top, but that is apt to cause a spill). More toothpicks are on the front desk. Record the color of these solutions, which are at equilibrium.

4. Add Na2SO4 to a well. With the pipet, add one drop of the Na2SO4 solution to well A1. Mix the solution with a toothpick. Probably not much change in color is observed, so add a drop more with mixing until a color change is observed. Record the color change.

For example, if the indicator had originally been pink, the color change might be “less pink” or “more pink”. If the color had been purple, the color change might be “redder” or “bluer”, or “more blue” or “less red”.

5. Analyze. Based on the color change, record on the data sheet the direction the reaction shifted on adding Na2SO4 solution to well A1. Explain the direction of the shift on the data sheet using Le Châtelier’s principle.

Here is a template for creating the explanation. “Adding/removing _______________ increases/decreases the concentration of ___________. To counteract this change, the reaction will shift to the left/right, which will increase/decrease the concentration of ____________, resulting in the indicator color becoming ________________________.”“First, what was done: adding/removing ___________ (the reagent) increases/decreases the concentration of _________________ (one of the species involved in the reaction) Second, how the system responded: The system will counteract this disturbance by shifting to the _______ (left or right); as a result, the concentration of (one of the species involved in the reaction) will decrease/increase, resulting in the indicator color becoming ________________________.”

6. Add NaHSO4 to a well. Since NaHSO4 is already in the well, to increase its concentration, solid NaHSO4 will be added. With a microspatula, add a crystal of NaHSO4 to well A3 (well A2 is the color with nothing added, for comparison). Mix the solution with a toothpick. Add crystals with mixing until a color change is observed. Record the color change.

7. Analyze. Based on the color change, record on the data sheet the direction the reaction shifted on adding NaHSO4 solution to well A3. As before, explain the direction of the shift on the data sheet using Le Châtelier’s principle.

8. Clean the pipets. Fill a beaker with tap water, and use it to rinse out the two pipets a couple of times, then rinse them with deionized water. The rinse water can be placed in the waste container in the hood that is labelled “Discarded HSO4

- mixtures”. Get as much of the liquid out of the pipets as possible.

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9. Place a mixture in the hot-water bath. Draw the solution in well A4 into a pipet. Invert the pipet, and tap it or swirl it to get all the liquid out of the stem and into the bulb. (Be careful: squeezing the pipet may expel a drop of the liquid in an unexpected direction, like towards your eye.) Place the pipet bulb-down into the hot-water bath.

10. Place a mixture in the cold-water bath. Draw the solution in well A5 into the other pipet. Likewise, get all the liquid out of the stem and into the bulb. Place the pipet bulb-down into the cold-water bath.

11. Equilibrate. Swirl the mixtures in the pipets for a minute or two to speed up the heat transfer. When the color has largely stopped changing, record the changes in color of the two solutions.

12. Reverse the process. Exchange the pipets in the two baths to see whether the color changes are permanent, or whether the process can be reversed.

13. Analyze. From the color change, determine whether the reaction shifted to the left or to the right on heating. Based on the direction the reaction shifted on heating, determine whether heat is a reactant or product in this reaction. Rewrite the net ionic equation, including heat in the reaction, as in Eq (3) and Eq (4). State whether this is an exothermic or endothermic reaction.

14. Clean up. Take the 24-well plate and a bottle of deionized water over to the hood containing the waste containers. Dump the contents of the plate into the “Discarded HSO4

- Reaction Mixtures” container. Rinse the plates with the deionized water, letting the water run into the waste container. Be careful to completely rinse out any crystals of NaHSO4 that may remain. Dry the plate with a “Lab Wipe” tissue paper.Empty the two test tubes into the waste container, and rinse with deionized water from a squirt bottle.Empty the two pipets into the waste container, and rinse them two or three times with deionized water. Get most of the liquid out of the pipets.Throw the toothpicks into the “Discarded Toothpicks” container.

Equilibrium of a Slightly Soluble Salt, Mg(OH)2

Mg(OH)2(s) Mg⇌ 2+(aq) + 2OH–(aq)

Eq (6) is the net ionic equation for this equilibrium.

1. Prepare storage solutions. Use a graduated cylinder to transfer about 10 mL of the provided 1.0 M MgCl2(aq) solution to the test tube previously labeled “MgCl2”. Transfer about 10 mL of the provided 0.5 M NaOH(aq) solution to the tube labeled “NaOH”. Place one of the disposable plastic transfer pipet that you just cleaned into each tube.

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Figure 7. Structure of EDTA (left), and a metal atom surrounded by an EDTA ion (right).


2. Transfer MgCl2 to the 24-well plate. Transfer 20 drops of the MgCl2 solution into wells B1, B2, B3, and B4. Place a toothpick in each well for future use.

Caution: As always, wear safety glasses while performing this experiment. Sodium hydroxide, NaOH is very hazardous. If you get NaOH on your skin, rinse with water until it stops feeling soapy. Likewise, HCl can cause burns. If you get HCl on your skin, rinse it off with water.

3. Transfer NaOH to the 24-well plate. Transfer 10 drops of the NaOH solution into wells B1, B2, B3, and B4. Mix with the toothpicks. Write a complete chemical reaction for what was observed in cells B1 and B2.

4. Net ionic equation for the equilibrium. In the data sheet, record the net ionic equation, Eq (6), for the equilibrium involving Mg(OH)2.

5. Add concentrated HCl to a well. The concentrated HCl is stored in the hood. Take the plate to the hood and add 1 drop of concentrated HCl to well B1. Mix the solution with a toothpick. If not much change is observed, add a drop more with mixing until a change is observed. Describe this change in the data sheet.

6. Analyze. State on the data sheet which component of the equilibrium reaction has its concentration directly changed by adding HCl(aq). Based on the observed change, record on the data sheet the direction the reaction shifted on adding conc. HCl solution to well B1. Explain the direction of the shift on the data sheet using Le Châtelier’s principle.

7. Add Na4EDTA to a well. Add 1 drop of Na4EDTA to well B2. Mix the solution with a toothpick. Probably not much change will be observed, so continue adding drops, mixing after each drop, until a distinct change is observed (this reaction may take a couple of minutes of stirring). Record the change on the data sheet.

EDTA stands for ethylene diamine tetraacetic acid, Fig. 7. It is available as a sodium salt, Na4EDTA. EDTA wraps around metal ions, like Ca2+, Mg2+, and Co2+, as shown in the structure on the right. Although the ion stays in solution, the EDTA prevents it from reacting with OH–, so in effect, EDTA lowers the concentration of Mg 2+ ion .

8. Analyze. State on the data sheet which component of the equilibrium reaction has its concentration changed by adding Na4EDTA(aq) to the mixture. Based on the observed change, record on the data sheet the direction the reaction shifted on adding Na4EDTA. Explain the direction of the shift on the data sheet using Le Châtelier’s principle.

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9. Add indicator. Add 1 drop of phenolphthalein indicator from the dropper bottle to the solutions in wells B3 and B4. Mix the solutions with the toothpicks. Record the color if this indicator solution, which is at room temperature, on the data sheet.

The acid-base indicator used in this part, phenolphthalein, becomes pink as a solution becomes more basic, and colorless as a solution becomes more acidic.

Also, below the net ionic equation for this equilibrium (step 4) write the indicator color change expected after the reaction has shifted to the left and to the right, as in H3PO4(aq) + H2O(aq) H3O+(aq) + H2PO4–(aq) Eq (2⇌ ).

10. Clean the pipets. Fill a beaker with tap water, and use it to rinse out the two pipets a couple of times, then rinse them with deionized water. The rinse water can be placed in the waste container in the hood that is labelled “Discarded Mg(OH)2 Reaction Mixtures”. Get as much of the liquid out of the pipets as possible.

11. Place a mixture in the hot-water bath. Draw the solution and solid in well B3 into a pipet. Carefully expel the liquid from the pipet back into the well to help suspend the solid in the liquid, then draw the liquid back into the pipet. Invert the pipet, and tap it or swirl it to get all the liquid out of the stem and into the bulb. (Be careful: squeezing the pipet may expel a drop of the liquid in an unexpected direction.) Place the pipet bulb-down into the hot-water bath.

12. Place a mixture in the cold-water bath. Likewise, draw the solution in well B4 into the other pipet. Place this pipet bulb-down into the cold-water bath.

13. Equilibrate. Swirl the mixtures in the pipets occasionally to speed up the heat transfer. After three or four minutes, record the changes in color of the two solutions.

14. Reverse the process. Exchange the pipets in the two baths to see whether the color changes are permanent, or whether the process can be reversed.

15. Analyze. From the color change, determine whether the reaction shifted to the left or to the right on heating. Based on the direction the reaction shifted on heating, determine whether heat is a reactant or product in this reaction. Rewrite the net ionic equation, including heat in the reaction, as in Eq (3) and Eq (4). State whether this is an exothermic or endothermic reaction.

16. Clean up. Take the 24-well plate and a bottle of deionized water over to the hood containing the waste containers. Dump the contents of the plate into the “Discarded Mg(OH)2 Reaction Mixtures” container. Rinse the plates with the deionized water, letting the water run into the waste container. Be careful to completely rinse out any solid Mg(OH)2 that may remain. Dry the plate with a “Lab Wipe” tissue paper.

Empty the two test tubes into the waste container, and rinse with deionized water from a squirt bottle.

Empty the two pipets into the waste container, and rinse them two or three times with water, and put them in the trash.

Throw the toothpicks into the “Discarded Toothpicks” container.

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Data Sheet Equilibrium of a Weak Acid, HSO4

–

Net ionic equation for this equilibrium, with change in indicator colors:

Color of the initial solution after adding the indicator: ____________________________________

The solution should be more orange if the equilibrium shifts to the _______(left or right).

The solution should be redder if the equilibrium shifts to the _______ (left or right).

Adding Na2SO4

Color change after adding Na2SO4 solution: _________________________________________

Based on the color change, the reaction shifted to the (circle one): left right

Explain the direction of the shift using Le Châtelier’s principle:

Adding NaHSO4

Color change after adding NaHSO4 solution: _________________________________________

Based on the color change, the reaction shifted to the (circle one): left right


Changing the TemperatureColor change on heating: _________________________________________

Color change on cooling: _________________________________________

On heating, the reaction shifted to the (circle one): left right

Net ionic equation for this equilibrium, with change in indicator colors, and with heat as a reactant or product:

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According to the reaction above, this process is (circle one): exothermic endothermic

Equilibrium of a Slightly Soluble Salt, Mg(OH)2

Adding HClChemical reaction for formation of a precipitate on adding NaOH(aq) to MgCl2(aq).

Net ionic equation for the Mg(OH)2 equilibrium.

Change on adding conc. HCl to a mixture containing Mg(OH)2(s): ________________________

Component of the equilibrium reaction whose concentration is most directly changed by adding HCl to the mixture: __________________________________

Direction the reaction shifted on adding HCl (circle one): left right


Adding Na4EDTAChange on adding Na4EDTA solution: _________________________________________

Component of the equilibrium reaction whose concentration is directly changed by adding Na4EDTA to the mixture: __________________________________

Direction the reaction shifted on adding Na4EDTA (circle one): left right


Changing the TemperatureColor of indicator in the room temperature solution: ___________________________________Color change on heating: _________________________________________Color change on cooling: _________________________________________On heating, the reaction shifted to the (circle one): left right

Net ionic equation for this equilibrium, with change in indicator colors, and with heat as a reactant or product:

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According to the reaction above, this process is (circle one): exothermic endothermic

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Determining the Ka of an Acid from pH MeasurementsTroy University Chemistry Faculty


IntroductionStrong acids, like HCl(aq), completely dissociate:

HCl(aq) + H2O(l) H3O+(aq) + Cl–(aq)

On the other hand, weak acids, like acetic acid, only partially dissociate; the dissociation can be written as an equilibrium:

HC2H3O2(aq) + H2O(l) C⇌ 2H3O2–(aq) + H3O+(aq)

Because this is an equilibrium, an equilibrium constant is associated with it:

K a=¿¿.

Determining the numerical value of Ka in this lab requires knowing the concentration of H3O+, [H+]. The range of hydrogen ion concentrations that are routinely encountered in the laboratory is enormous: from around 18 M for concentrated acids to 1 10-15 M for concentrated bases. To handle such a wide range of values, [H+] is commonly reported as “pH”, which is defined as the negative logarithm of [H+]:

pH = –log([H+])The pH of a solution can be measured with a pH meter that uses “glass electrodes” to measure a voltage in a solution. The voltage is proportional to the logarithm of [H+], so it was convenient to define a pH scale in terms of logarithms.

The logarithm is the exponent needed to express a number as a power of 10. For example, 0.001 M can be written as 1 10-3 M. The logarithm of 0.001 is –3.0, so the pH of a solution that is 0.001 M is 3.0 (pH has no units). Because fractional exponents are possible, nearly any positive number can be written as an exponent. For example, 0.025 can be written as 10–1.60, so a solution that is 0.025 M has a pH of 1.60. To show this with a calculator, enter .025 and push the log button (not the ln button), and the display should show –1.602. Then, push the 10x key, and the display should show the original number of 0.025. (The key strokes are a little different if you have a graphing calculator.)

In this lab pH will need to be converted back to [H+]. The formula for that is

[H+] = 10-pH




Determining Ka from pH Measurements

Determining Ka from the pH of an Acid Solution of Known ConcentrationThe chemical reaction for this equilibrium is:

HA(aq) + H2O(l) H⇌ 3O+(aq) + A–(aq) Ka=¿¿

Example 1: Suppose a solution is prepared by dissolving an unknown acid in water such that the molarity of the acid is 1.00 M. The pH of this solution is measured and found to be 2.36. The equilibrium concentrations of all species will be determined using an ICE table. The initial molarity of HA is 1.00 M. No information is given about the initial concentrations of H3O+ and A–, so their initial concentrations are assumed to be zero. The equilibrium molarity of H3O+ is found using Eq (2) to be 0.0044 M. Although it is not obvious, the equilibrium concentration of A- is also 0.0044 M. To see this, consider the H3O+: it was initially 0, but increased to 0.0044 M; it had to come from somewhere. It came from dissociation of the HA. But every HA that dissociates also produces an A-, so the equilibrium concentrations of H3O+ and A– must be equal, since they were both zero initially.

Table 3. An ICE TableHA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M: 1.00 0 0

Change in conc. at equilibrium / M:

Equilibrium conc. / M: 0.0044

0.0044

The ICE table can then have the changes entered. The change in reactant HA is opposite in sign of the change in the two products. That is, if [A–] increased, then [HA] must have decreased.

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M: 1.00 0 0


–0.0044 +0.0044

+0.0044

Equilibrium conc. / M: 0.9956 0.0044

0.0044

The equilibrium concentration of HA can then be calculated by adding the change to the initial concentration, giving 0.9956 M.

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Now that all the equilibrium concentrations are known, the value of Ka can be calculated from the three equilibrium concentrations:

Ka=¿¿

Ka from Partial NeutralizationThe above determination of Ka is not very accurate, because a small error in the

measurement of pH makes a large error in the value of Ka. Another approach is to start with varying initial concentrations of both HA and A–, which are obtained by partially neutralizing some of the initial HA. If, say, four different solutions were prepared by neutralization, then four values of Ka would be obtained. These four values could be averaged, which would reduce random errors and give a better estimate of the actual value of Ka than a single value would.

The neutralization involves adding some sodium hydroxide to the unknown acid, HA, which converts some of the acid to its deprotonated form, A–:

HA(aq) + OH–(aq) A–(aq) + H2O(l) Moles before reaction: 6 2 0Moles after reaction: 4 0 2

As an example, if 2 moles of NaOH and 6 moles of HA were combined, only 4 moles of HA would be left, the hydroxide would be used up, and the number of moles of A– formed would be equal to the number of moles of hydroxide added.

Example 2: Suppose 50.00 mL on 1.00 M unknown acid is dispensed from a buret into a 100 mL volumetric flask. Then, 12.50 mL of 0.500 M NaOH is dispensed from another buret to the volumetric flask. The flask is filled to the mark with deionized water, and its pH is measured and found to be 4.46. Determine the Ka of the acid.

The first entry in an ICE table is the initial concentrations. No A– was added to the mixture, but some was formed from the HA reacting with NaOH, as shown in eq (3). The moles of A– initially present is equal to the moles of OH– added. The moles of OH– added is

calculated using the definition of molarity, M= molvolume∈L , which, on solving for moles

gives mol = M × volume in L. Applying that formula gives

mol of OH– added = M NaOH × L of NaOH added

¿ 0.500 molL

× 12.5 mL× 1 L1000mL

=0.00625 molof OH−¿¿

mol of A– = mol of OH– added = 0.00625 mol A–

The initial molarity of A– is the moles of A– initially present divided by the total volume of the solution, which was 100 mL, which is 0.100 L, so the molarity is as follows:

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molarity=molL

=0.00625 mol0.100 L

=0.0625 M A−¿¿

Next, the initial molarity of HA is needed. The number of moles of HA transferred from the buret to the volumetric flask is

mol of HA added = M HA × L of HA added

¿ 1.00 molL

×50.00 mL× 1L1000 mL

=0.0500 mol

According to Eq (3), the number of moles of HA present is reduced by the number of moles of NaOH added (the 0.00625 mol calculated above)

mol of HA remaining = starting mol of HA – mol of NaOH added

= 0.0500 mol – 0.00625 mol = 0.0438 mol

(The result was rounded to 3 significant digits.)

To get the molarity of the remaining HA, divide the moles by the volume:

molarity of HA=molL

=0.0438 mol0.100L

=0.438 M HA

This is the “Initial” molarity of HA used in the ICE table.

The last value needed is the equilibrium concentration of [H+]. This is obtained from the pH using Eq (2): [H+] = 10–pH = 10-4.46 = 3.5 10-5.

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M: 0.438 0 0.0625


Equilibrium conc. / M: 3.5 10-5

Now that initial concentrations after neutralization have been calculated, the changes in concentrations can be determined. And, from those, the equilibrium concentrations of HA and A– are determined. (Because the change in concentrations of those two are so small, their equilibrium concentrations are essentially equal to their initial concentrations.)

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

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Initial conc. / M: 0.438 0 0.0625


–3.5 10-

5

3.5 10-5

3.5 10-5

Equilibrium conc. / M: 0.438 3.5 10-5

0.0625

Finally, the equilibrium constant can be calculated.

Ka=¿¿This value should be somewhat close to the value obtained in the previous section (they are not very close because all the numbers are made up).

ProcedureCalibration of the pH meter

1. Obtain pH 4 and pH 7 standardization buffers in 50-mL beakers by pouring from the bottles into the beakers until the beaker is about half full. Obtain a 100-mL beaker to hold rinse water.

2. Plug the pH electrode USB cable into the computer. Start the LoggerPro software by double clicking on the icon.

3. Leave the pH electrode in the electrode holder that is attached to an iron stand (leaving it in the holder frees up a hand). The electrode is stored in a storage solution. Remove the cap from the electrode. Place the 100-mL rinse water beaker under the electrode and rinse the tip of the electrode with deionized water.

The tip of the electrode is made of glass and is fragile. This glass tip produces a tiny voltage that is inversely proportional to the pH of the solution. Do not touch the tip of the electrode; oil from the fingers may contaminate the glass.

4. Blot the electrode dry with a lab tissue. (That way, the extra liquid won’t dilute the solution being examined.) Place the electrode in the pH 4 buffer. Wait for the pH reading (at the lower left corner on the screen) to stabilize. Swirling the beaker in a circle seems to help stabilize the reading.

5. Click “Experiment” >>> “Calibrate” >>> “Go!Link: 1 pH” as shown in the following figure. A window will pop up.

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Figure 6. pH Electrode in a solution. The electrode holder frees the hands. The storage bottle has an o-ring that stays with the bottle.


6. In the pop-up window, click “Calibrate Now” button. Then, enter “4.0” in the blank below “Reading 1”, and click the “Keep” button below it.

7. Rinse the tip of the pH electrode with deionized water, dry it with a lab tissue, and place the electrode in the pH 7 buffer. The pH will not update, but the voltage will display in the dialog box. Swirl the beaker in a circle until the voltage reading stops changing.

8. Then, enter “7.0” in the blank below “Reading 2”, and click the “Keep” button below it.

9. Click the “Done” button. Now, you’ve finished the calibration of the pH electrode. Typically, a pH electrode only need to be calibrated once a day, which means the calibration you just did will be good for the whole lab and you don’t need to calibrate the pH electrode again in this lab period.

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Initial Preparations1. Obtain the Unknown acid. Into a clean, dry 250 mL Erlenmeyer flask dispense

about 180 mL of the unknown acid from the carboy labeled “Unknown Acid”. On the report sheet record the concentration of this unknown, which is also written on the carboy. Label the flask “X” (for unknown) (labeling tape may be on the front desk).

2. Obtain sodium hydroxide. Into a clean, dry 250 mL Erlenmeyer flask dispense about 140 mL of the NaOH solution from the carboy labeled “NaOH”. On the report sheet record the concentration of this solution, which is also written on the carboy. Label the flask “NaOH”.

Part I, Ka from the pH of an Acid Solution of Known Concentration3. Measure the pH of the unknown acid. Fill a 50 mL beaker about half-full with the

solution of 1.0 M unknown acid. Record the actual molarity of the acid, which is written on the carboy. Read the pH and record it on the data sheet. Remove the electrode from the solution and thoroughly rinse it with distilled water using the wash bottle collecting the waste into the beaker. Discard the solution in the beaker into the waste container. Rinse the beaker with water.

4. Calculate Ka. Following the example calculation in the introduction, fill in the ICE table on the data sheet, and calculate the Ka of the unknown acid.

Preparation of Solutions5. Clean two 50 mL burets and a 250 mL volumetric flask by rinsing them out a few

times with around 10 mL of deionized water. 6. Rinse buret with unknown acid solution. Rinse one of the burets out with around

5 mL of the unknown acid solution. Tip the buret around so that the liquid contacts most of the buret wall. Pour the liquid out. Repeat with two more 5 mL portions.

7. Adjust liquid level. Place the buret in the buret holder. Place a waste beaker under the buret to catch any liquid runoff if the buret is overfilled (which is easy to do). Using a funnel, fill the buret to above the 0 mL level with the unknown acid solution. Remove any air bubbles from the buret tip (to do this, give the buret a little downward “jerk” while the stopcock is open). Dispense liquid until the liquid level is below or at the 0 mL mark.

Do not try to get the liquid level exactly at the 0 mL mark. It is more accurate to record the actual initial level, rather than to try to get it exactly at zero, and call it “close enough”.

Remove the hanging droplet from the buret tip by touching the tip to the inside wall of the waste beaker. Determine this initial liquid level to 0.01 mL and record the buret reading on report sheet 1.

The size of the droplet on the tip of the buret is supposed to be the same at the start and at the end of dispensing liquid, so after dispensing liquid from

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the buret to another container, touch the tip to the inside wall of that container.

Caution: 0.5 M NaOH is caustic. If you get it on your hands, rinse them with water for at least a couple of minutes, until the skin no longer feels soapy. Wear goggles all the time when in the lab.

8. Rinse buret with NaOH; adjust liquid level. Rinse and fill the other buret with the NaOH solution following the same procedure as before. Determine the initial liquid level to the nearest 0.01 mL and record this value in the report sheet.

Part II, Ka from Partial Neutralization9. Prepare solution 1. Prepare solution 1 by adding about 40 mL of the unknown

acid solution from the buret to the 250 mL volumetric flask (cleaned in step 1). Determine the actual buret reading after delivering the liquid and record that level on the report sheet to the nearest 0.01 mL.

Then add about 12.5 mL of the NaOH solution from the other buret. Read the buret to the nearest 0.01 mL and record the reading on the report sheet.

Dilute the contents of the 250 mL flask in two steps: First, dilute to the base of the neck with deionized water, then swirl to mix (or invert a few times). Second, fill up to the mark with deionized water from a wash bottle so that the bottom of the meniscus is at the same level as the mark. Stopper the flask. Invert the flask, swirl, turn back upright; repeat a total of three times.

You might wonder why the flask was not filled to the mark in one step. The reason is that the final volume when water and a concentrated solution are mixed may be more or less than the sum of the volumes. Suppose water was poured on top of a concentrated solution, and the flask was filled to the mark. On mixing the solution, the volume may increase (or decrease) a little: the flask may suddenly be above the mark. The behavior is avoided by filling with water in two steps: the first step dilutes the solution, so that water is added to a dilute solution in the second step, so the volume does not noticeably change on mixing.

Next, the pH of this solution will be measured, then three more solutions will be prepared.

10. pH of solution 1. Pour about 50 mL of solution 1 into a clean, dry 100-mL beaker. Place the glass electrode in this solution. Read the pH and record it on Data Sheet 1. Remove the electrode from the solution and thoroughly rinse it with distilled water using the wash bottle collecting the waste into the beaker. Discard the solution in the beaker and in the volumetric flask into the waste container. Thoroughly wash the beaker and flask with distilled water. Keep the electrode from drying out by putting it in water in another beaker.

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11. Prepare solution 2. Use the amounts of reactants given below to prepare solution 2.Solution Number

Volume of 1.0 M unknown weak acid /

mL

Volume of 0.5 M NaOH / mL

1 40.0 12.52 40.0 25.03 40.0 37.54 40.0 50.0

12. Measure the pH of solution 2 as in Step 10. Record this pH on Data Sheet13. Prepare solutions 3 and 4, and measure their pHs. Repeat the previous step two

more times to prepare solutions 3 and 4.14. Calculate the Ka as shown in the example in the “Ka from Partial Neutralization”

section of the introduction.15. Clean up. Discard all solutions into the waste container and rinse all glassware

used. Replace the glass electrode into the buffer solution. Log off of the computer.

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Data SheetConcentration of unknown acid stock solution ____________________ MConcentration of NaOH stock solution ____________________ M

Part I Ka from the pH of an Acid Solution of Known ConcentrationpH of the unknown acid stock solution ____________________

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M:


Equilibrium conc. / M:

Value of Ka ____________________

Part II, Ka from Partial Neutralization Run

Unknown Acid 1 2 3 4Initial buret reading

mL

mL

mL

mL

Final buret reading mL

mL

mL

mL

Volume dispensed into flask

mL

mL

mL

mL

NaOH

Initial buret reading mL

mL

mL

mL

Final buret reading mL

mL

mL

mL

Volume dispensed into flask

mL

mL

mL

mL

Volume of volumetric flask in which the acid

m

m

m

m

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and base were diluted L L L L

pH of the mixture

Initial Concentrations after neutralizationRun

1 2 3 4Moles of NaOH added

mol

mol

mol

mol

Moles of A– present mol

mol

mol

mol

A– molarity M M M M

Moles of HA added mol

mol

mol

mol

Moles of HA present after neutralization

mol

mol

mol

mol

HA molarity M M M M

H3O+ molarity (from pH) M M M M

ICE Tables

Run 1

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M:



Value of Ka ____________________Run 2

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M:

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Value of Ka ____________________Run 3

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M:



Value of Ka ____________________Run 4

HA(aq) + H2O(l)

⇌ H3O+

(aq)

+ A–(aq)

Initial conc. / M:



Value of Ka ____________________Average Value of Ka ____________________

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Change in pH on Adding Acid or Base to a BufferTroy University Chemistry Faculty


Introduction A buffer is a solution that resists change in pH when acid or base is added. That is, if

acid or base is added to a buffered solution, the pH only changes slightly. (A buffered solution also resists change in pH when it is diluted.)

A buffer consists of a weak acid and its conjugate base. To understand how a buffer works, consider the dissociation of a weak acid, represented as HA:

HA(aq) + H2O(l) H⇌ 3O+(aq) + A–(aq) Eq (1)acid conjugate base

Suppose a buffer is prepared by dissolving both HA and A– in a beaker containing water. (The A– is typically added as a salt, e.g., NaA.) To this buffer solution is added the strong acid, HCl, which dissociates to form H+(aq) and Cl–(aq). The H+(aq) from the strong acid then reacts with the A–(aq) from the buffer, forming HA(aq), which is a weak acid.

A–(aq) + H+(aq) HA(aq)⇌So, the strong acid HCl is converted to the weak acid, HA. And, since weak acids dissociate less than strong acids, the solution is less acidic than it would be without the A– present.

Likewise, if the strong base, NaOH, is added to the buffer solution, the hydroxide (OH–) reacts with the HA(aq) from the buffer to form A–(aq), which is a weak base.

HA(aq) + OH–(aq) A⇌ –(aq) + H2O(l)So, the strong base NaOH is converted to the weak base, A–(aq). And, since weak bases produce less hydroxide ion than strong bases do, the solution will be less basic than it would be without the HA present.

A–(aq) + H2O(l) OH⇌ –(aq) + HA(aq)The pH of a buffer is determined by the molarity ratio of the weak acid and its

conjugate base, and can be calculated by the Henderson-Hasselbalch equation:

p H=p Ka+log ¿where Ka is the dissociation constant of the weak acid.

In this lab, the resistance of a buffer to change in pH will be demonstrated by measuring the pH of two sets of solutions: one set will be prepared by adding acid and base to water; the other set by adding acid and base to a buffer.




Change in pH on Adding Acid or Base to a Buffer

ProcedureCalibration of the pH meter

1. Obtain standardization buffers. Obtain two 50 mL beakers. Fill one half-full with the pink pH 4 standardization buffer (in a bottle, probably located on the side table), and fill the other one half-full with the blue pH 10 standardization buffer. Also, obtain a 100 mL beaker to hold waste rinse water.

2. Start LoggerPro. Plug the pH electrode USB cable into the computer. Start the LoggerPro software by double clicking on the icon.

The tip of the pH electrode contains a glass bulb that produces a tiny voltage that is inversely proportional to the pH of the solution. This glass tip has to be kept wet, so it is stored in a buffer solution. If the glass tip dries out, it stops responding to pH until it is rehydrated by soaking in a buffer solution overnight. The glass tip should not be touched so that oil from the fingers does not contaminate the glass. The glass tip is very fragile, so it is protected by a plastic guard tip that should not be removed. PLEASE BE VERY CAREFUL!

3. Setting up the pH electrode. The pH electrode is held in a plastic “electrode holder” that is attached to a ring stand. The electrode should be in the storage bottle. Loosen the lid and pull it and the bottle down to remove them from the electrode. When you pull the bottle lid off of the electrode, try to leave the plastic guard tip attached to the electrode. Leave the pH electrode in the electrode holder that is attached to a ring stand (leaving it in the holder frees up a hand). Lift the electrode up (it slides freely in the electrode holder) and place the empty waste rinse water beaker under the electrode. Lower the electrode into the beaker. Adjust the electrode holder on the ring stand so that the tip of the electrode is close to but not touching the bottom of the beaker. After that, leave the electrode holder firmly attached to the ring stand; to raise the electrode, just lift it up in the holder.

4. Place the electrode in first buffer solution. Lift the electrode up an inch or two, and rinse its tip with deionized water (the waste rinse water beaker collects this water). Blot the electrode dry with a lab tissue. (That way, the extra liquid won’t dilute or contaminate the solution being examined.) Place the electrode in the pH 4 buffer. (No need to move the electrode holder!) LoggerPro should display the pH at

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Figure 7. pH electrode in a solution. The electrode holder frees the hands.


the lower left corner of the screen. The pH should be about 4.0. Wait for the pH reading to stabilize. Swirling the beaker in a circle seems to help stabilize the reading.

5. Calibrate with the pH 4 buffer. Click “Experiment” >>> “Calibrate” >>> “Go!Link: 1 pH” as shown in the following figure. A window will pop up. In the pop-up window, shown below, click the “Calibrate Now” button. Then, enter “4.0” in the blank below “Reading 1”, and click the “Keep” button below it.

6. Calibrate with the pH 10 buffer. Rinse the tip of the pH electrode with deionized water, dry it with a lab tissue, and place the electrode in the pH 10 buffer. The pH will not update, but the voltage will display in the dialog box. Swirl the beaker in a circle until the voltage reading stops changing. Then, enter “10.0” in the blank below “Reading 2”, and click the “Keep” button below it. Click the “Done” button. Now, you’ve finished the calibration of the pH electrode. Typically, a pH electrode only needs to be calibrated every few hours, which means the calibration you just did will be good for the whole lab.

Prepare a buffer solution1. Label a 250 mL Erlenmeyer flask as “buffer stock solution”.2. Using a clean 10 mL graduated cylinder, transfer 10.0 mL of the 1.0 M acetic acid

stock solution to the buffer flask. Rinse the graduated cylinder after use. 3. Using a clean 100 mL graduated cylinder, transfer 50.0 mL of the 0.20 M sodium

acetate stock solution to the buffer flask. Thoroughly rinse the graduated cylinder after use.

4. Using the rinsed 100 mL graduated cylinder, transfer 40.0 mL of deionized water into the buffer flask, making the total volume of the buffer 100 mL.

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A fine point: because the buffer is made from three solutions (10 mL, 50 mL, and 40 mL) having different concentrations, the final volume of the buffer will be very slightly different from the sum of the volumes of its component solutions. But the difference is typically small and is negligible in this lab. Therefore, the total volume of the buffer you just made is assumed to be 100 mL.

5. Mix the solution by swirling the flask (the shape of Erlenmeyer flasks makes them easy to swirl without spilling).

6. Using the dilution equation, calculate the molarity of acetic acid in this buffer, and record this on the data sheet under “Buffer Composition”. Likewise, calculate and record the molarity of sodium acetate in the buffer.

Prepare acid and base solutions1. Prepare 0.10 M HCl. Label a 250 mL Erlenmeyer flask as “0.10 M HCl”. Using a

clean 25 mL graduated cylinder, transfer 21.0 mL of 1.0 M HCl stock solution into the HCl flask. Thoroughly rinse the graduated cylinder after use. Using that rinsed 100 mL graduated cylinder, transfer 100 mL of deionized water into the flask, then add an additional 89 mL of deionized water. Carefully swirl the solution to mix it. The total volume of 0.10 M HCl is now approximately 210 mL. Record the molarity of this solution (0.10 M) on the data sheet below “Effect of adding acid to a buffer”.

2. Prepare 0.10 M NaOH. Label another 250 mL Erlenmeyer flask as “0.10 M NaOH”. Make this solution using 21.0 mL of 1.0 M NaOH stock solution, and 189 mL of deionized water (this solution is make like the previous solution was made). Record the molarity of this solution (0.10 M) on the data sheet below “Effect of adding base to a buffer”.

Effect of adding acid to a buffer1. Prepare starting water and buffer solutions. Obtain two clean 250 mL beakers.

Label one beaker “water”, and, using a clean 50 mL graduated cylinder, transfer 50.0 mL of deionized water into it. Label the other beaker “buffer”, and, using the 50 mL graduated cylinder, transfer 50.0 mL of the buffer solution you just made into it. Thoroughly rinse the graduated cylinder after use.

2. Add indicators. Add 5 drops of 0.03% methyl orange indicator solution and 5 drops of 0.1% malachite green indicator solution to each beaker. (Yes, two indicators in each solution.)

3. Measure the pH. Use the pH meter to measure the pH of the solution in the “buffer” beaker. Record the pH on Data Sheet 1 in the “pH meter reading, buffer” column. Rinse the pH electrode with deionized water into the waste rinse water beaker and blot the electrode dry. Then, measure the pH of the solution in the “water” beaker. Record that pH on the Data Sheet in the “pH meter reading, water” column.

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4. Solution color. Record the color of each solution on the Data Sheet in the “Solution Color” columns.

5. Effect of adding HCl. Using a clean 10 mL graduated cylinder, transfer 5.0 mL of your “0.10 M HCl” solution into the “water” beaker, and 5.0 mL into the “buffer” beaker. Mix the solutions well. Measure and record the pH of each solution, rinsing the electrodes between measurements. Also record the solution color.

6. Effect of adding more HCl. Repeat the previous step for the following 0.10 M HCl solution additions: 5.0 mL (for a second addition), 10 mL, 30 mL and 50 mL. Use a 50 mL graduated cylinder for the last two additions.

7. Clean up. When finished, discard the contents of the “buffer”, “water”, and “waste rinse water” beakers into the waste container in the front fume hood. Rinse the “buffer” and “water” beakers with tap water, then, with deionized water. Dry the two beakers for reuse in the next section.

Note: Do NOT discard the solutions in the Erlenmeyer flasks. Those solutions will be used in the next section.

Effect of adding base to a buffer1. Prepare starting water and buffer solutions. Repeat step 1 in the previous

section.2. Add indicators. Add 4 drops of methyl purple and 2 drops of 1% phenolphthalein

to each solution.3. Measure the pH. Measure the pH of each solution as was done in step 3 in the

previous section. Record all data on data sheet in the “Effect of adding base to a buffer” section.

4. Solution color. Record the color of each solution. 5. Effect of adding NaOH. Repeat step 5 in the previous section, except use 0.10 M

NaOH, instead of 0.10 M HCl.6. Effect of adding more NaOH. Repeat step 6 in the previous section, except use

NaOH instead of HCl.7. Clean up. Discard all solutions into the waste container in the front fume hood.

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CalculationsEffect of adding acid to a buffer

Note: the shaded boxes do not need to be filled in.1. Fill in the column “Total mL HCl added”. This number is the sum of the amounts

previously added, plus the amount added in the current run, given in the first column.

2. Fill in the column “Total mL of solution”. To the initial volume in a beaker, 50 mL, add the total mL of HCl added. This is the total volume of the solution in the “water” beaker. (As can be seen, the buffer gets more dilute as the acid additions are made.)

3. The Ka of acetic acid is 1.74 10-5. Calculate the pKa, which is the negative log of Ka: pKa = –log(Ka). Enter this value below the first table on the data sheet.

4. The column, “Moles remaining in buffer”, gives how many moles of HA and A– remain after adding HCl. However, the first row is before any HCl is added, so this is the initial number of moles of each component in the buffer. Use the molarity of HA (acetic acid) that you previously wrote on the top of the data sheet to calculate how many moles of HA are in the 50 mL of buffer. Likewise, calculate how many moles of A– are in the 50 mL of buffer. (The other rows are filled in later.)

5. Calculate the pH expected of the buffer initially (before any HCl was added). The pH expected for a buffer is given by the Henderson-Hasselbalch equation, p H=p Ka+log ¿. In the last two steps you calculated the value of pKa for acetic acid, and the number of moles of A– and HA in the buffer solution. To get concentrations, divided the moles of A– and HA by the total volume of the buffer solution (50 mL) converted to liters. Your expected pH should be somewhat close to the pH that was actually measured.

Shortcut: for buffer problems, both HA and A– are in the same solution, so they have the same volumes, which cancel in the Henderson-Hasselbalch equation. The ratio of concentration in that equation can be written as ¿. So, the mole ratio can be used instead of the concentration ratio.

6. Calculate the pH expected of the water initially (before any HCl was added). (Hint: what is the pH of neutral water?)

Row 4 calculations:7. Calculate the “Moles H+ added”. Calculate this value using the volume in the “total

mL HCl added” column, and the “molarity of HCl solution that was prepared” previously recorded above the table.

8. The column, “Moles remaining”, gives how many moles of HA and A– remain after adding HCl. The moles of HA that are present was increased by the number of moles of H+ that were added, because the HCl converts the A– to HA.

H+(aq) + A–(aq) HA(aq)

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Therefore, to the moles of HA initially in the buffer add the moles of HCl (i.e., H+) transferred to the buffer beaker. Similarly, from the moles of A– initially in the buffer, subtract the moles of HCl added.

9. Using the Henderson-Hasselbalch equation and the moles found in the previous step, calculate the expected pH of the buffer solution

10. Calculate the expected pH of the solution in the “water” beaker. First, calculate the molarity: divide the “moles H+ added” by the “total mL of solution” (in liters). Second, convert to pH: pH = –log[H+]

11. Calculate the molarity of H+ in the water beaker from the observed pH using the following: [H+] = 10(-pH).

Last row calculations:12. Calculate the “Moles H+ added” as in step 7. Notice that this number of moles is

greater than the number of moles of A– initially present in the buffer, so every bit of the initial A– is converted to HA. Enter the number of moles of A– remaining (0), and the number of moles of HA remaining.

13. Calculate the pH of the “buffer” solution. This is not really a buffer solution any longer, because all of the A– has been used up. First, calculate how many moles of H+

remain by subtracting the initial moles of A– from the moles of H+ added. Second, calculate the concentration of H+ by dividing this number by the “total mL of solution mL”. Finally, calculate the pH using pH = -log([H+]).

14. Calculate the expected pH of the solution in the “water” beaker as in step 10. Calculate the molarity of H+ in the water beaker as in step 11.

Effect of adding base to a buffer15. The first row (“0 mL HCl added”) of the previous table and the first row (“0 mL

NaOH added”) of the table in this section have the same calculated values (because they both refer to the solutions before anything was added to them), so copy the values in the first row from the previous table to the table in this section. Also, the two columns, “Total mL NaOH added”, and “Total mL of solution”, are calculated using the same values as were used in the previous table to calculate “Total mL HCl added”, and “Total mL of solution”, so copy the values in those two columns to the analogous columns in this table.

16. Fill in the column “Moles OH–added”. Calculate this value using the volume in the “total mL NaOH added” column, and the “molarity of NaOH solution that was prepared” previously recorded above the table.

Row 4 calculations:17. The column, “Moles remaining”, gives how many moles of HA and A– remain after

adding OH–. The moles of HA that are present are decreased by the number of moles of NaOH that are added, because the OH– converts the HA that is present to A–:

OH–(aq) + HA(aq) A–(aq) + H2O(l)

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Therefore, from the moles of HA initially in the buffer, subtract the moles of NaOH transferred to the buffer beaker. Similarly, to the moles of A– initially in the buffer, add the moles of NaOH added.

18. Using the Henderson-Hasselbalch equation and the moles found in the previous step, calculate the expected pH of the buffer solution

19. Calculate the expected pH of the solution in the “water” beaker. First, calculate the molarity of OH–: divide the “moles OH- added” by the “total mL of solution” (in liters). Second, convert to pOH: pOH = –log[OH–]. Third, convert from pOH to pH using pH + pOH = 14. (Yes, a lot of new formulas for you to use here.)

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Data SheetBuffer composition: Molarity of acetic acid ____________; Molarity of sodium acetate: ____________.

Effect of adding acid to a buffer

Molarity of HCl solution that was prepared ____________

pKa = _____________

mL HCl

added each time

Solution color pH meter reading

Total mL HCl

added

Total mL of solu-tion

Moles H+

added

Molesremaining in

buffer

H+ molarity in water from pH reading

Expected pH

Buffer Water Buffer Water HA A– Buffer Water

0 50 ―(initial) (initial)

5

5

10

30

50

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Effect of adding base to a bufferMolarity of NaOH solution that was prepared ____________

mL NaOH added each time

Solution color pH meter reading

Total mL

NaOH added

Total mL of solu-tion

Moles OH–

added

Molesremaining in

bufferOH–

molarity in water

Expected pH

Buffer Water Buffer Water HA A– Buffer Water

0 50 ―(initial) (initial)

5

5

10

30

50

100

Post Lab Questions1. Based on your results in the “Effect of adding acid to a buffer” section of the data

sheet, compare the abilities of the buffer solution and the water to resist pH changes when HCl solution was added.

2. Based on your results in the “Effect of adding base to a buffer” section of the data sheet, compare the abilities of the buffer solution and the water to resist pH changes when NaOH solution was added.

3. Buffers have limits. If enough acid or base is added, the buffer “capacity” will be exceeded. In which runs did the buffer capacity nearly get reached? (Hint: this would be where the moles of either HA or A– nearly reaches zero, and the pH starts to change a lot.)

4. The measured pHs were not exactly the same as the expected pHs you calculated. What are some reasons that could cause this experimental errors?

5. Would you trust using the indicator color changes to determine pH?