Today is Tuesday,June 2nd, 2015

Pre-Class:Earlier in the year I used hydrogen peroxide in

a lab but mentioned it was different from normal peroxide. How come?

Also: Please find a worksheet and have your solubility curve packets out.

Stuff You Need:Calculator

Periodic TablePolyatomic Ion

ListWorksheet

In This Lesson:Molarity

(Lesson 2 of 4)

Today’s Agenda

• Concentrations• Molarity• Percent solutions• Dilutions– In other words, today we’re going to talk about the

strength of solutions.• Colligative Properties

• Where is this in my book?– P. 480 and following…

By the end of this lesson…

• You should be able to calculate the concentration of a solution in several different ways.

• You should be able to calculate freezing-point depression and boiling-point elevation.

Practice

• Molarity Practice Chart and Problems– First three rows.– Use the first column as a sample!

Molarity Practice Chart and ProblemsRest of key online…

Name Calcium iodide

Sodium chloride

Potassium carbonate

Aluminum hydroxide Glucose Sulfuric

acid

Formula CaI2 NaCl K2CO3 Al(OH)3 C6H12O6 H2SO4

Electrolyte or non-

electrolyteElectrolyte Electrolyte Electrolyte Electrolyte Non-

electrolyte Electrolyte

Ions in solution Ca2+, 2 I- Na+, Cl- 2 K+, CO3

2- Al3+, 3 OH- N/A 2 H+, SO42-

Let’s start with the easy, shall we?

• Recently we talked about solutions, and today we started by discussing the difference between 3% and 30% H2O2.

• How are those numbers calculated?

http://www.acne.org/images/products/img-442.jpg

Percent Solutions

• There are two types of “__% Solutions.”– Percent by volume.– Percent by mass.

• Decide which to use by looking at the denominator’s units.

• Let’s start with volume, and here’s your practice problem:

• What is the percent by volume of 5 g of CaCl2 salt in 250 mL solution?

Percent by Volume

• What is the percent by volume of 5 g of CaCl2 salt in 250 mL solution?

• This is a percent by volume problem, since we’re looking at a 250 mL solution.

• 5 g out of 250 mL is 2%.• (5 g / 250 mL) * 100 = 2%

• If the solute is a solid or gas:

100solvent of volume

solute of or volume massby volumePercent

Percent by Volume

• If the solute is a liquid, you have to use this formula:

100solution of volume

solute of volumeby volumePercent

Percent By Mass

• Percent by mass is very similar.• Note, however, that the denominator is the

total mass of solution (and not solvent).– This is because volume is not meaningfully

impacted by a solid or gas solute, but mass is.

100solution of mass

solute of massmassby Percent

Okay. Why?

• Why the differences in calculations?• Imagine a glass of 50 mL of water.– It’s also 50 g.

• If I shake 10 g of sodium chloride into it, does the water level change?– Nope.

• But does the mass change?– Yep.

• Hence, volume of solvent but mass of solution.

Practice

• Molarity and Solution Percent worksheet– #1, 4, 5 (#1 is challenging)

Molarity

• At some point this year you may have seen me label acids or other liquids with something like:– “6 M HCl”

• The big capital M stands for Molar, as in not your teeth.– 6 M HCl is pronounced “Six molar hydrochloric acid.”

• Molarity [that’s the name of this concept] is another way to measure the concentration or dilution of a solution.– The higher the molarity, the stronger the solution.

Calculating Molarity

• Calculating molarity is fairly simple.• Here’s the equation:– M is the symbol for Molarity

solution of liters

solute of molesM

Need a Reminder?

g

mol molar mass

mol

M L

Molarity Equation

Mole Conversion

Molarity Practice Problem

• Intravenous (IV) saline solutions are often administered to patients in the hospital. One saline solution contains 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution?• 0.90 g NaCl = 0.0154 mol NaCl• 100 mL = 0.100 L• (0.0154 / 0.100) = 0.154 M

Molarity Practice Problem

• Household laundry bleach is a dilute aqueous solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5 L of 0.70 M NaClO?• 0.7 M = moles of NaClO / 1.5 L• Solve for “moles of solute”• 1.05 mol NaClO

Practice

• Molarity and Solution Percent worksheet– #2, 3

Practice

• Molarity Practice Chart and Problems– Shaded four rows.– Use the first column as a sample!

• Molarity Practice Chart and Problems– Bottom two rows.– For cation/anion molarity, multiply the coefficient

in front of each ion in the “Ions in Solution” row by the molarity in the “Molarity of Solution” row.

– Use the first column as a sample!

Molarity Practice Chart and ProblemsRest of key online…

Name Calcium iodide

Sodium chloride

Potassium carbonate

Aluminum hydroxide Glucose Sulfuric

acid

Ions in Solution Ca2+, 2 I- Na+, Cl- 2 K+, CO3

2- Al3+, 3 OH- N/A 2 H+, SO42-

Molarity of

Solution0.045 M 1 M 0.50 M 0.75 M 1.5 M 0.51 M

Molarity of Cation

1 x .045 M = 0.045 M 1 M Na+ 2 x 0.5 M

= 1 M K+0.75 M

Al3+ -2 x 0.51 M = 1.02 M

H+

Molarity of Anion

2 x .045 M = 0.090 M 1 M Cl- 0.5 M

CO32-

3 x 0.75 M = 2.25 M

OH--

0.51 M SO4

2-

Computer Practice

• Okay, now it’s time to do a little reinforcement using the computers.

• Head to Quia and open the quiz called:– Salts and Solubility

• From there, it’s a pretty self-explanatory quiz that will have you visiting another page at the same time (PhET, an awesome website).

• You will be looking at various salts dissolving into water and making molarity calculations based on them.– Username: [firstnamelastname]832– Password: Gleicher[house/apt. #]

Dilutions of Molar Solutions

• Sometimes chemists need to create dilutions using known molarities and volumes.

• For example, during our Baking Soda/HCl lab, I used 6 M HCl. However, the school receives hydrochloric acid in big jugs of 12 M HCl (nasty stuff).

• I needed to dilute it down to 6 Molar, and here’s how I did…

Molar Dilutions

• Because we’re not changing how much solute is around during a dilution (we’re just adding water, the solvent), we can use the following formula:

• M1V1 = M2V2

• In other words, like with gas laws, when molarity goes up, volume typically goes down.– This is Boyle’s Law, except with molarity instead of

pressure.

Molar Dilution Practice Problem

• How many milliliters of aqueous 2.00 M MgSO4 solution must be diluted with water to prepare 100.0 mL of aqueous 0.500 M MgSO4?• M1V1 = M2V2

• 2.00 M * V1 = 0.5 M * 100 mL

• V1 = 25 mL

Molar Dilution Practice Problem

• You put 2 moles of HCl into 312 mL of water. If you wanted to make a 1 M dilution, to what volume would the original solution need to be diluted?• Molarity of the solution = 6.41 M• 6.41 M * 312 mL = 1 M * V2

• V2 = 2000 mL

Practice

• Molarity Practice Chart and Problems– Reverse side (all)

Finally…

• …our last topic of the lesson. Let’s introduce it with a photo:

http://www.bceo.org/images/SaltBrineSystem_e.JPG

So why use salt on a road?

• Salt on a road lowers the freezing point of water, thus reducing icing.– All those dissolved salt ions interfere with the water’s ability to make

an orderly crystalline arrangement and thus freeze.• You can also use this phenomenon to make ice cream.

– Put your ingredients in a plastic bag and put the bag in a bucket of ice water.• The ingredients (milk, sugar) have lots of dissolved solutes.

– Add lots of salt to the ice water.– The ice water temperature will drop as the ice absorbs heat from the

water, but the water can’t freeze.– In turn, the ice water absorbs heat from the bag of ingredients, which

needs to drop significantly below freezing to…freeze.– Eventually you get that nice softish ice cream texture.

Colligative Properties

• In fact, this is one of three main colligative properties we’re going to learn:1. Freezing-Point Depression2. Boiling-Point Elevation3. Vapor-Pressure Lowering

• A colligative property is one that depends only on how much solute you have, not what kind of solute it is.– Like osmosis. You remember biology, right?

1. Freezing-Point Depression

• Every mole of dissolved solute particles lowers the freezing point of 1000 g of water by 1.86°C.

• Example Problem:• What is the freezing point of these two

solutions?A. 1000 g of 1 M CaCl2 solution.

B. 58.44 g NaCl in 1000 g water.C. 180.15 g C6H12O6 in 1 kg water.

• Well A is easy: -1.86°C.

1. Freezing-Point Depression

• Wait a sec…maybe A isn’t so easy.• When a unit of CaCl2 dissolves, it creates three

ions: Ca2+, Cl-, and Cl-.– In other words, 1 mol CaCl2 dissolves to 3 mol of

particles.• 3 mol * 1.86°C/1 mol = -5.58°C.• Similarly, for Part B, 58.44 g NaCl = 1 mol NaCl.– That’s 2 mol of particles, so -3.72°C.

1. Freezing-Point Depression

• Now for Part C: 180.15 g C6H12O6 in 1 kg water.• That’s 1 mol of sugar and it’s in one kg of

water, so it should be a freezing point of -1.86°C.

• But what about dissociation?– Sugar doesn’t dissociate since it’s not an

electrolyte, so -1.86°C is correct.

1. Freezing-Point Depression

• One last freezing-point depression question:• Suppose you’re the person in charge of Doylestown’s

road department, and therefore are in charge of the de-icing process that happens usually every winter.

• You have two salt choices to spread on the roads: NaCl or CaCl2.

• Assuming they’re equal in cost, which do you choose?– CaCl2 – every one mole of calcium chloride dissociates into

three moles of ions, whereas only two are formed from sodium chloride.

2. Boiling-Point Elevation

• Similarly, for every mole of solute in 1000 g of water, the boiling point increases by 0.512°C.– This is NOT why some recipes call for “salting”

water before boiling.– Think of how much salt you’d have to add. It’s just

for taste.• Our last colligative property is closely related.

3. Vapor-Pressure Lowering

• Recall that vapor pressure is the pressure of a gas above a liquid in a container.

• Dissolved solutes will lower this pressure, since lots of solutes will cause solvent molecules to bind to them, making it harder for evaporation to occur.– Don’t worry about numbers here.

So…

• After all of this, there’s one more (confusingly-named) calculation we should know how to do: molality.– Molarity is mol/Lsolution

– Molality is mol/kgsolvent

• These are equal when water is the solvent, but different when it’s not.

• Molality is particularly useful for calculations of colligative properties, especially for non-H2O stuff.– Molality, oddly, uses the symbol m (like meters) or b.

Colligative Property Equation

• ΔT = K * m * i• ΔT = change in temperature• K = constant

• Kb for boiling-point elevation

• Kf for freezing-point depression

• m = molality• i = number of dissociated ions

• K is different based on the solvent and is usually in units of °C/m.

• For water:• Kb = 0.512°C/m

• Kf = 1.86°C/m

Solvent Kb (°C/m)Water 0.512

Ethanol 1.19

Benzene 2.53

Cyclohexane 2.79

Acetic Acid 3.07

Phenol 3.56

Nitrobenzene 5.24

Camphor 5.95

Solvent Kf (°C/m)Water 1.86

Acetic Acid 3.90

Benzene 5.12

Nitrobenzene 7.00

Phenol 7.40

Cyclohexane 20.2

Camphor 37.7

Example

• What is the freezing-point depression of a glucose solution containing 20.0 g of C6H12O6 in 50.0 g H2O? (Kf = 1.86°C/m)

• 20 g C6H12O6 = 0.111 mol C6H12O6

• 0.111 mol C6H12O6 / 0.05 kg H2O = 2.22 m

• ΔTf = Kf * m * i

• ΔTf = 1.86 * 2.22 * 1 (since it’s molecular)

• ΔTf = 4.13°C (realistically it’s -4.13°C)

Example Deux• 500 kg of CaCl2 is spread across Rt. 611 in winter. If we

assume that it comes in contact with 1000 kg of H2O, what will the freezing point of that water become?

• Wow. Where to begin?• We have 500 kg of CaCl2 (solute). Let’s make it moles.

• 500 kg CaCl2 = 500,000 g CaCl2 = 4481.09 mol

• 4481.09 mol in 1000 kg H2O is 4.48 m.

• BUT! CaCl2 splits into three ions in solution, so the real molality is 4.48 * 3.

• ΔT = Kf * m * i• ΔT = 1.86°C/m * 4.48 m * 3• ΔT = 24.99°C• So the new freezing point is -24.99°C.

Practice

• Colligative Properties Worksheet (except #3)• Note: Some of these are challenging. Skip #3

for now.

• About #3:– “25.0 mass percent solution” means that 25% of

the total solution mass is, in this case, ethylene glycol.

– You’ll have to make up a “kg” value for water, but be careful with it.

Closure – TEST QUESTION

• Determine the molar mass of a molecular compound that will raise acetic acid’s boiling point to 121°C when 25.8 grams of solute are added to 135.0 g acetic acid. Kb = 2.9°C/m, acetic acid’s boiling point is 118.0°C.

• ΔTb = Kb * m

• ΔTb = 3°C• 3°C = 2.9°C/m * m• m = 1.03

Closure – TEST QUESTION

• Okay, so the molality of the compound is 1.03.• We know that there were 135.0 g acetic acid.• m = mol / kg• 1.03 = mol / 0.135 kg• mol = 0.13905• g / molar mass = mol.• We know 25.8 g = 0.13905 mol.• 0.13905 = 25.8 g / molar mass• Molar mass = 185.54 g/mol

Another way to think of this…• Colligative property problems almost always come

down to the two main equations:

• From only reading the problem, I can fill in this much:

• Start with the left equation and you’re off!

m KT b

solution

solute

kg

molm

mm C/9.2118121 kg 0.135

molsolutem

Closure

• One last little thing.• Once you have that solution, now freezing at a

lower temperature than normal, you have yourself a eutectic mixture.

• You don’t need that word for a test, but it’s fun nonetheless.

Closure Deux

• An enormously wide-ranging review video in under five minutes:– TED: Cynthia Chubbuck – The Chemistry of Cars