Transcript
Page 1: Honors Chemistry: Chapter 12 SOLUTIONS

SOLUTIONS(Made of solute and solvent)

Honors Chemistry: Chapter 12

Page 2: Honors Chemistry: Chapter 12 SOLUTIONS

SOLUTION AND SOLUBILITIES 2

Learning Targets - Unit 12: Solutions (PLEASE READ)

1. Identify solute and solvent in a given solution.

2. Use solubility curves and laboratory methods to determine whether

solutions are unsaturated, saturated, or supersaturated.

3. Differentiate between various ways of expressing concentration,

including appropriate units. (i.e. molarity, molality, percent, mole

fraction).

4. Solve concentration problems.

5. Determine how to prepare molarity solutions.

6. Perform dilution problems.

7. Define colligative properties.

8. Explain how concentration and colligative properties are related.

9. List and describe two types of colligative properties: boiling point,

freezing point.

10. Calculate changes expected (in freezing point, boiling point) at

various concentrations.

11. Perform solution stoichiometry problems.

Page 3: Honors Chemistry: Chapter 12 SOLUTIONS

TERMS YOU NEED TO KNOW!

Solution: a homogeneous mixture. Examples are air, brass, saline, bronze.

Colloid: a homogeneous mixture containing slightly larger particles. Examples include fog, smoke, dust.

Suspensions: are mixtures with even larger particles, but they are not considered true solutions because they separate upon standing. An example is Italian Dressing that eventually settles out after sitting.

Solute: the dissolved substance in a solution

Solvent: the major component in a solution

Page 4: Honors Chemistry: Chapter 12 SOLUTIONS

A solution is saturated when no additional solute can be dissolved at a particular temperature

A supersaturated solution can form when more than the equilibrium amount of solute is dissolved at an elevated temperature, and then the supersaturated solution is slowly cooled.

An unsaturated solution is formed when more of the solute can dissolve in it at a particular temperature.

4

TERMS (cont.)

Page 5: Honors Chemistry: Chapter 12 SOLUTIONS

How to Make a

Supersaturated Solution(this is on your chapter 12 test!)

5

https://youtu.be/FcxZ9DyOaUk

See how unstable a supersaturated solution is:

1. Place a beaker of measured water onto a hotplate and add MORE than the

amount of solute than would normally dissolve into the solvent. Though

you stir and stir you will not get it all to go into solution and will see it

swirling around.

2. Heat the solution until all of the solute dissolves. By heating it you FORCE

the solute to dissolve into the solvent.

3. This step is important as it is not yet a supersaturated solution. You must

first cool it back down to room temperature before we consider it

supersaturated. The solution is very unstable at this point and agitating it

just a bit will cause all of the solute to come back out.

Be sure you understand these steps as how to make this is on your test. See

the link below for what happens when you agitate a supersaturated solution.

Page 6: Honors Chemistry: Chapter 12 SOLUTIONS

Now watch the video describing different

characteristics of solutions, but before you do,

these are the things you will want to look for

on the video because every one of these

questions is also on your chapter 12 test!

6

1. What are solutes and solvents?

2. How do ionic cpds “dissociate” in solvents?

3. How do molecular (covalent) cpds dissolve in

solvents?

4. What do the terms solubility and concentration

mean?

https://youtu.be/i6EDUQSWWgo

Page 7: Honors Chemistry: Chapter 12 SOLUTIONS

WHAT IS SOLUBILITY?

Solubility is the degree to which a

substance dissolves in a solvent to

make a solution (usually expressed

as grams of solute per liter of

solvent).

Page 8: Honors Chemistry: Chapter 12 SOLUTIONS

SOLUBILITY GRAPH OF SALTS IN WATER

SOLUTION AND SOLUBILITIES 8

Page 9: Honors Chemistry: Chapter 12 SOLUTIONS

Answer these questions from

the previous slide…

1. What is the solubility of NH4Cl at 20 C? _________

2. Ce2(SO4)3 is the only gas on the chart. What can you

gather from the curve of this substance as T increases. This

is actually true for all gases. __________________________

3. Which substance has the least change in solubility as the

temperature increases? __________________________

4. At 60 C, which substance has a solubility of

approximately 40 g/100 mL of water? ________________

Answers on next slide…

Page 10: Honors Chemistry: Chapter 12 SOLUTIONS

Solutions

1. 20 g/ 100 mL

2. As the temperature increases,

the solubility of GASES

decreases.

3. NaCl has the smallest slope.

4. CuSO4

How did you do????

Page 11: Honors Chemistry: Chapter 12 SOLUTIONS

Things we Need to Know About

SOLUBILITY The amount of solute per unit solvent required to

form a saturated solution is called the solute's Solubility.

When two liquids are completely soluble in each other they are said to be Miscible.

Solubility is effected by Temperature. With increase in temperature solubility of most of the substances increases.

Most gases become less soluble in water as the temperature increases. See the graph on the next slide.

Page 12: Honors Chemistry: Chapter 12 SOLUTIONS

SOLUBILITY GRAPH OF GASES IN WATER

Pressure has little effect on the solubility of liquids and

solids. The solubility of gases is strongly influenced by

pressure. Gases dissolve more at high pressure.

Page 13: Honors Chemistry: Chapter 12 SOLUTIONS

Concentration

In chemistry, a solution’s concentration is how much

of a dissolvable substance, known as a solute, is

mixed with another substance, called the solvent.

There are many ways to do this and the following

slides will go over each method and include

examples which will help you with your homework

problems. We will learn 6 of the ways to do this!

Page 14: Honors Chemistry: Chapter 12 SOLUTIONS

1. Percent Composition by Mass

Mass solute (g) x 100

Mass Solution (g)

Example:

Determine the percent composition by mass of a

100 g salt solution which contains 20 g salt.

Solution:

20 g NaCl / 100 g solution x 100 = 20% NaCl solution

X 100

Page 15: Honors Chemistry: Chapter 12 SOLUTIONS

2. Volume Percent (% v/v) Volume

Volume solute

Volume solutionX 100

Example:

What is the percent of rubbing alcohol bought

in the store that is prepared by taking 700 mL

of isopropyl alcohol and adding sufficient

water to obtain 1000 mL of solution.

Solution:

700 mL

1000 mLX 100 = 70%

Page 16: Honors Chemistry: Chapter 12 SOLUTIONS

3. Parts per Hundred, per Thousand,

per Million (can be in any unit V/V, m/m, m/V)All of these are simply exactly what you do to calculate your grade. They are:

little number (solute)

big number (solution)

Part per Hundred is really just “percent”! The others are done the same way but

parts per thousand (ppt) and parts per million (ppm) are just multiplied by those

numbers!

Example: If 5.4 g of solute is dissolved in 1300 grams of solvent,

calculate the concentration in ppm.

Solution: 5.4 g solute

(1300 g solvent + 5.4 g solute)

***remember, the solution is made of the solute and the solvent!

= 4137 ppm = 4100 ppm (with sig figs)

X 100 or 1,000 or 1,000,000

X 1,000,000

Page 17: Honors Chemistry: Chapter 12 SOLUTIONS

4. Mole Fraction (X)Mol solute 1

Total moles of solutes and solvents

***Keep in mind, the sum of all mole fractions in a

solution always equals 1.

Example: What are the mole fractions of the components of the solution

formed when 92 g glycerol is mixed with 90 g water? (molecular weight

water = 18; molecular weight of glycerol = 92)

Solution:

90 g water = 90 g x (1 mol / 18 g) = 5 mol water

92 g glycerol = 92 g x (1 mol / 92 g) = 1 mol glycerol

total mol = 5 + 1 = 6 mol

xwater = 5 mol / 6 mol = 0.833

x glycerol = 1 mol / 6 mol = 0.167

X 100

Page 18: Honors Chemistry: Chapter 12 SOLUTIONS

5. Molarity (M) Molarity is probably the

most commonly used unit of concentration.

Example: What is the molarity of a solution made

when water is added to 11 g CaCl2 to make 100 mL

of solution? (The molecular weight of CaCl2 = 110)

Moles solute

L solution

Solution:

1. 11 g CaCl2 x (1 mol CaCl2/110 g CaCl2) = 0.10 mol CaCl22. 100 mL x (1 L / 1000 mL) = 0.10 L

3. molarity = 0.10 mol / 0.10 L = 1.0 M

Page 19: Honors Chemistry: Chapter 12 SOLUTIONS

6. Molality (m) Molality is the number of

moles of solute per kilogram of solvent.

m = moles solute

Kg solvent

Example:What is the molality of a solution of 10 g

NaOH in 500 g water? (Molecular weight of NaOH is 40)

Solution:

1. 10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol

NaOH

2. 500 g water x (1 kg/1000 g) = 0.50 kg water

3. molality = 0.25 mol / 0.50 kg = 0.05 mol/kg = 0.50 m

Page 20: Honors Chemistry: Chapter 12 SOLUTIONS

Summin’ It Up…a

You must make notecards of all of these

formulas. The bulk of this chapter deals with

these units of concentration and many of your

homework problems have you calculate them.

So be sure to write them all down and learn

them. They are simply plug-n-chug so if you

know the formulas you will have no issues!!!a

HINT: KNOW THE DIFFERENCE

BETWEEN SOLUTE, SOLVENT AND

SOLUTION!!!Now let’s practice a few of them from your HW packet

Page 21: Honors Chemistry: Chapter 12 SOLUTIONS

HW 12.2 (Note: some of the answers for 12.2 in the packet are incorrect

so ignore them!)

1. 5.0 g = 4.8 pph or %(100 g + 5.0 g)

4. (done on slide #16)

5. 2.8 % = X g Solve for x = 1.4 g

50 mL

9a. M = moles solute …first convert g to moles

1 L solution

(54 g NaOH) (1mole NaOH) = 1.35 moles NaOH1 40 g NaOH

M = 1.35 moles NaOH = 1.35 M1.0 L soln.

X 100

X 100

Page 22: Honors Chemistry: Chapter 12 SOLUTIONS

Continue to work through these problems using

the formulas given from the slides. We will go

over them in our Monday Zoom session or you

can join tomorrow’s office hours if you need

help with a question.

The rest of this powerpoint will be uploaded

as soon as I finish it. I wanted to get you

what I had completed. Sorry it is taking

so long. It takes me FOREVER to

write using superscripts and

subscripts!


Top Related