Solutions
Solutions, Colloids, and Suspensions
Solution Formation and Factors Affecting Solubility
Solubility Curves
Molarity and Solubility Product Constant (Ksp)
Colligative Properties
Key
In Your Textbook
Solutions, Suspensions, and Colloids: pp. 482–83 and pp. 490-493
Factors Affecting Solubility: pp. 501 - 508
Solubility Curves: pp. 504 – 505
Molarity and Ksp: pp. 509–512 and pp. 630-637
Colligative Properties: pp. 517 - 524
Assignment 1: Defining Solutions
Compare and contrast the contents of the three test tubes. Record observations in the
chart below.
A
B
C
Green clear
liquid
White
cloudy
liquid
Brown
cloudy liquid
with particles
at the bottom.
Read pp. 482-483, pp. 490-494, pp. 502-508, and pg354 in your textbook.
Which of the three test tubes contains a solution. Explain why.
The first tube has a solution because the particles are dissolved as small as they can get.
It is not at all cloudy.
How could I test to see if you are correct?
I could look for a Tyndall effect by shining a light through the mixture and
see if the light is scattered (you can see the beam of light in the mixture as it
passes through). If you can, it is not a solution.
Which of the following would also be a solution?
Blood no (colloid)
Mayonnaise no (colloid)
Italian Salad Dressing no (suspension)
Coca-Cola yes
Air yes
Ocean Water no/yes The water itself would be a solution, but if you scooped up
a sample, it would have all kinds of things in it (a suspension).
Define: Solubility, Solute, Solvent
Solvent: dissolves the solute
Solute: is dissolved by the solvent
Solubility: the amount of solute that will dissolve in a given amount of water at a certain
temperature.
Tonight, find 10 things in your home that are classified as solutions. Make a list of these
solutions. (What is the solute and the solvent in each solution?)
Some examples: windex, coke, Listerine, vinegar…..
Assignment 2: Solutions and Molarity Review.
Solubility: Predict which of the following will dissolve in water and why. Look up the
structures and consider the bonding, structure, and the intermolecular forces.
Sodium chloride (table salt)
NaCl is ionic so is likely to dissolve.
Sucrose (sugar)
C12H22O11 is covalent, but is polar so will dissolve to some degree. These molecules
have dipole-dipole interactions.
Vegetable oil
CH3(CH2)16COOH is a long carbon chain and they tend to be nonpolar so not soluble. It
would have only dispersion forces.
Isopropyl alcohol (rubbing alcohol)
C4H9OH is polar covalent and would dissolve in water. It has dipole-dipole interactions,
specifically H-bonding.
Acetic Acid (vinegar)
HC2H3O2 is polar covalent (actually ionizes as an acid in water) so would dissolve. It
would have dipole-dipole attractions.
What does the rule “like dissolves like” state? This rule says that polar solvents dissolve
solutes that are polar or ionic (have a + and a -) and solvents that are non-polar dissolve
solutes that are non-polar (turpentine dissolves non-polar solutes).
Test your hypothesis by dissolving a small amount of the substance in water. Are there
any that do not follow the rule “like dissolves like”? Are any of these substance already
in solution when you buy them from the grocery store?
These all follow the rule. The alcohol and vinegar are already in water when you buy
them.
Molarity Review:
1.
A salt solution has a volume of 250 mL and contains 0.70 mol of sodium chloride.
What is the molarity of the solution?
0.70 moles NaCl = 2.8 M
0.25 L
2.
A solution of glucose has a volume of 2.0 L and contains 36.0 g of solute. What
is the molarity of the solution?
36.0 g C6H12O6 x
3.
1 mole
180.16 g
= 0.20 moles
M = 0.20 moles = 0.10 M
2.0 L solution
How many moles of solute are in 250 mL of 2.0 M calcium chloride solution?
How many grams would that be?
2.0 M =
x moles
0.25 L
x = 0.50 moles Ca Cl2
0.50 moles CaCl2 x
4.
110.98 g CaCl2 = 55 g CaCl2
1 mole CaCl2
Describe how you would make a 0.25 M solution of calcium hydroxide. (Use
sketch to illustrate.)
Put 0.25 moles (19 grams) of Ca(OH)2 in a volumetric flask and add water to the
1 L mark.
5.
What is the molarity of the sugar solution in a can of coke? (Assume sugar to be
sucrose)
39 g C12H22O11 x
1 mole
= 0.11 mole
M = 0.11 mole =
0.31M
342 g
.355 L
Assignment 3: Factors Affecting Solubility:
Effect
Temperature
Change of
Solids
Temperture
Change of
Gases
Stirring
Particle Size
Pressure
Of
Gases
Reason
Does Solubility
Change?
Assignment 4: Ksp Problems:
1.
2.
Write the equation for the dissolving and the solubility product constant
expression for the following solids:
NaCl
Ksp = [Na] [Cl]
Ag2SO4
Ksp = [Ag]2 [SO4]
Pb(OH)2
Ksp = [Pb] [OH]2
Al2(SO4)3
Ksp = [Al]2 [SO4]3
The solubility of PbSO4 in water is 0.038 grams per liter. Calculate the solubility
product constant.
PbSO4 (s)  Pb+2 (aq) + SO4-2 (aq)
Ksp = [Pb+2][SO4-2]
0.038g x
1 mole = 1.25 x 10-4 moles in 1 liter
303 g
Ksp = [1.25 x 10-4] [1.25 x 10-4] = 1.6 x 10-8
3.
The solubility of silver(I) chromate in water is 0.024 g per liter. Determine the
solubility product constant.
Ag2CrO4 (s)  2 Ag+1 (aq) + CrO4-2 (aq)
Ksp = [Ag+1]2 [CrO4-2]
0.024g x
1 mole = 7.2 x 10-5 moles in 1 liter
332 g
Ksp = [2(7.2 x 10-5)]2 [7.2 x 10-5] = 1.0 x 10-8
4.
5.
If the Ksp of barium chromate is 8.3 x 10-11, calculate the solubility in moles per
liter.
BaCrO4 (s)  Ba+2 (aq) + CrO4-2 (aq)
Ksp = [Ba+2][CrO4-2]
I
x
0
0
C
-x
+x
+x
E
0
x
x
8.3 x 10-11 = x2
x = 9.2 x 10-6 M
If the Ksp for strontium chromate is 3.6 x 10-5, would a solution in which 1.02
grams is dissolved in one liter be saturated?
SrCrO4 (s)  Sr+2 (aq) + CrO4-2 (aq)
Ksp = [Sr+2][CrO4-2]
I
x
0
0
C
-x
+x
+x
E
0
x
x
3.6 x 10-5 = x2
x = 0.0060 M
0.0060 moles x 203.6 g
1 mole
= 1.2 g will dissolve in
1 liter (unsaturated)
6.
What happenes if the product of the concentrations of A+ ion and B- ion is greater
than the Ksp value?
The solution would be supersaturated and would not be stable. It would
precipitate is shaken or given a seed crystal.
7.
Will calcium carbonate precipitate if the Ca+2 concentration in a solution is
0.0002M and the CO-2 concentration is the same. The Ksp is 4.8 x 10-9.
CaCO3 (s)  Ca+2 (aq) + CO3-2 (aq)
Ksp = [Ca+2][CO3-2]
I
x
0
0
C
-x
+x
+x
E
0
x
x
4.8 x 10-9 = x2
x = 6.0 x 10-5 M
8.
Seawater is saturated with AgCl. The [Cl-] in seawater is 0.53M and the Ksp for
AgCl is 1.8 x 10-10. Calculate the [Ag+]
AgCl (s)  Ag+1 (aq) + Cl-1 (aq)
Ksp = [Ag+1][Cl-1]
1.8 x 10-10 = [Ag+1] [0.53]
x = 3.4 x 10-10 M
9.
yes, 0.0002 is larger than 0.000060
which is a saturated solution.
*There are lots of
salts in seawater, so
not all Cl-1 is from
AgCl. You shift the
reaction
When 12 M HCl is added to a saturated solution of NaCl, a precipitate forms.
Explain why?
NaCl (s)  Na+1 (aq) + Cl-1 (aq)
Ksp = [Na+1][Cl-1]
By adding Cl-1 ions, the reaction is shifted toward the reactants (the solid) so a
precipitate forms.
Challenge:
The Ksp for PbF2 is 3.7 x 10-8. Calculate the grams of each ion dissolved on one
liter of a saturated solution.
PbF2 (s)  Pb+2 (aq) + 2 F-1 (aq)
Ksp = [Pb+2][F-1]2
I
x
0
0
C
-x
+x
+2x
E
0
x
2x
3.7 x 10-8 = (x)(2x)2
3.7 x 10-8 = 4x3
x = 2.1 x 10-3 moles Pb+2 in one liter and 2(2.1 x 10-3) moles
of F-1 in one liter.
2.1 x 10-3 x 1 mole PbF2 = 0.51 grams PbF2 in 1 liter
245 g
Assignment 5: Solubility Curves
1. Does heat increase the solubility of all compounds the same amount? no
2. How many grams of solute are needed to saturate 100.0 g of water at 80.0oC to
prepare a saturated solution of:
Potassium bromide 98g
Sodium chloride 40g
3. At what temperature are sodium chloride and potassium nitrate equally soluble?
At about 22oC
4. Is sodium chloride of potassium nitrate more soluble at 20.0oC? NaCl
5. What is the molality of a saturated solution of potassium bromide at 50.0oC?
80g of KBr will dissolve in 100g of water at 50oC
80g x
1 mole
119 g KBr
= 0.67 moles
m = 0.67 moles KBr = 6.7m
0.1 kg water
6. What is the molality of a saturated solution of sodium chlorate at 10.0oC?
95g of NaClO3 will dissolve in 100g of water at 10oC
95g x
1 mole
= 0.89 moles
106.5 g NaClO3
m = 0.89 moles NaClO3 = 8.9m
0.1 kg water
7. Is a 2.0M solution of sodium chloride possible at 30.0oC?
2M = x moles = 0.2 moles NaCl
0.1 L
0.20 moles NaCl x
58.5 g NaCl
= 11.7 grams NaCl in 0.1 L
water (about 0.1kg or 100g)
The chart indicates that 40g NaCl will dissolve in 100g H2O at 30.0oC, so it is
possible to make a 2.0 M solution.
8. Are the following solutions saturated, unsaturated, or supersaturates?
40.0g of potassium bromide in 100 g of water at 20.0oC
The chart indicates that 70g of KBr will dissolve in 100g water at 20oC, so
40g in 100g water would be unsaturated.
40.0g of potassium bromide in 50.0g of water at 50.0oC
The chart indicates that 80g of KBr will dissolve in 100g of water at 50oC,
so half that amount would dissolve in 50g water:
80g KBr = x g KBr
100g H2O
50g H2O
x = 40g
The solution would be saturated.
20.0g of sodium chloride in 50.0g of water at 10.0oC
The chart indicates that 35g of NaCl will dissolve in 100g water at 10oC,
so half that amount will dissolve in 50g water.
35g NaCl = x g NaCl
100g H2O
50g H2O
x = 17.5g
So, if 20.0g of NaCl was dissolved in 50g of water, it would be
supersaturated.
Book Problems:
25.
I
C
E
PbS (s)  Pb+2 (aq) + S-2 (aq)
x
0
0
-x
+x
+x
0
x
x
Ksp = [Pb+2][S-2]
3 x 10-28 = x2
x = 2 x 10-14 M PbS is a saturated solution
27.
I
C
E
Ag2S (s)  2 Ag+1 (aq) + S-2 (aq)
x
0
0
-x
+2x
+x
0
2x
x
Ksp = [Ag+1]2 [S-2]
8 x 10-51 = (2x)2(x)
8 x 10-51 = 4x3
x = 1 x 10-17
so, [Ag+] = 2 (1 x 10-17) = 2 x 10-17 M
[S-2] = 1 x 10-17 M
(or M 1 x 10-17 Ag2S is a saturated solution
29.
FeS (s)  Fe+2 (aq) + S-2 (aq)
x
0.04
x
-x
+x
+x
0
0.04 + x
x
)
Ksp = [Fe+2][S-2]
*Because the Ksp is so small, you know the x in 0.04 – x above will be
negligible in comparison to 0.04, so you can leave the x out.
8 x 10-19 = [0.04] x
x = [2 x 10-17]
31.
Ca(NO3)2 (s)   Ca+2 (aq) + 2 NO3-1 (aq)
0.001
0
0
-0.001
+0.001
+0.002
0
0.001
0.002
Na2CO3 (s)   2 Na+1 (aq) + CO3-1 (aq)
0.0008
0
0
-0.0008
+0.0016
+0.0008
0
0.0016
0.0008
*This solution would
provide 0.001M Ca-2
*This solution would
provide
0.0008M CO3-2
so:
CaCO3 (s)  Ca+2 (aq) + CO3-2 (aq)
0.001
0.0008
Ksp = [0.001] [0.0008]
= 8 x 10-7
This number is higher than the actual Ksp of 4.5 x 10-9 so a ppt would
form!
Assignment 6: More Solubility Curves
1. Most substances on this graph show increased solubility as the temperature
increases. What are the exceptions?
NH3 (gas) and Ce2(SO4)3
2. Each curve shows how solubility for that substance changes as temperature
changes. The solubilities of substances whose curves show greater (steeper)
slopes are mores (more/less) affected by temperature changes than those that have
more gradual slopes.
NaCl 3. Which salt has solubility values that are least affected by changes in
temperature?
96oC
4. At what temperature do potassium chlorate and potassium chloride have the
same solubility in water?
KClO3 5. Which compound is least soluble in water at 12oC?
KNO3 6. A saturated solution of which compound contains 130 grams of solute per
100 grams of water at 70oC?
40 g
7. How many grams of sodium chloride are required to saturate 100 grams of
water at 100oC?
200g 8. How many grams of sodium chloride are required to saturate 500 grams of
water at 100oC?
40g NaCl = x g NaCl
x = 200g NaCl
100g H2O
500g H2O
160g 9. How many grams of sodium nitrate are required to saturate 200 grams of water
at 10oC?
80g NaNO3 = x g NaNO3 = 160g NaNO3
100g H2O
200g H2O
13g 10. At 50oC, 100 grams of water is saturated with cerium(III) sulfate. How many
grams of cerium(III) sulfate must be added to saturate the solution if the
temperature is changed to 0oC? 5g Ce2(SO4)3 at 50oC, 18g Ce2(SO4)3 at 0oC
18g – 5g = 13g need to be added
20g 11. At 50oC, 100 grams of water is saturated with potassium nitrate. How many
grams of potassium nitrate will precipitate when the solution is cooled to
40oC? 82g KNO3 at 50oC, 62g KNO3 at 40oC
82g – 62g = 20g will fall out.
13% 12. What is the percent by mass of a saturated solution of potassium chlorate at
40oC?
15g KNO3 x 100 = 13%
115g total
10.8m 13. What is the molarity of a saturated solution of NaNO3 at 25oC?
92g NaNO3 x 1 mole NaNO3 = 1.08 moles
85.0g NaNO3
m =
117
1.08 moles NaNO3
0.100 kg H2O
= 10.8m
14. What is the Ksp of the Na NO3 at 25oC?
NaNO3 (s)   Na+1 (aq) + NO3-1 (s)
Ksp = [Na+1] [NO3-1]
[10.8] [10.8] = 117
Assignment 7: Yet More Solubility Curves
20g
5g
20g
1. At 80oC, 100 grams of water is saturated with potassium chloride. How many
grams of solute will precipitate when the solution is cooled to 45oC
82g at 50oC, 62g at 40oC
82g – 62g = 20g ppt.
2. If 50 grams of water saturated with potassium chlorate at 23oC is slowly
evaporated to dryness, how many grams of salt will be recovered?
10gKClO3 = xg KClO3
x = 5g
100g H2O
50g H2O
3. If 30 grams of potassium chloride is dissolved in 100 grams of water, how
many additional grams would be needed to make the solution saturated at
80oC?
50g will go in 100g H2O at 80oC; 50g – 30g = 20g
51g
4. What is the smallest mass of water required to dissolve completely 23 grams of
ammonium chloride at 40oC?
45g NH4Cl = 23g NH4Cl
x = 51g H2O
100g H2O
x g H2O
1g/oC 5. A saturated solution of sodium nitrate in 100 grams of water at 40oC is heated
to 50oC. What is the rate of increase in solubility in grams per degree??
105g NaNO3 at 40oC
115g NaNO3 at 50oC
NH4Cl 6. Which chloride has the greatest percent by mass at 60oC? What is the percent?
55g NH4Cl x 100 = 35%
155g total
0.18M 7. What is the molarity (molality) of a saturated solution of cerium(III) sulfate at
25oC?
10g Ce2(SO4)3 x 1 mole Ce2(SO4)3 = 0.018 moles
568g Ce2(SO4)3
M = 0.018 moles Ce2(SO4)3 = 0.18M
0.100 L solution
0.020 8. What is the Ksp of a saturated solution of cerium(III) sulfate at 25oC?
Ce2(SO4)3 (s)   2 Ce+3 (aq) + 3 SO4-2 (aq)
[0.18]
0
0
- 0.18
+ 0.36
+ 0.54
0
[0.36]
[0.54]
Ksp = [Ce+3]2 [SO4-2]3 = [0.36]2 [0.54]3 = 0.020
-30 C 9. A solution of KNO3 is saturated at 50oC. At what temperature would the
solution freeze
82g KNO3 dissolve in 100g H2O at 50oC.
o
82g KNO3
x
1 mole KNO3 = 0.81 moles
101g KNO3
m = 0.81 moles = 8.1m
0.100 kg H2O
KNO3 (s)   K+1 (aq) + NO3-1 (aq)
[8.1]
0
0
-8.1
+8.1
+8.1
0
[8.1]
[8.1] = 16.2 m total particles
tf = 1.86oC/m x 16.2m = - 30oC
3.3m 10. A solution of KNO3 freezes at -12oC. At what temperature would the solution
be saturated?
-12oC = 1.86oC/m x m
m = 6.5m total particles
Because KNO3 breaks into 2 particles and the total is 6.5, it must be half that for
each ion or 3.3m each
KNO3 (s)   K+1 (aq) + NO3-1 (aq)
[3.3]
0
0
-3.3
+3.3
+3.3
0
[3.3]
[3.3]
= 6.5m total
Assignment 8: Molarity, molality, and colligative properties
1. What is the molarity of a solution in which 82.0g of calcium nitrate is dissolved in
enough water to make 500.0 mL of solution?
82.0g Ca(NO3)2 x 1 mole = 0.4998 moles
164.09g
M = 0.4998 mol = 1.00M
0.5000 L soln
2. What is the molality of a solution in which 50.0g of copper(II) sulfate is dissolved
in 250.0mL of water
50.0g CuSO4 x 1 mole CuSO4 = 0.313 moles
159.5 g CuSO4
m = 0.313 moles CuSO4 = 1.25 m
0.2500 kg H2O
3. Calculate the mass of solute in 250.0mL of sodium sulfate solution that is 2.00M
(molar).
2.00M = x moles Na2SO4 = 0.500 moles
0.2500 L soln
0.500 moles Na2SO4 x
142.04g Na2SO4 = 71.0g Na2SO4
1 moles Na2SO4
4. Calculate the mass of solute in 250.0mL of sodium sulfate solution that is 2.00m
(molal).
2.00M = x moles Na2SO4 = 0.500 moles
~ 0.2500 kg H2O
0.500 moles Na2SO4 x
142.04g Na2SO4 = 71.0g Na2SO4
1 moles Na2SO4
5. The Ksp of silver iodide is 8.3 x 10-17 at 25oC; what is the molarity of a saturated
solution? Write the reaction for the dissolving.
AgI (s)   Ag+1 (aq) +
x
0
0
-x
+x
+x
0
x
x
Ksp = [Ag+1] [I-1]
8.3 x 10-17 = x2
x = 9.1 x 10-9 M
I-1 (aq)
6. A saturated solution of lead(II) chloride at 25oC contains 2.2 grams of solute in
500.0 mL. What is the Ksp? Write the reaction for the dissolving.
2.2g PbCl2 x
1 mole PbCl2
278.1g PbCl2
= 7.9 x 10-3 moles
M = 7.9 x 10-3 moles = 0.016 M
0.5000 L
PbCl2 (s)   Pb+2 (aq)
[0.016]
0
- 0.016
+ 0.016
0
[0.016]
Ksp = [Pb+2] [Cl-1]2
+ 2 Cl-1 (aq)
0
+ 0.032
[0.032]
= [0.016] [0.032]2 = 1.6 x 10-5
Is the lead(II) chloride or the silver iodide in the previous problem more soluble?
17
The lead(II) chloride is more soluble as the Ksp is higher (1.6 x 10-5 vs 8.3 x 10)
7. What is the freezing point of a 0.85 molal solution of sugar? Kf = 1.86oC/m.
C12H22O11 (s)   C12H22O11 (aq)
0.85
0
- 0.85
+ 0.85
0
0.85
tf = 1.86oC/m x 0.85m = 1.6oC
so the freezing pt. is – 1.6oC
8. What is the freezing point of a solution that contains 68.5 grams of sucrose,
C12H22O11, dissolved in 100.0 grams of water?
68.5g C12H22O11 x
1 mole
342g
= 0.200 moles
m = 0.200 moles = 2.00 m
0.1000 kg
C12H22O11 (s)   C12H22O11 (aq)
2.00
0
- 2.00
+ 2.00
0
2.00
tf = 1.86oC/m x 2.00m = 3.72oC
so the freezing pt. is – 3.72oC
9. What is the freezing point of a solution that contains 68.5 grams of salt, NaCl,
dissolved in 100.0 grams of water?
68.5g NaCl x 1 mole = 1.17 moles
58.5g
m = 1.17 moles = 11.7 m
0.1000 kg
NaCl (s)   Na+1 (aq) + Cl-1 (aq)
11.7
0
0
- 11.7
+ 11.7
+11.7
0
11.7m
11.7m
23.5m total
tf = 1.86oC/m x 23.4m = 43.5oC
so the freezing pt. is – 43.5oC
Why is CaCl2 used on the roads instead of NaCl or sucrose?
When CaCl2 dissolves, 3 particles are produces. When NaCl dissolves, only 2
particles are produced.
10. What is the boiling point of the solution in problem #9?
68.5g NaCl x 1 mole
58.5g
= 1.17 moles
m = 1.17 moles = 11.7 m
0.1000 kg
NaCl (s)   Na+1 (aq) + Cl-1 (aq)
11.
0
0
- 11.7
+ 11.7
+11.7
0
11.7m
11.7m
23.5m total
tb = 0.512oC/m x 23.5m
tb = 12oC
so the boiling pt. is 112oC
Assignment 10: Solutions Review
Define:
1.
Solution
2.
Solvent
3.
Solute
4.
Homogeneous
5.
Heterogeneous
6.
Soluble
7.
Insoluble
8.
Solvation
9.
Colligative property
10.
Saturated solution (how do you tell?)
11.
Unsaturated solution (how do you tell?)
12.
Supersaturated solution (how do you tell?)
Questions:
Write the reaction for the dissolving of sucrose in water. Does sucrose
dissociate?
C12H22O11 (s) ------- C12H22O11 (aq) Does not dissociate. Molecules stay together.
Write the reaction for the dissolving of potassium bromide in water? Does
KBr dissociate?
KBr (s) ------ K+ (aq) + Br- (aq) Does dissociate into ions.
Write the Ksp expression for the above equations. What does Ksp describe?
What does a large Ksp mean?
Ksp = [K+ ] [Br- ]
Ksp = [C12H22O11]
The solubility product constant describes how far the reaction goes toward the products
which is the dissolved for of the compound. A large Ksp means the compound is more
soluble.
What happens to the freezing point of a solvent when a solute is added? The
boiling point?
The freezing point is lowered because the solute gets in the way of crystal
formation. The boiling point is increased due to a lowered vapor pressure. The water
molecules are attracted to the solute and are less likely to evaporate causing a lowered
vapor pressure. Therefore, more heat must be added to get the vapor pressure to equal
the atmospheric pressure.
Would the addition of sodium chloride or the addition of calcium chloride
raise the boiling point of water more? Explain.
Calcium chloride would raise the B.P. more because there are more ions released
upon dissolving of calcium chloride that sodium chloride.
Is it safer to ice skate on a frozen lake or a frozen oceanic bay in the same
conditions? Why?
It would be safer to skate on the lake because the salt dissolved in the oceanic water
would depress the freezing point, so that ice would not be as thick.
List 4 factors that effect the dissolving of solutes. What effect do each of
these factors have? For which factor(s) would the Ksp change? Would the
Ksp go up or down?
Temperature change-heating speeds up molecules so dissolve faster and more.
Ksp increases with heat. The opposite is true for gases. Increased temperature
causes the dissolved gas to "boil" out.
Stirring-brings solute in contact with solvent so speeds up dissolving. Ksp
unchanged.
Particle Size-more surface area, so more contact of solvent with solute, so speeds
up dissolving. Ksp unchanged.
Pressure of gases-pushes molecules closer together so more dissolves. Ksp goes
up.
Describe some techniques that could be used to separate the components of a
solution.
Your should be able to describe:
Distillation (look at your lab)
Reverse Osmosis
Chromatography
Dialysis
Problems:
What is the molarity of a solution in which 45.0g of sodium nitrate is added
to enough water to make 500.0mL of solution?
45.0 g NaNO3 x 1 mole NaNO3
85.01 g NaNO3
0.529 mole NaNO3 =
.5000 L soln
=
0.529 mole NaNO3
1.06 M
What is the molality of a solution in which 65.0g of potassium chloride is
added to 600.0mL of water?
65.0 g KCl x 1 mole KCl
74.5 g KCl
=
0.872 mole KCl
0.872 mole KBr = 1.45 m
.6000 kg water
How many grams of sodium chloride are needed to make 100.0mL of a
2.0M solution?
.1000 L x 2.0 moles NaCl
1 L solution
= 0.20 moles x 58.5 g NaCl = 12g NaCl
1 mole
How many moles of ammonium chloride are used with 2.0L of water
to make a 0.50m solution?
2.0 kg x 0.50 moles = 1.0 moles NH4Cl x 53.5 g NH4Cl = 53.5 g NH4Cl
1 kg
1 mole NH4Cl
What is the expression for the solubility product constant (Ksp) for the
dissolving of calcium chloride?
CaCl2 (s)   Ca+2 (aq) + 2 Cl-1 (aq)
Ksp + [Ca+2] [Cl-1]2
The solubility of copper(II) hydroxide is 3.4 x 10-7 moles per liter of
solution; what is the solubility product constant (Ksp)?
Cu(OH)2 (s)   Cu+2 (aq) + 2 OH-1 (aq)
3.4 x 10-7
0
0
-7
-7
- 3.4 x 10
+ 3.4 x 10
+ 6.8 x 10-7
0
3.4 x 10-7
6.8 x 10-7
Ksp = [Cu+2] [OH-1]2
= [3.4 x 10-7] [6.8 x 10-7]2
= 1.6 x 10-19
If 1.2g of SrCrO4 will dissolve in 1 liter of solution, calculate the Ksp.
1.2g SrCrO4 x
1 mole SrCrO4
203.6g SrCrO4
= 5.9 x 10-3 moles/L = Molarity
SrCrO4 (s)   Sr+2 (aq) + CrO4-2 (aq)
[5.9 x 10-3]
0
0
- 5.9 x 10-3
+ 5.9 x 10-3
+ 5.9 x 10-3
0
[5.9 x 10-3]
[ 5.9 x 10-3]
Ksp = [Sr+2] [CrO4-2]
= [5.9 x 10-3] [5.9 x 10-3]
= 3.5 x 10-5
If 0.85g of MgF2 will dissolve in 0.50 liters of solution, calculate the Ksp.
0.85g MgF2 x
1 mole MgF2
62.27g MgF2
= 0.14 moles
M = 0.14 moles MgF2
0.50 L solution
MgF2 (s)   Mg+2 (aq)
[0.027]
0
- 0.027
+0.027
0
[0.027]
= 0.027M
2 F-1 (aq)
0
+ 0.054
[0.054]
+
Ksp = [Mg+2] [F-1]2
= [0.027] [0.054]2
= 7.8 x 10-5
Is the SrCrO4 or the MgF2 more soluble? How do you know?
The MgF2 is more soluble because it has a higher Ksp.
If the Ksp is 2 x 10-16 at 250, what is the molarity of a saturated solution at
this temperature?
PbCrO4 (s) ------ Pb+2 (aq) + CrO4 -2 (aq)
x
0
0
-x
+x
+x
0
x
x
2 x 10-16 = [Pb+2 ] [CrO4-2 ]
2 x 10-16 = x2
1.4 x 10-8 M = [PbCrO4 ]
If the Ksp for the dissolving of CoS in water is 4.9 x 10-22, calculate the
solubility in moles per liter of solution (maximum molarity).
CoS (s)   Co+2 (aq)
x
0
-x
+x
0
x
+ S-2 (aq)
0
+x
x
4.9 x 10-22 = [Co+2] [S-2]
4.9 x 10-22 = [x] [x] or x2
x = 2.2 x 10-11 M
The solubility product constant for the dissolving of AgI in water is
8.5x 10-17 . Calculate the solubility of AgI in grams per one liter of
solution.
AgI (s)   Ag+1 (aq)
x
0
-x
+x
0
x
+
I-1 (aq)
0
+x
x
Ksp = [Ag+1] [I-1]
8.5 x 10-17 = [x][x]
x = 9.2 x 10-9 M
9.2 x 10-9 moles x 235 g AgI = 2.2 x 10-6 grams in 1L
1 mole AgI
The Ksp for the dissolving of ZnS is 1.3 x 10-22. Would it be possible to
dissolve 0.045g in 780mL of water? If so, is the solution saturated or
unsaturated?
ZnS (s)   Zn+2 (aq)
x
0
-x
+x
0
x
+
S-2 (aq)
0
+x
x
Ksp = [Zn+2] [S-2]
1.3 x 10-22 = [x][x]
x = 1.1 x 10-11 M
1.1 x 10-11 moles x
1 mole ZnS
97.4 g ZnS
1.07 x 10-9 g ZnS
1000 ml
=
= 1.07 x 10-9 grams in 1L (1000mL)
x g ZnS
780 mL
x = 8.3 x 10-10 grams
so….0.045g will not dissolve in 780mL of water.
What is the freezing point of an NaCl solution that contains 21.2 grams
solute in 135g water?
21.2g NaCl x
1 mole NaCl
58.5g NaCl
= 0.368 moles NaCl
m = 0.368 moles NaCl = 2.72m
0.135 kg H2O
NaCl (s)   Na+1 (aq)
2.72
0
- 2.72
+ 2.72
0
2.72
+ Cl-1 (aq)
0
+ 2.72
2.72
= 5.44m total particle
tf = 1.86oC/m x 5.44m = 10.1oC
so freezing point would be – 10.1oC
Which of the following solutions has (a) the higher boiling point and (b) the
higher melting point: 0.35m calcium chloride or 0.9m sucrose
CaCl2 (s)   Zn+2 (aq) + 2 Cl-1 (aq)
0.35
0
0
- 0.35
+ 0.35
+ 0.70
0
0.35
0.70 = 1.05m total particles
C12H22O11 (s)   C12H22O11 (aq)
0.9
0
- 0.9
+ 0.9
0
0.9
= 0.9 total particle
therefore CaCl2 causes the larger freezing point depression and the larger boiling
point elevation. So, CaCl2 has the higher boiling point and sucrose has the higher
melting (freezing) point.
What are the boiling points of the above solutions if the Kb for water is
0.512oC/m?
For CaCl2
tb = 0.512oC/m x 1.05m = 0.54oC
So it boils at 100.54oC.
For Sucrose tb = 0.512oC/m x 0.9m = 0.46oC
So it boils at 100.46oC
Arrange the following solutions in order of decreasing freezing point:
0.10m Na3PO4
dissociates to 4 particles so 0.40m total
0.35m NaCl
dissociates to 2 particles so 0.70m total
0.20m MgCl2
dissociates to 3 particles so 0.60m total
0.15m C12H22O11 does not dissociate, so 0.15m total
C12H22O11, Na3PO4, MgCl2, NaCl = lowest freezing point
Use the chart on assignments 6 and 7 to answer the following:
Determine if the solutions above are saturated, unsaturated, or supersaturated
if prepared at 25oC. (hint: you have to express as grams in 100g water)
9 g of NaNO3 in 100 g of water is unsaturated at any temp.
45g NaNO3 = x g NaNO3
500g H2O
100g H2O
10.8 g of KBr in 100 g of water is unsaturated at any temp.
10.8g KCl = x g KCl
600g H2O
100g H2O
12 g NaCl in 100 g of water is unsaturated at any temp.
2.68 g of NH4Cl in 100 g of water is unsaturated at any temp.
53.5g NH4Cl = x g NH4Cl
2000g H2O
100g H2O
Calculate the Ksp of a saturated solution of potassium nitrate at 25oC.
KNO3 (s) ------- K+ (aq) + NO3 (aq)
Ksp = [K+ ] [NO3]
At 25 degrees, about 40 grams of potassium nitrate will dissolve in 100 g of water:
40 g x 1 mole KNO3 = 0.396 moles KNO3
101 g KNO3
0.396 moles KNO3 = 4.0 M (about)
.100 L
so…. Ksp = [4.0 ] [4.0] = 16
How many grams of sodium chloride will dissolve in 100g of water at 30 oC?
About 38 g NaCl will dissolve in 100g H2O
How many grams of ammonium chloride will dissolve in 122g of water at
15oC?
35 g NH4Cl = x g NH4Cl
100 g water
122 g water
= 42.7 g NH4Cl
How many grams will precipitate out of 100mL of a solution of potassium
chloride that is saturated at 50oC if the solution is cooled to 10oC?
At 50 degrees, about 40 g will dissolve. At 10 degrees, only about 30 grams will
dissolve. About 10 grams will precipitate out.
A solution of NaCl is saturated at 25oC. At what temperature would the
solution freeze.
At 25oC, about 38g of NaCl will dissolve in 100g water.
38g NaCl x
1 mole NaCl
58.5g NaCl
= 0.65 moles NaCl
m = 0.65 moles NaCl = 6.5m
0.100 kg H2O
NaCl (s)   Na+1 (aq)
6.5
0
- 6.5
+ 6.5
0
6.5
+ Cl-1 (aq)
0
+ 6.5
6.5
= 13m total particle
tf = 1.86oC/m x 13m = 23.9oC
so freezing point would be – 23.9oC