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
1.5x 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