Date:__________________________
Block:____
Name:_________________________
Chemistry 12 – Solubility Equilibrium Unit Test
/ 71 marks
Multiple Choice Section: Choose the best answer. (16 marks)
1. Molecular solutions do not conduct electricity because they contain:
a)
b)
c)
d)
molecules only
cations and anions
molecules and anions
molecules and cations
2. To determine the solubility of a solute in water, a solution must be prepared that is:
a)
b)
c)
d)
saturated
unsaturated
concentrated
supersaturated
3. From the list of salts below, how many are considered soluble at 25°C?
CuCl2
a)
b)
c)
d)
CaSO4
PbS
Ag3PO4
none
one
two
three
4. Consider the following equilibrium:
NH4Cl (s)  NH4+ (aq) + Cl- (aq)
ΔH = positive
Which of the following will increase the solubility of ammonium chloride?
a)
b)
c)
d)
stirring the solution
adding water
adding NH4Cl
heating the solution
5. Na2SO4 solution is slowly added to a solution which contains 0.10 M Ba2+ and 0.10 M
Pb2+. Which of the following statements describes the result of the addition of Na2SO4?
a)
b)
c)
d)
BaSO4 precipitates first because it is more soluble.
PbSO4 precipitates first because it is more soluble.
BaSO4 precipitates first because it is less soluble.
PbSO4 precipitates first because it is less soluble.
Date:__________________________
Block:____
Name:_________________________
6. Identify the most soluble sulphide.
a)
b)
c)
d)
HgS
PbS
FeS
MnS
Ksp = 1.6 x 10-54
Ksp = 7.0 x 10-29
Ksp = 3.7 x 10-19
Ksp = 2.3 x 10-13
7. Four samples of a solution were analyzed and the following data were collected:
Anion added
S2SO42OHCO32-
Observation
nothing
precipitate
nothing
precipitate
Which one of the following Group II cations is found in the unknown solution?
a)
b)
c)
d)
Mg2+
Ca2+
Sr2+
Ba2+
8. The [OH-] is measured to be 3.3 x 10-3 mol/L in a 100 mL sample of a saturated solution
of Al(OH)3. What is the solubility of Al(OH)3?
a)
b)
c)
d)
1.1 x 10-4 mol/L
3.3 x 10-4 mol/L
1.1 x 10-3 mol/L
3.3 x 10-3 mol/L
9. Which of the following salts has the lowest solubility?
a)
b)
c)
d)
Copper (I) chloride
Ammonium sulphide
Potassium hydroxide
Mercury (II) sulphate
10. The mixture that could produce a precipitate of two compounds is:
a)
b)
c)
d)
HgSO4 and FeCl2
AgNO3 and MgCl2
K2CO3 and CuSO4
ZnSO4 and Ba(OH)2
Date:__________________________
Block:____
Name:_________________________
11. In a saturated solution of zinc hydroxide, at 40°C, the [Zn2+] = 1.8 x 10-5 M. What is
the Ksp of the compound?
a)
b)
c)
d)
5.8 x 10-15
2.3 x 10-14
1.8 x 10-14
6.5 x 10-10
12. When equal volumes of 0.060 M AgNO3 and 0.00090 M Ba(BrO3)2 are mixed, the trial
ion product is:
a)
b)
c)
d)
Less than Ksp and a precipitate forms.
Greater than Ksp and a precipitate forms.
Less than Ksp and no precipitate forms.
Greater than Ksp and no precipitate forms.
13. What is the maximum amount of sodium sulphate that will dissolve in 1.0 L of 0.10 M
Pb(NO3)2 without forming a precipitate?
a)
b)
c)
d)
1.8 x 10-8 mol
1.8 x 10-7 mol
1.3 x 10-4 mol
1.0 x 10-1 mol
14. Which one of the following equilibrium systems is described by a Ksp?
a)
b)
c)
d)
CaCO3 (s)  CaO (s) + CO2 (g)
CaCO3 (s)  Ca2+ (aq) + CO32- (aq)
Ca2+ (aq) + CO32- (aq)  CaCO3 (s)
Ca(OH)2 (aq) + H2CO3 (aq)  CaCO3 (s) + 2H2O (l)
15. In an experiment, a student mixes equal volumes of 0.0020 M Pb2+ ions with 0.0040 M Iions. What is the trial ion product?
a)
b)
c)
d)
4.0 x 10-9
3.2 x 10-8
1.3 x 10-7
8.0 x 10-6
16. A 0.50 L solution of CuBr2 contains 0.30 mol Br- ions. What are the ionic
concentrations in the solution?
a)
b)
c)
d)
[Cu2+]
[Cu2+]
[Cu2+]
[Cu2+]
=
=
=
=
0.15 M
0.30 M
0.60 M
0.60 M
[Br-]
[Br-]
[Br-]
[Br-]
=
=
=
=
0.30 M
0.60 M
0.60 M
1.20 M
Date:__________________________
Block:____
Name:_________________________
Written Section: Answer the following questions showing all your work! (55 marks)
1. Write an equation that describes the equilibrium present in a saturated solution of
Cu3PO4. (1 mark)
2. 53 g of Na2CO3 are dissolved in sufficient water to make 5.0 L of solution.
a) Write the equation for the dissolution of Na2CO3 into its aqueous ions. (1 mark)
b) Calculate the concentration of each ion in solution. (3 marks)
c) Describe the change in entropy as the Na2CO3 dissolves. (1 mark)
3. Write an equilibrium equation and a Ksp expression for only those salts that have low
solubility: (5 marks)
a) Cu2S
b) PbI2
c) Al2(SO4)3
4. A 1.0 M solution of sodium sulphide is added to a 1.0 M solution of copper (II) chloride
resulting in the formation of a precipitate.
a) Write the formula of the precipitate. (1 mark)
b) Write the formula equation for the reaction. (1 mark)
c) Write the net ionic equation for the reaction. (1 mark)
5. Calculate the concentration of each ion in the following saturated solutions: (12 marks)
a) Al(OH)3
Ksp = 3.0 x 10-33
b) Pb(IO3)2
Ksp = 3.7 x 10-13
6. A suspension of barium sulphate is used to improve the quality of X-rays in the digestive
system. If the patient is required to drink 0.400 L of this suspension, calculate the actual
mass in grams of the dissolved BaSO4. (5 marks)
7. Calculate the Ksp for Mg(OH)2 if the solubility of Mg(OH)2 is 7.6 mg/L. (5 marks)
8. What maximum [F-] exists in a solution in which the [Sr2+] = 4.4 x 10-3 M? (4 marks)
Date:__________________________
Block:____
Name:_________________________
9. Show by calculation if a precipitate forms when 10.0 mL of 0.010 M AgNO3 are mixed
with an equal volume of 0.10 M Na2CO3. (5 marks)
10. A solution may contain Ba2+ and/or Al3+ ions. Describe a procedure to confirm the
presence or absence of these ions. (4 marks)
11. A 25.0 mL sample of saturated PbCl2 solution is titrated to the endpoint with 48.1 mL of
0.015 M AgNO3 solution. Calculate the Ksp of PbCl2. (6 marks)
Date:__________________________
Block:____
Name:_________________________
Multiple Choice Answers:
1.
2.
3.
4.
a
a
b
d
5.
6.
7.
8.
c
d
c
c
9. a
10. d
11. b
12. c
13. b
14. b
15. a
16. b
Written Answers:
1. Cu3PO4 (s)  3Cu+ (aq) + PO43- (aq)
2. a) Na2CO3(s)  2Na+ (aq) + CO32- (aq)
b) Molar Mass Na2CO3 = 106.0 g / mol.
[Na2CO3] = 53 g x (1 mol / 106.0 g) x (1 / 5.0 L) = 0.10 M
[Na+] = 2 x 0.10 M = 0.20 M
[CO32-] = 0.10 M
c) Entropy increases as ions are formed.
3. a) Cu2S (s)  2Cu+ + S2b) PbI2 (s)  Pb2+ + 2I-
Ksp = [Cu+]2 [S2-]
Ksp = [Pb2+] [I-]2
c) Soluble (Therefore, no rxn)
4. a) CuS
b) Na2S (aq) + CuCl2 (aq)  CuS (s) + 2NaCl (aq)
c) Cu2+ (aq) + S2- (aq)  CuS (s)
5. a) Al(OH)3 (s)  Al3+ + 3OHx
3x
Let x = [Al3+] ; 3x = [OH-]
Ksp = [Al3+] [OH-]3 = x (27x3) = 27x4 = 3.0 x 10-33
x = 3.25 x 10-9 M
Therefore, [Al3+] = x = 3.3 x 10-9 M and [OH-] = 3x = 9.7 x 10-9 M.
Date:__________________________
Block:____
b) Pb(IO3)2 (s)  Pb2+ + 2IO3x
2x
Name:_________________________
Let x = [Pb2+] ; 2x = [IO3-]
Ksp = [Pb2+] [IO3-]2 = x (4x2) = 4x3 = 3.7 x 10-13
x = 4.52 x 10-5 M
Therefore, [Pb2+] = x = 4.5 x 10-5 M and [IO3-] = 2x = 9.0 x 10-5 M.
6. BaSO4 (s)  Ba2+ + SO42x
x
x
Let x = solubility of BaSO4 = [Ba2+] = [SO42-]
Ksp = [Ba2+] [SO42-] = x2 = 1.1 x 10-10
x = 1.05 x 10-5 M
Molar mass BaSO4 = 233.4 g / mol
Mass BaSO4 = 1.05 x 10-5 mol / L x 0.400 L x (233.4 g / 1 mol) = 9.8 x 10-4 g.
7. Mg(OH)2 (s)  Mg2+ + 2OHx
x
2x Let x = solubility of Mg(OH)2 = [Mg2+]; 2x = [OH-]
Molar mass Mg(OH)2 = 58.3 g / mol
Mass Mg(OH)2 = 7.6 mg / L x (1 g / 1000 mg) x (1 mol / 58.3 g) = 1.30 x 10-4 M
[Mg2+] = x = 1.30 x 10-4 M
[OH-] = 2x = 2.61 x 10-4 M
Ksp = [Mg2+] [OH-]2 = (1.30 x 10-4 M) (2.61 x 10-4 M)2 = 8.8 x 10-12
8. SrF2 (s)  Sr2+ + 2FKsp = [Sr2+] [F-]2 = [4.4 x 10-3] [F-]2 = 4.3 x 10-9
[F-] = 9.9 x 10-4 M
Date:__________________________
Block:____
Name:_________________________
9. Dilution calculations required!
[Ag+] = (10.0 mL x 0.010 M) / 20.0 mL = 0.0050 M
[CO32-] = (10.0 mL x 0.010 M) / 20.0 mL = 0.0050 M
Ag2CO3 (s)  2Ag+ + CO32Ksp = [Ag+]2 [CO32-] = 8.5 x 10-12
Q = [Ag+]2 [CO32-] = (0.0050)2 (0.0050) = 1.25 x 10-6
Q > Ksp (Therefore, a ppt forms!)
10. Multiple answers exist. A possible answer:
Step 1: Add Na2SO4 to ppt out Ba2+ as BaSO4. Filter if ppt forms.
Step 2: Add Na2SO3 to ppt out Al3+ as Al2(SO3)3. Filter if ppt forms.
11. Moles of Ag+ added = 0.0481 L x 0.015 mol / L = 0.000722 mol Ag+
Moles of Ag+ = moles of Cl- in solution
[Cl-] = (0.000722 mol) / 0.0250 L = 2.89 x 10-2 M
PbCl2 (s)  Pb2+ + 2Cl[Pb2+] = 0.5 x [Cl-] = 0.5 x (2.89 x 10-2 M) = 1.44 x 10-2 M
Ksp = [Pb2+] [Cl-]2 = (1.44 x 10-2 M) (2.89 x 10-2 M)2 = 1.2 x 10-5