Dynamic Equilibrium – Le Chatelier’s Principle
1.
Some solid radioactive iodine-131 is added to a saturated solution of iodine in a beaker. The beaker also contains
some non-radioactive solid iodine crystals. Eventually, the initially non radioactive solution becomes radioactive.
How do you account for this observation? Explain what is happening at the atomic scale.
(6 marks)
Addition of the radioactive iodine does not cause the equilibrium to shift because the solution is already saturated
in iodine. However, since dissolving and recrystallization continues, the radioactive iodine becomes part of the
existing equlibrium, dissolving and recrystallizing with the regular iodine so that the radioactive atoms are spread
throughout the beaker.
2.
The solubility of a solid is 0.00500 g/100. mL. Would a solute / solution equilibrium be established if
5.00 mg of this solid were dissolved in 10.0 litres of solution? Explain (ie. what are the conditions needed for a
solute/solution equilibrium ?). Show your math. (6 marks)
The solubility of the substance is 0.00500 g/100. mL x (1000 mL/ 1L)
= 0.0500 g/L x 10.0 L
= 0.500 g of the solid can dissolve in the given water
5.00 mg x (1g / 1000mg)
= 0.00500 g of the solid is available
Since more of the solid can dissolve than is available, the solution will be unsaturated.
Equilibrium would not be established.
3.
For each of the following predict whether the given change will cause the equilibrium to shift in favour of the
products (to the right) or the reactants (to the left), or neither. Explain your answers. (15 marks)
a)
2 Cl2 (g) + 2 H2O
i)
ii)
iii)
iv)
v)
b)
 4 HCl
(g)
+ O2 (g) + energy
Increase temperature
Decrease total pressure
Add more O2 at constant pressure
Add a catalyst
Decrease volume of the container
reactants
products
reactants
neither
reactants
Ag2CrO4(s) + heat  2 Ag1+(aq) + CrO42-(aq)
i)
ii)
iii)
iv)
v)
c)
(g)
Add more Ag2CrO4(s)
Increase temperature
Add AgNO3(aq)
Increase pressure
Add NaCl(aq) (which precipitates AgCl)
neither
products
reactant (common ion effect)
neither
products
4 CuO (s) + CH4 (g)  CO2 (g) + 4 Cu (s) + 2 H2O (g)
i)
Add CuO(s)
neither
ii)
Increase temperature
reactant
iii)
Add Cu(s)
neither
iv)
Add CO2(g) at constant volume.
reactant
v)
Increase volume of the container.
product
ΔH = - 342 kJ
d)
CoCl42-(aq) + 6 H2O
i)
ii)
iii)
iv)
v)
e)
3 Fe
i)
ii)
iii)
iv)
v)
(l)
 Co(H20)62+
(aq)
+ 4 Cl1- (aq) + heat
Increase temperature
Remove CoCl42- ions
Add more chloride ions at constant pressure
Add a catalyst
Decrease volume of the container
(s)
+ 4 H2O
(g)
 Fe3O4
(s)
+ 4 H2
Add Fe(s)
Decrease temperature
Increase volume of the container.
Add Fe3O4 (s)
Add H2(g) at constant volume.
(g)
reactant
reactant
reactant
neither
neither
ΔH = - 149 kJ
neither
product
neither
neither
reactant
Equilibrium Constants - Calculations
1.
Write the correct Keq expression for the following chemical reaction equations. (2 marks each)
i)
4 CuO
(s)
+ CH4
(g)
 CO2
+ 4 Cu
(g)
(s)
+ 2 H2O
(g)
Keq = [CO2][H2O]2
[CH4]
ii)
PbI4 (s)  Pb4+(aq) + 4 I1-(aq)
Keq = [Pb4+][I1-]4
iii)
2 Bi3+(aq) + 3 H2S(g)  Bi2S3 (s) + 6 H1+(aq)
Keq =
iv)
4 NO(g) + 6 H2O(g)  4 NH3(g) + 5 O2(g)
Keq =
v)
2 Cl2
(g)
[NH3]4[O2]5
[NO]4[H2O]6
+ 2 H2O
Keq =
vi)
[H1+]6
[Bi ] [H2S]3
3+ 2
(g)
 4 HCl
(g)
+ O2
(g)
[HCl]4[O2]
[Cl2]2[H2O]2
Zn(s) + 2 Ag1+(aq)  Zn2+(aq) + 2 Ag(s)
[Zn2+]
[Ag1+]2
CH3COOH(aq) + C2H5OH(aq)  CH3COC2H5
Keq =
vii)
Keq =
(aq)
+ H2O (l)
[CH3COC2H5]
[CH3COOH ][C2H5OH]
viii) Ca3(PO4)2 (s)  3 Ca2+(aq) + 2 PO43-(aq)
Keq = [Ca2+]3[PO43-]2
ix)
8 H1+(aq) + 5 Fe2+(aq) + MnO41- (aq)  Mn2+ (aq) + 5 Fe3+(aq) + 4 H2O
Keq =
x)
NaHCO3
(s)
[Mn2+][Fe3+]5
[H ] [Fe2+]5[MnO41-]
1+ 8
 NaOH
Keq = [CO2]
(s)
+ CO2
(g)
(l)
xi)
2 Mn(s) + 3 Ca2+(aq)  2 Mn3+(aq) + 3 Ca(s)
Keq =
2.
[Mn3+]2
[Ca2+]3
Acetic acid is a weak acid that dissociates into the acetate ion and a proton in aqueous solution:
HC2H3O2
(aq)
 C2H3O21-
(aq)
+ H1+
(aq)
At equilibrium at 25 °C a 0.100 mol/L solution of acetic acid has the following concentrations:
[HC2H3O2] = 0.0990 mol/L, [C2H3O21-] = 1.33 x 10-3 mol/L and [H1+] = 1.33 x 10-3 mol/L.
Calculate the equilibrium constant for this reaction. (4 marks)
Keq = [C2H3O21-][H1+] = (1.33 x 10-3 mol/L)(1.33 x 10-3 mol/L)
[HC2H3O2]
(0.0990 mol/L)
3.
Write an equilibrium expression for the following reaction:
= 1.79 x 10-5
2 A(g) + B(g)  A2B(g)
Then calculate the value of Keq, given that [A] = 1.0 x 10-6 mol/L, [B] = 2.2 x 10-4 mol/L,
and [A2B] = 6.5 x 10-1 mol/L. (4 marks)
Keq = [A2B] =
[A]2[B]
4.
(6.5 x 10-1 mol/L)
(1.0 x 10-6 mol/L)2(2.2 x 10-4 mol/L)
= 3.0 x 1015
Write an equilibrium expression for the following reaction:
2 A(g) + 3 B(g)  A2B3(s)
Then calculate the value of Keq, given that [A] = 4.60 x 10-3 mol/L and [B] = 1.50 x 10-5 mol/L.
(4 marks)
Keq =
1
=
1
= 1.40 x 1019
[A]2[B]3
(4.60 x 10-3 mol/L)2(1.50 x 10-5 mol/L)3
5.
The value of Keq for the equilibrium
H2
(g)
+ I2
(g)
 2 HI
(g)
is 54.0 at 427 °C. If the [H2] at equilibrium is 0.145 mol/L and the [I2] = 0.0811 mol/L what is the equilibrium
[HI] ? (4 marks)
Keq =
[HI]2
[H2][I2]
[HI]
= √ (54.0)(0.145 mol/L)(0.0811 mol/L)
= 0.797 mol/L
6.
Write an equilibrium expression for the following reaction:
A2(g) + B(s)  A(g) + AB(s)
Then calculate the concentration of A(g), given that Keq = 1.5 x 10-3 and [A2] = 2.5 x 10-4 mol/L.
(4 marks)
Keq =
[A]
[A2]
[A] = (1.5 x 10-3)(2.5 x 10-4 mol/L) = 3.8 x 10-7 mol/L
7.
Write an equilibrium expression for the following reaction:
2 A(g) + B2(g)  A2B(g) + B(g)
Then calculate the concentration of A 2B(g), given that Keq = 7.1 x 104, [A] = 1.9 x 10-2 mol/L,
[B2] = 4.1 x 10-3 mol/L and [B] = 8.4 x 10-3 mol/L. (4 marks)
Keq =
[A2B][B]
[A]2[B2]
[A2B] = (7.1 x 104)(1.9 x 10-2 mol/L)2(4.1 x 10-3 mol/L) = 13 mol/L
8.4 x 10-3 mol/L
8.
Consider the following reaction:
2 NOBr
(g)
 2 NO
(g)
+ Br2
(g)
At equilibrium at 25 °C the concentration of [NOBr] = 0.1990 mol/L, [NO] = 6.23 x 10-5 mol/L and
[Br2] = 3.12 x 10-5 mol/L. Calculate the equilibrium constant for this reaction. (4 marks)
Keq =
[NO]2[Br2]
[NOBr]2
=
(6.23 x 10-5 mol/L)2(3.12 x 10-5 mol/L)
(0.1990 mol/L)2
= 3.06 x 10-12
9.
Write an equilibrium expression for the following reaction:
3 A(g) + B(g)  A3B(g)
Then calculate the value of Keq, given that [A] = 1.00 x 10-6 mol/L, [B] = 2.20 x 10-4 mol/L,
and [A2B] = 6.50 x 10-1 mol/L. (4 marks)
Keq = [A3B] =
[A]3[B]
(6.50 x 10-1 mol/L)
= 2.95 x 1021
-6
3
-4
(1.00 x 10 mol/L) (2.20 x 10 mol/L)
10. Write an equilibrium expression for the following reaction:
2 A(s) + 3 B(g)  A2B3(s)
Then calculate the value of Keq, given that [B] = 1.50 x 10-5 mol/L. (4 marks)
Keq =
1
[B]3
=
1
(1.50 x 10-5 mol/L)3
= 2.96 x 1014
11. The value of Keq for the equilibrium
2 HI
(g)
 H2
(g)
+ I2
(g)
is 0.0185 at 427 °C. If the [H2] at equilibrium is 0.0451 mol/L and the [I2] = 0.00118 mol/L what is the
equilibrium [HI] ? (4 marks)
Keq = [H2][I2]
[HI]2
[HI]
= √ (0.0451 mol/L)(0.00118 mol/L) / 0.0185
= 0.0536 mol/L
12. Write an equilibrium expression for the following reaction:
A2 (g) + 2 B(g)  A
(g)
+ AB2 (s)
Then calculate the concentration of A(g), given that Keq = 1.5 x 10-3, [A2] = 2.5 x 10-4 mol/L, and
[B] = 4.27 x 10-5 mol/L. (4 marks)
Keq =
[A]
[A2][B]2
[A] = (1.5 x 10-3)(2.5 x 10-4 mol/L)(4.27 x 10-5)2 = 6.8 x 10-16 mol/L
11. Write an equilibrium expression for the following reaction:
2 Cl2 (g) + 2 H2O (g)  4 HCl
(g)
+ O2 (g)
Then calculate the concentration of H2O(g), given that Keq = 0.0752, [Cl2] = 1.9 x 10-2 mol/L,
[HCl] = 4.1 x 10-3 mol/L and [O2] = 8.4 x 10-3 mol/L. (4 marks)
Keq =
[HCl]4[O2]
[Cl2]2[H2O]2
[H2O] = √ (8.4 x 10-3 mol/L)(4.1 x 10-3 mol/L)4 = 3.0 x 10-4 mol/L
(1.9 x 10-2 mol/L)2(0.0752)
ICE box problems
1.
Carbon monoxide reacts with water in the following reaction:
CO
(g)
+ H2O
(l)
 CO2
(g)
+ H2
(g)
Initially 0.0150 moles of each of the substances are added to a 1.00 L flask. At equilibrium
[CO] = 0.0201 mol/L. What is the value of the Keq? ? (6 marks)
CO
+
(g)
H2O

(l)
CO2
I
0.0150 mol/L
-----
0.0150 mol/L
C
+ 0.0051 mol/L
-----
1/1(-0.0051 mol/L)
E
0.0201 mol/L
-----
0.0099 mol/L
Keq =
[CO2][H2]
[CO]
=
+
(g)
H2
(g)
0.0150 mol/L
1/1(-0.0051 mol/L)
0.0099 mol/L
(0.0099 mol/L)( 0.0099 mol/L)
(0.0201 mol/L)
= 4.9 x 10-3
2.
A mixture of 0.75 mol of N2 and 1.20 mol of H2 are placed in a 3.0 liter container. When the reaction
N2 (g) + 3H2 (g)  2NH3 (g)
reaches equilibrium, [H2] = 0.100 mol/L. What is the value of [N2] and [NH3] at equilibrium? Calculate the value
of the Keq. (6 marks)
N2 (g)
+
I
0.75/3.0 mol/L
=0.25 mol/L
C
1/3(-0.300mol/L)
= -0.100 mol/L
E
0.15 mol/L
Keq =
3 H2 (g)
2 NH3 (g)
1.20/3.0 mol/L
= 0.40 mol/L
-0.300 mol/L
0
2/3( +0.300 mol/L)
= +0.200 mol/L
0.100 mol/L
[NH3]2
=
[N2][H2]3
= 270

(0.200 mol/L)2
(0.15 mol/L)(0.100 mol/L)3
0.200 mol/L
3.
A mixture of 2.5 moles H2O and 100 g of C are placed in a 50-L container and allowed to come to equilibrium
subject to the following reaction:
C(s) + H2O(g)  CO(g) + H2 (g)
The equilibrium concentration of Hydrogen is found to be [H2] = 0.040 mol/L. What is the equilibrium
concentration of water, [H2O] ? What is the value of the Keq ?
C
+
(s)
H2O
CO

(g)
I
-----
C
-----
1/1(-0.040 mol/L)
1/1(+0.040 mol/L)
E
-----
0.010 mol/L
0.040 mol/L
Keq =
2.5/50. mol/L
= 0.050 mol/L
(g)
[CO][H2]
[H2O]
=
0
+
H2
(g)
0
1/1(+0.040 mol/L)
0.040 mol/L
(0.040 mol/L)(0.040 mol/L)
(0.010 mol/L)
= 0.16
4.
Consider the following reaction:
2 NO
(g)
+ Br2 (g)  2 NOBr
(g)
If 0.800 mol/L NO and 0.400 mol/L Br2 are placed in a flask and allowed to come to equilibrium,
0.216 mol/L NO remains. Calculate the value of the K eq. (6 marks)
2 NO (g)
+
Br2 (g)
I
0.800 mol/L
C
-0.584 mol/L
1/2(-0.584 mol/L)
= -0.292 mol/L
E
0.216 mol/L
0.108 mol/L
Keq =
[NOBr]2
[NO]2[Br2]
= 67.7
0.400 mol/L
=

2 NOBr (g)
0
2/2(+0.584 mol/L)
= 0.584 mol/L
0.584 mol/L
(0.584 mol/L)2
(0.216 mol/L)2(0.108 mol/L)
5.
The Keq for the reaction H2(g) + I2(g)  2 HI(g) has a value of 64.0 at 425°C. If 1.00 mole of I2 and 1.00
mole of H2 is placed in a 5.00 L flask and allowed to come to equilibrium at this temperature, calculate the
equilibrium concentrations of all species. (6 marks)
H2(g)
I
C
E
+
(1.00 mol/5.00 L)
= 0.200 mol/L
I2(g)
(1.00 mol/5.00 L)
= 0.200 mol/L
-x
-x
0.200 mol/L - x
Keq =
0.200 mol/L - x
2 HI(g)
0
+2x
2x
[HI]2
[H2][I]
64.0 =
8.00 =
x

(2x)2
(0.200 mol/L - x)2
2x
0.200 mol/L - x
(take square root of both sides)
= 0.160 mol/L
[H2] = [I2] = 0.200 mol/L – 0.160 mol/L = 0.040 mol/L
[HI] = 2(0.160 mol/L) = 0.320 mol/L
6.
At 90°C, the equilibrium constant is 14.7 for the reaction
H2S (g)  H2 (g) + S (s)
If 0.150 mol H2S is heated to 90.0°C in a 1.00 L container, what will be the concentration of all three species at
equilibrium? (6 marks)
H2S (g)

H2 (g)
+
S (s)
I
0.150 mol/L
0
----
C
-x
+x
----
E
0.150 mol/L - x
x
----
Keq =
14.7 =
[H2]
[H2S]
(S is a solid and is not included in the Keq expression)
x
0.150 mol/L -x
x = 0.140 mol/L = [H2]
[H2S] = 0.150 mol/L – 0.140 mol/L = 0.010 mol/L
7.
Given this equation:
COCl2 (g)  CO (g) + Cl2 (g)
Calculate all three equilibrium concentrations when K eq = 0.680 with initial [CO] = 0.500 mol/L and
[Cl2] = 1.00 mol/L. (6 marks)
COCl2 (g)
I
0
C
+x
E
x
Keq =
0.680 =
CO (g)

+
0.500 mol/L
Cl2 (g)
1.00 mol/L
-x
0.500 mol/L - x
-x
1.00 mol/L - x
[CO][Cl2]
[COCl2]
(0.500 mol/L - x)(1.00 mol/L - x)
x
1 x2 + (2.18)(x) - 0.500 = 0
a
b
c
x = -b ± √ b2 - 4ac = -2.18 ± √ (2.18)2 – 4(1)(-0.500)
2a
2(1)
x = 0.209 mol/L = [COCl2]
[CO] = 0.500 mol/L - 0.209 mol/L = 0.291 mol/L
[Cl2] = 1.00 mol/L - 0.209 mol/L = 0.791 mol/L
8.
Nitrosyl bromide decomposes according to the following equation.
2NOBr (g)  2NO (g) + Br2 (g)
A sample of NOBr (0.64 mol) was placed in a 1.00-L flask containing no NO or Br2. At equilibrium the flask
contained 0.46 mol of NOBr. How many moles of NO and Br2, respectively, are in the flask at equilibrium? What
is the value of the Keq? ? 6 marks
2NOBr (g)
2NO (g)

+
I
0.64 mol/L
C
-0.18 mol/L
2/2(+0.18 mol/L)
E
0.46 mol/L
0.18 mol/L
Keq =
[NO]2[Br2]
[NOBr]2
= 1.4 x 10-2
0
=
Br2 (g)
0
½(+0.18 mol/L)
= 0.090 mol/L
0.090 mol/L
(0.18 mol/L)2(0.090 mol/L)
(0.46 mol/L)2
9.
0.04000 mol of NO and 0.00300 mol of O 2 are introduced into a 1.000 L flask, and the reaction
4 NO(g) + O2(g)  2 N2O3(g)
occurs. At equilibrium, it is determined that [N2O3] = 0.00350 mol/L. What is the value of K eq for the reaction?
6 marks
4 NO(g)
I
+
O2(g)
0.04000 mol/L
2 N2O3(g)

0.00300 mol/L
0
C 4/2(-0.00350 mol/L)
½(-0.00350 mol/L)
+0.00350 mol/L
= -0.00700 mol/L
= -0.00175 mol/L
E
0.03300 mol/L
Keq =
0.00125 mol/L
[N2O3]2
[NO]4[O2]
=
0.00350 mol/L
(0.00350 mol/L)2
(0.03300 mol/L)4(0.00125 mol/L)
= 8260
10. At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide:
H2
(g)
+ Br2
(g)
 2 HBr
(g)
A mixture of 0.682 mol of H2 and 0.440 mol of Br2 is combined in a reaction vessel with a volume of 2.00 L. At
equilibrium at 700 K, there are 0.566 mol of H2 present. What is the value of the equilibrium constant ?
6 marks
H2
I
C
E
+
(g)
(0.682 mol/2.00 L)
= 0.341 mol/L
-0.058 mol/L
0.283 mol/L
Keq =
Br2
(g)

2 HBr
(0.440 mol/2.00 L)
= 0.220 mol/L
1/1(-0.058 mol/L)
0.162 mol/L
[HBr]2
=
(0.116 mol/L)2
[H2][Br2]
(0.283 mol/L)(0.162 mol/L)
= 0.294 (0.31)
(g)
0
2/1(+0.058 mol/L)
= +0.116 mol/L
0.116 mol/L (0.12 mol/L)
11.
In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction:
CO
(g)
+ H2O
(g)
 CO2
(g)
+ H2
(g)
In an experiment, 0.35 mol of CO and 0.40 mol of H2O were placed in a 1.00-L reaction vessel. At
equilibrium, there were 0.19 mol of CO remaining. What is the Keq at the temperature of the experiment ?
6 marks
CO
+
(g)
H2O
(g)
I
0.35 mol/L
C
-0.16 mol/L
1/1(-0.16 mol/L)
E
0.19 mol/L
0.24 mol/L
Keq =
CO2

0.40 mol/L
[CO2][H2]
[CO][H2O]
=
+
(g)
0
H2
(g)
0
1/1(+0.16 mol/L)
0.16 mol/L
1/1(+0.16 mol/L)
0.16 mol/L
(0.16 mol/L)(0.16 mol/L)
(0.19 mol/L)(0.24 mol/L)
= 0.56
12. The Keq for the reaction 2 HI(g)  H2(g) + I2(g) has a value of 1.85 x 10-2 at 425°C. If 0.180 mol of HI is placed
in a 2.00 L flask and allowed to come to equilibrium at this temperature, what will be the equilibrium [I 2] ? 6
marks
2 HI(g)
I
C
E

(0.180 mol/2.00 L)
= 0.0900 mol/L
-2x
0.0900 mol/L - 2x
H2(g)
0
+
I2(g)
0
+x
+x
x
x
Keq = [H2][I]
[HI]2
1.85 x 10-2 =
0.136 =
x
x2
(0.0900 mol/L - 2x)2
x
0.0900 mol/L - 2x
= 9.59 x 10-3 mol/L = [I2]
(take square root of both sides)
13. Gaseous phosphorus pentachloride decomposes to gaseous phosphorus trichloride and chlorine at a temperature
where Keq = 1.00 x 10-3. Suppose 2.00 mol of phosphorus pentachloride in a 1.00 L vessel is allowed to come to
equilibrium. Calculate the equilibrium concentrations of all species. 6 marks
PCl5 (g)
I
PCl3 (g)

2.00 mol/L
C
-x
E
2.00 mol/L - x
Keq =
+
Cl2 (g)
0
0
+x
+x
x
x
[PCl3][Cl2]
[PCl5]
1.00 x 10-3 =
(x)(x)
2.00 mol/L -x
(convert to form of quadratic equation)
1 x2 + (1.00 x 10-3)(x) - (2.00 x 10-3) = 0
a
b
c
x = -b ± √ b2 - 4ac = -1.00 x 10-3 ± √ (1.00 x 10-3)2 - 4(-2.00 x 10-3)
2a
2(1)
x = 0.0442 mol/L = [PCl3] = [Cl2]
[PCl5] = 2.00 mol/L - 0.0442 mol/L = 1.96 mol/L
14. At 90°C, the equilibrium constant is 6.8 x 10-2 for the reaction
H2 (g) + S (s)  H2S (g)
If 0.150 mol hydrogen and 1.00 mol sulfur are heated to 90.0°C in a 1.00 L container, what will be the
concentration of H2S at equilibrium? 6 marks
H2 (g)
+
S (s)
H2S (g)
I
0.150 mol/L
C
-x
+x
E
0.150 mol/L - x
x
Keq =
6.8 x 10-2 =
[H2S]
[H2]
-

0
(S is a solid and is not included in the Keq expression)
x
0.150 mol/L -x
x = 9.6 x 10-3 mol/L = [H2S]
15. At a particular temperature, assume that Keq = 1.00 x 102 for the reaction
H2 (g) + F2 (g)  2 HF (g)
a)
In an experiment, 2.00 mol H2 and 2.00 mol F2 are introduced into a 1.00 L flask. Calculate the
concentration of all species when equilibrium is reached. 6 marks
H2 (g)
+
F2 (g) 
2 HF (g)
I
2.00 mol/L
2.00 mol/L
0
C
-x
-x
+2x
E
2.00 mol/L - x
2.00 mol/L - x
2x
Keq =
[HF]2
[H2][F2]
1.00 x 102 =
10.0 =
(2x)2
(2.00 mol/L - x)2
2x
2.00 mol/L - x
x = 1.67 mol/L
[H2] = [F2] = 2.00 mol/L - 1.67 mol/L = 0.33 mol/L
[HF] = 2(1.67 mol/L) = 3.33 mol/L
b)
To the equilibrium mixture in part a), an additional 0.500 mol of H2 is added. Calculate the new
equilibrium concentrations of H2, F2 and HF. 6 marks, bonus
start question with equilibrium concentrations calculated in a); add 0.0500 mol/L to the [H2]
H2 (g)
+
F2 (g) 
2 HF (g)
I
0.33 + 0.500 mol/L
0.33 mol/L
3.33 mol/L
C
-x
-x
+2x
E
0.83 mol/L - x
0.33 mol/L - x
3.33 mol/L + 2x
Keq =
[HF]2
[H2][F2]
1.00 x 102 =
(3.33 mol/L + 2x)2
(0.83 mol/L - x)(0.33 mol/L - x)
1.00 x 102 =
11.1 + 13.3x + 4x2
0.274 - 1.16x + x2
27.4 - 116x + 100.x2 = 11.1 + 13.3x + 4x2
96x2 - 129.3x + 16.3 = 0
a
b
c
x = -b ± √ b2 - 4ac = 129.3 ± √ (-129.3)2 - 4(96)(16.3)
2a
2(96)
=
129.3 ± 102.3
192
= 1.21 mol/L or 0.141 mol/L
[H2] = 0.83 mol/L - 0.141 mol/L = 0.69 mol/L
[F2] = 0.33 mol/L - 0.141 mol/L = 0.19 mol/L
[HF] = 3.33 mol/L + 2(0.141 mol/L) = 3.61 mol/L
16. For the equilibrium reaction CO(g) + H2O(g)  CO2(g) + H2(g) the Keq value at 690°C is 10.0. A mixture of 0.300
mol of CO, 0.300 mol of H2O, 0.500 mol of CO2 and 0.500 mol of H2 is placed in a 1.00 L flask.
a)
Calculate the Keq based on the above concentrations. Is this reaction at equilibrium ? 2 marks
Trial Keq =
[CO2][H2] = (0.500 mol/L)2
[CO][H2O]
(0.300 mol/L)2
= 2.78
2.78 is not the same as the actual Keq so the reaction is not at equilibrium
b)
Which direction will the reaction shift to reach equilibrium ? How do you know ? 2 marks
For the reaction to reach equilibrium the concentrations will change so that the K eq ratio will rise to
10.0 from 2.78. For the ratio to rise the amount of product will increase and the reactant will
decrease.
c)
Calculate the equilibrium concentrations of all four substances. 6 marks
CO(g)
I
0.300 mol/L
C
E
-x
+
H2O(g)
[CO2][H2]
[CO][H2O]
CO2(g)
0.300 mol/L
+
0.500 mol/L
-x
0.300 mol/L - x 0.300 mol/L - x
Keq =

+x
0.500 mol/L + x
= (0.500 mol/L + x)2
(0.300 mol/L - x)2
= 10.0
x = 0.108 mol/L
[CO] = [H2O] = 0.300 mol/L - 0.108 mol/L = 0.192 mol/L
[CO2] = [H2] = 0.500 mol/L + 0.108 mol/L = 0.608 mol/L
H2(g)
0.500 mol/L
+x
0.500 mol/L + x
17. A 1.00 L vessel contains at equilibrium 0.300 mol of N 2, 0.400 mol H2, and 0.100 mol NH3. If the temperature is
maintained constant, how many moles of H 2 must be introduced into the vessel in order to double the equilibrium
concentration of NH3? (6 marks)
Keq =
[NH3]2 =
[N2][H2]3
N2 (g)
(0.100 mol/L)2
(0.300 mol/L)(0.400 mol/L)3
+
3 H2 (g)

= 0.521
2 N2 (g)
I
0.300 mol/L
0.400 mol/L
0.100 mol/L
C
½(-0.100 mol/L)
= -0.050 mol/L
0.250 mol/L
3/2(-0.100 mol/L)
= -0.150 mol/L
x
+0.100 mol/L
E
0.200 mol/L
Use x at equilibrium to get E concentration of H 2. Use the change later, to get how much hydrogen
must be added.
Keq =
0.521
[NH3]2
[N2][H2]3
(0.200 mol/L)2
(0.250 mol/L)(x)3
x = 0.675 mol/L
Since the loss of hydrogen would be 0.150 mol/L as the equilibrium shifts that would give
0.250 mol/L. The final hydrogen concentration needs to be 0.675 mol/L to keep the K eq constant.
0.675 mol/L – 0.250 mol/L = 0.425 mol/L
18. For the equilibrium reaction CO(g) + H2O(g)  CO2(g) + H2(g) the Keq value at 690°C is 10.0. A mixture of 0.300
mol of CO, 0.300 mol of H2O, 0.500 mol of CO2 and 0.500 mol of H2 is placed in a 1.00 L flask.
a)
Calculate the Keq based on the above concentrations. Is this reaction at equilibrium ? (4 marks)
Trial Keq =
[CO2][H2] = (0.500 mol/L)2
[CO][H2O]
(0.300 mol/L)2
= 2.78
2.78 is not the same as the actual Keq so the reaction is not at equilibrium
b)
Which direction will the reaction shift to reach equilibrium ? How do you know ? (2 marks)
For the reaction to reach equilibrium the concentrations will change so that the K eq ratio will rise to
10.0 from 2.78. For the ratio to rise the amount of product will increase and the reactant will
decrease.
c)
Calculate the equilibrium concentrations of all four substances. (6 marks)
CO(g)
I
0.300 mol/L
C
E
-x
+
H2O(g)
[CO2][H2]
[CO][H2O]
CO2(g)
0.300 mol/L
+
0.500 mol/L
-x
0.300 mol/L - x 0.300 mol/L - x
Keq =

+x
0.500 mol/L + x
= (0.500 mol/L + x)2
(0.300 mol/L - x)2
= 10.0
x = 0.108 mol/L
[CO] = [H2O] = 0.300 mol/L - 0.108 mol/L = 0.192 mol/L
[CO2] = [H2] = 0.500 mol/L + 0.108 mol/L = 0.608 mol/L
H2(g)
0.500 mol/L
+x
0.500 mol/L + x
Ksp problems
1.
Given the following solubilities calculate the solubility product for each substance. Remember that the given
solubilities must be converted to moles per litre before the Ksp can be calculated.
a)
CuSCN, 5.00 mg/L 6 marks
63.55 g/mol + 32.07 g/mol + 12.01 g/mol + 14.01 g/mol = 121.64 g/mol
n = (5.00 mg/L)(1 g / 1000 mg) = 4.11 x 10-5 mol/L
121.64 g/mol
CuSCN(aq)

Cu1+(aq)
+
-5
4.11 x 10 M
1(4.11 x 10-5 M)
SCN1-(aq)
1(4.11 x 10-5 M)
Ksp = [Cu1+][SCN1-] = (4.11 x 10-5 M)2 = 1.69 x 10-9
b)
Mg(OH)2, 8.16 mg/L 6 marks
24.30 g/mol + 2(16.00 g/mol) + 2(1.01 g/mol) = 58.32 g/mol
n = (8.16 mg/L)(1 g / 1000 mg) = 1.40 x 10-4 mol/L
58.32 g/mol
Mg(OH)2 (aq)
1.40 x 10-4 M

Mg2+(aq)
+
2 OH1-(aq)
-4
1(1.40 x 10 M) 2(1.40 x 10-4 M)
= 2.80 x 10-4 M
Ksp = [Mg2+][OH1-]2 = (1.40 x 10-4 M)(2.80 x 10-4 M)2 = 1.10 x 10-11
c)
Al(OH)3, 2.63 x 10-9 mol/L 3 marks
Al(OH)3 (aq)
2.63 x 10-9 M

Al3+(aq)
+
1(2.63 x 10-9 M)
3 OH1-(aq)
3(2.63 x 10-9 M)
= 7.89 x 10-9 M
Ksp = [Al3+][OH1-]3 = (2.63 x 10-9 M)(7.89 x 10-9 M)3 = 1.29 x 10-33
d)
BaCO3, 7.07 x 10-5 mol/L (4 marks)
BaCO3 (aq)

Ba2+(aq)
+
-5
7.07 x 10 M
1(7.07 x 10-5 M)
CO31-(aq)
(7.07 x 10-5 M)
Ksp = [Ba2+][CO32-] = (7.07 x 10-5 M)( 7.07 x 10-5 M) = 5.00 x 10-9
e)
CaF2, 16.6 mg/L (8 marks)
40.08 g/mol + 2(19.00 g/mol) = 78.08 g/mol
n = (16.6 mg/L)(1 g / 1000 mg) = 2.13 x 10-4 mol/L
78.08 g/mol
CaF2 (aq)
2.13 x 10-4 M

Ca2+(aq)
+
2 F1-(aq)
-4
1(2.13 x 10 M) 2(2.13 x 10-4 M)
= 4.26 x 10-4 M
Ksp = [Ca2+][F1-]2 = (2.13 x 10-4 M)(4.26 x 10-4 M)2 = 3.84 x 10-11
f)
La(IO3)3, 0.458 g/L (8 marks)
138.91 g/mol + 3(126.90 g/mol) + 9(16.00 g/mol) = 663.61 g/mol
n =
0.458 g/L
663.61 g/mol
= 6.90 x 10-4 mol/L

La3+(aq)
+
3 IO31-(aq)
1(6.90 x 10-4 M) 3(6.90 x 10-4 M)
= 2.07 x 10-3 M
La(IO3)3 (aq)
6.90 x 10-4 M
Ksp = [La3+][IO31-]3 = (6.90 x 10-4 M)(2.07 x 10-3 M)3 = 6.12 x 10-12
2.
Given the following ksp values, calculate the solubility of each compound, then convert the solubility to the units
given in parenthesis.
a)
MgCO3, Ksp = 3.5 x 10-8 (g/L) 6 marks
MgCO3
x
(aq)

Mg2+(aq)
x
+
CO32-(aq)
x
Ksp = [Mg2+][CO2-]
3.5 x 10-8 = x2
x = 1.9 x 10-4 mol/L
(solubility of the substance in solution)
24.30 g/mol + 12.01 g/mol + (3)(16.00 g/mol) = 84.31 g/mol
mass
= (1.9 x 10-4 mol/L)(84.31 g/mol)
= 0.0.16 g/L
b)
BaF2, Ksp = 1.7 x 10-6 (mol/L) 3 marks
BaF2
x

(aq)
Ba2+(aq)
x
+
2 F1-(aq)
2x
Ksp = [Ba2+][F1-]2
1.7 x 10-6 = (x)(2x)2 = 4x3
x = 7.5 x 10-3 mol/L
c)
(solubility of the substance in solution)
Fe2S3, Ksp = 1.4 x 10-85 (μg/L) 6 marks
Fe2S3
x

(aq)
2 Fe3+(aq)
2x
+
3 S2-(aq)
3x
Ksp = [Fe3+]2[S2-]3
4.2 x 10-85 = (2x)2(3x)3 = 108x5
x = 4.2 x 10-18 mol/L
(solubility of the substance in solution)
(2)(55.85 g/mol + (3)(32.07 g/mol) = 207.91 g/mol
mass
d)
= (4.2 x 10-18 mol/L)(207.91 g/mol)(106 μg/g)
= 8.7 x 10-10 μg/L
PbS, Ksp = 3.0 x 10-28 (mol/L) (4 marks)
PbS (aq)
x

Pb2+(aq)
x
+
S2-(aq)
x
Ksp = [Pb2+][S2-]
3.0 x 10-28 = x2
x = 1.7 x 10-14 mol/L
(solubility of the substance in solution)
e)
Ag2SO4, Ksp = 1.5 x 10-5 (g/mL) (8 marks)
Ag2SO4
x
(aq)

2 Ag1+(aq)
2x
+
SO42-(aq)
x
Ksp = [Ag1+]2[SO42-]
1.5 x 10-5 = (2x)2(x) = 4x3
x = 1.6 x 10-2 mol/L
(solubility of the substance in solution)
2(107.87 g/mol) + 32.06 g/mol + (4)(16.00 g/mol) = 311.80 g/mol
mass
f)
= (1.6 x 10-2 mol/L)(311.80 g/mol)
= 4.9 g/L (1 L / 1000 mL)
= 4.9 x 10-3 g/mL
Ca3(PO4)2 , Ksp = 2.0 x 10-29 (μg/L) (8 marks)
Ca3(PO4)2
x
(aq)

3 Ca2+(aq)
3x
+
2 PO43-(aq)
2x
Ksp = [Ca2+]3[PO43-]2
2.0 x 10-29 = (3x)3(2x)2 = 108x5
x = 7.1 x 10-7 mol/L
(solubility of the substance in solution)
(3)(40.08 g/mol) + (2)(30.97 g/mol) + (8)(16.00 g/mol) = 310.18 g/mol
mass
= (7.1 x 10-7 mol/L)(310.18 g/mol)(106 μg/g)
= 220 μg/L
3. If you mix 20.0 mL of 1.50 x 10-4 mol/L ZnCl2 with 30.0 mL of 2.25 x 10-6 mol/L (NH4)2CO3 will a precipitate form ?
ZnCO3 has a Ksp of 1.00 x 10-10. (10 marks)
ZnCl2 (aq)
1.50 x 10-4 M

Zn2+(aq)
1.50 x 10-4 M
+
2 Cl1-(aq)
(NH4)2CO3 (aq)
2.25 x 10-6 M
 2 NH41+(aq) +
[Zn2+] = (1.50 x 10-4 M)(20.0 mL) = 6.00 x 10-5 mol/L
(20.0 mL + 30.0 mL)
[CO32-] = (2.25 x 10-6 M)(30.0 mL) = 1.35 x 10-6 mol/L
(20.0 mL + 30.0 mL)
ZnCO3 (aq)

Zn2+(aq)
+
CO32- (aq)
Trial Ksp = [Zn2+][CO32-] = (6.00 x 10-5 mol/L)( 1.35 x 10-6 mol/L) = 8.10 x 10-11
Trial Ksp is lower than the actual Ksp; No Precipitate will be formed
CO32-(aq)
2.25 x 10-6 M
4.
If you mix 160.0 mL of 2.50 x 10-6 mol/L Cr(NO3)3 with 90.0 mL of 1.50 x 10-8 mol/L Ca(OH)2 will a precipitate
form ? Cr(OH)3 has a Ksp of 1.60 x 10-30. (10 marks)

Ca(OH)2 (aq)
1.50 x 10-8 M
Ca2+(aq) + 2 OH1-(aq)
2(1.50 x 10-8 M)
3.00 x 10-8 M
Cr(NO3)3 (aq)
 Cr3+(aq) +
-6
2.50 x 10 M
2.50 x 10-6 M
3 NO31-(aq)
[Cr3+] = (2.50 x 10-6 M)(160.0 mL) = 1.60 x 10-6 mol/L
(160.0 mL + 90.0 mL)
[OH1-] = (3.00 x 10-8 M)(90.0 mL) = 1.08 x 10-8 mol/L
(160.0 mL + 90.0 mL)
Cr(OH)3

(aq)
Cr3+(aq)
+
3 OH1-(aq)
Trial Ksp = [Cr3+][OH1-]3 = (1.60 x 10-6 mol/L)(1.08 x 10-8 mol/L)3 = 2.02 x 10-30
Trial Ksp is higher than the actual Ksp; A Precipitate will be formed
5. If you mix 20.0 mL of 1.00 x 10-4 mol/L NaCl with 20.0 mL of 2.50 x 10-6 mol/L AgNO3 will a precipitate form ? AgCl
has a Ksp of 1.80 x 10-10. 8 marks
NaCl (aq)
1.00 x 10-4 M

Na1+(aq)
+
Cl1-(aq)
-4
1.00 x 10 M
1.00 x 10-4 M
AgNO3 (aq)
2.50 x 10-6 M

Ag1+(aq) +
2.50 x 10-6 M
[Ag1+] = (2.50 x 10-6 M)(20.0 mL) = 1.25 x 10-6 mol/L
(20.0 mL + 20.0 mL)
[Cl1-] = (1.00 x 10-4 M)(20.0 mL) = 5.00 x 10-5 mol/L
(20.0 mL + 20.0 mL)
AgCl (aq)

Ag1+(aq)
+
Cl1-(aq)
Trial Ksp = [Ag1+][Cl1-] = (1.25 x 10-6 mol/L)(5.00 x 10-5 mol/L) = 6.25 x 10-11
Trial Ksp is lower than the actual Ksp; No Precipitate will be formed
NO31-(aq)
2.50 x 10-6 M
6.
If you mix 40.0 mL of 2.50 x 10-3 mol/L Pb(NO3)2 with 60.0 mL of 1.50 x 10-3 mol/L NaI will a precipitate
form ? PbI2 has a Ksp of 1.40 x 10-8. 8 marks
NaI (aq)

1.50 x 10-3 M
Na1+(aq)
+
I1-(aq)
1.50 x 10-3 M
Pb(NO3)2 (aq)
2.50 x 10-3 M
 Pb2+(aq) +
2.50 x 10-3 M
2 NO31-(aq)
[Pb2+] = (2.50 x 10-3 M)(40.0 mL) = 1.00 x 10-3 mol/L
(40.0 mL + 60.0 mL)
[I1-] = (1.50 x 10-3 M)(60.0 mL) = 9.00 x 10-4 mol/L
(40.0 mL + 60.0 mL)
PbI2 (aq)

Pb2+(aq)
+
2 I1-(aq)
Trial Ksp = [Pb2+][I1-]2 = (1.00 x 10-3 mol/L)(9.00 x 10-4 mol/L)2 = 8.10 x 10-10
Trial Ksp is lower than the actual Ksp; No Precipitate will be formed
7.
A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba 2+. When the concentration of F1- exceeds
__________ M, BaF2 will precipitate. Neglect volume changes. For BaF 2, Ksp = 1.7 × 10-6.
4 marks, bonus
BaF2 (aq)

Ba2+(aq)
+
2 F1-(aq)
Ksp = [Ba2+][F1-]2
1.7 x 10-6 = (0.0144 mol/L) [F1-]2
[F1-]2 = 1.7 x 10-6 / 0.0144 mol/L
[F1-] = 1.1 x 10-2 mol/L
If the fluorine concentration exceeds that from the Ksp a precipitate will occur.