Download AP Chemistry Unit 7- Homework Problems Equilibrium and Ksp

AP Chemistry
Unit 7- Homework Problems
Equilibrium and Ksp
Nature of the Equilibrium State
1. Draw on this graph where equilibrium has been reached.
equilibrium
[X]
time
2. What are three qualities of any equilibrium equation?
a. Reversible
b. Dynamic
c. Occur when there are a stable ratio of products: reactants
3. For a general equation: aA + bB cC + dD , write the equation for Kc.
Kc = [C]c[D]d
[A]a[B]b
Developing Keq
1. For each of the equations below, write the expression for K c:
a. 2 H2S (g)  2 H2 (g) + S2 (g)
Kc = [H2]2[S2]
[H2S]2
b. HCN (aq) + H2O (l)  H3O+ (aq) + CN-1 (aq)
Kc = [H3O+] [CN-1]
[HCN]
c. PbCl2 (s)  Pb+2 (aq) + 2 Cl-1 (aq)
Kc = [Pb+2] [Cl-1]2
2. For each of the equations below, write the expression for K p:
a. SO2Cl2 (g)  SO2 (g) + Cl2 (g)
Kp = PSO2 PCl2
PSO2Cl2
b. CO (g) + H2O (g)  CO2 (g) + H2 (g)
Kp = PCO2 PH2
PCO PH2O
c. C6H12O6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (g)
Kp = P6CO2 P6H2O
P6 O2
3. Put the following K values in order of increasing product-favored ability.
a. K = 4x10-5
b. K = 2x10-9
c. K = 7x10-5
d. K = 3x10-3
b<a<c<d
Equilibrium Mathematics
1. The equation:
C (s) + 2 H2O (g)  CO (g) + H2 (g) has a value of Kc = 2.5 x10-6
What is the value of Kc for:
CO (g) + H2 (g)  C (s) + 2 H2O (g)?
Kc = (2.5x10-6)-1 = 4x105
What is the value of Kc for:
2 C (s) + 4 H2O (g)  2 CO (g) + 2 H2 (g)
-6 2
-12
Kc = (2.5x10 ) = 6.25x10
2. The equation:
H2O (g)  H2 (g) + ½ O2 (g)
has a value of Kp = 4.9 x10-3
What is the value of Kp for:
2 H2O (g)  2 H2 (g) + O2 (g)?
-3 2
-5
Kp = (4.9x10 ) = 2.4x10
What is the value of Kp for:
H2 (g) + ½ O2 (g)  H2O (g)?
Kp = (4.9x10-3)-1 = 204
3. The equation:
2 NH3 (g)  N2 (g) + 3 H2 (g)
has a value of Kc = 2.7 x10-4
At STP, What is the value for Kp ?
Kp = Kc (RT)n = (2.7 x10-4)[(0.0821)(273 K)]2 = 0.136
At STP, What is the value of Kc for:
Kc = (2.7 x10-4)-1/2 = 60.9
½ N2 (g) + 3/2 H2 (g)  NH3 (g)?
4. The equation:
N2O4 (g)  2 NO2 (g)
has a value of Kp = 3.2x10-4
What is the value for Kc at 300 K??
Kp = Kc (RT)n
3.2x10-4 = Kc [(0.0821)(300 K)]1 = 1.30x10-5
What is the value for Kc for: NO2 (g)  ½ N2O4 (g)
Kc = (1.30x10-5)-½ = 277
5. Given the following equations:
H2O (g) + CO (g)  H2 (g) + CO2 (g)
FeO (s) + CO (g)  Fe (s) + CO2 (g)
Kc = 4.8
Kc = 0.48
Calculate the Kc value for:
Fe (s) + H2O (g)  FeO (s) + H2 (g)
Kc = ???
H2O (g) + CO (g)  H2 (g) + CO2 (g)
Fe (s) + CO2 (g)  FeO (s) + CO (g)
Fe (s) + H2O (g)  FeO (s) + H2 (g)
Kc = 4.8
Kc = (0.48)-1 = 2.08
Kc = (4.8)(2.08) = 9.984
6. Given the following equations:
S (s) + O2 (g)  SO2 (g)
2 SO3 (g)  2 SO2 (g) + O2 (g)
Kp = 48.2
Kp = 0.075
Calculate the Kp value for:
S (s) + 3/2 O2 (g)  SO3 (g)
Kp = ???
S (s) + O2 (g)  SO2 (g)
SO2 (g) + ½ O2 (g)  SO3 (g)
S (s) + 3/2 O2 (g)  SO3 (g)
Kp = 48.2
Kp = (0.075)-1/2 = 3.65
Kp = 176
7. Which of the following equations has Kc = Kp
a. PCl5 (g)  PCl3 (g) + Cl2 (g)
b. 2 NOCl (g)  2 NO (g) + Cl2 (g)
c. CaCO3 (s)  CaO (s) + CO2 (g)
d. H2O (g) + CO (g)  H2 (g) + CO2 (g)
e. 2 NO (g)  N2 (g) + O2 (g)
No; gas = +1
No; gas = +1
No; gas = +1
Yes; gas = 0
Yes; gas = 0
Kc and Kp Calculations
At Equilibrium
1. For the reaction:
2 NO2 (g)  N2O4 (g)
At equilibrium [N2O4] = 0.25 M & [NO2] = 0.175 M. Calculate Kc
Kc = [N2O4]
[NO2]
2
Kc = [0.25]
[0.175]
Kc = 8.16
2
2. For the reaction:
2 NH3 (g)  N2 (g) + 3 H2 (g) Kp = 32
At equilibrium PNH3 = 0.64 atm & PN2 = 1.18 atm. Calculate PH2
Kp = PN2P3H2
P
2
NH3
32 = (1.18)P3H2
(0.64)
2
PH2 = 2.23
Dissociation
3. For the reaction:
PCl5 (g)  PCl3 (g) + Cl2 (g)
If the initial pressure of PCl5 is 2 atm and at equilibrium it is 15% dissociated, what is Kp?
PCl5 (g) 
PCl3 (g) +
Cl2 (g)
Kp = PPCl3PCl2
Po 2
0
0
PPCl5
 -0.30
+0.30
+0.30
Kp =
[0.30]2
Peq 1.7
0.30
0.30
1.7
Kp = 0.0529
4. For the reaction:
2 NO (g)  N2 (g) + O2 (g)
If the initial [NO] = 0.50 M and at equilibrium it is 5% dissociated, what is K c?
2 NO (g) 
N2 (g) +
O2 (g)
Kc = [N2] [O2]
[]o 0.50
0
0
[NO]2
 -0.025
+0.0125
+0.0125
Kc =
[0.0125]2
Kc = 6.9x10-4
[]eq 0.475
0.0125
0.0125
2
[0.475]
5. For the equation:
NH4I (s)  NH3 (g) + HI (g)
The total pressure at equilibrium is 4.2 atm. What is K p?
If Ptot = 4.2 atm
2x = 4.2
x = 2.1 atm
Kp = PNH3PHI = (2.1)2 = 4.41
6. For the equation:
(NH4)(H2NCO2) (s)  2 NH3 (g) + CO2 (g)
The total pressure at equilibrium is 0.33 atm. What is K p?
If Ptot = 0.33 atm
3x = 0.33
x = 0.11
Kp = P2NH3PCO2 = (0.22)2(0.11) = 5.32x10-3
7. For the equation:
N2 (g) + O2 (g)  2 NO (g), you start with 2 M of each of the reactants. They react away to
an extent of 27% to reach equilibrium. Calculate the value of K c.
N2 (g) +
[]o 2
 -0.54
[]eq 1.46
O2 (g) 
2
-0.54
1.46
2 NO (g)
0
+1.08
1.08
[NO]2
[N2] [O2]
Kc =
[1.08]2
[1.46] [1.46]
Kc =
Kc = 0.547
8. For the equation:
2 NOBr (g)  2 NO (g) + Br2 (g) , you start with 0.75 M of the NOBr. At equilibrium,
the NOBr has reacted away by 89%. Calculate the value of Kc.
2 NOBr (g) 
[]o 0.75
 -0.668
[]eq 0.082
2 NO (g) +
0
+0.668
0.668
Br2 (g)
0
+0.334
0.334
Kc =
Kc =
[NO]2[Br2]
[NOBr]2
[0.668]2[0.334]
[0.082]2
Kc = 22.2
Q vs. K
9. For the reaction:
2 NOCl (g)  2 NO (g) + Cl2 (g)
Kc = 1.2x10-3
If the initial [NOCl]o = 0.15 M, [NO]o = 0.75 M, and [Cl2]o = 0.05 M, is the system at equilibrium?
If not, which way will the reaction shift, left or right?
Kc =
Q=
[NO]2[Cl2]
[NOCl]2
[0.75]2[0.05]
[0.15]2
Q = 1.25 > 1.2x10-3 so the reaction goes left
10. For the reaction:
NH3 (aq) + H2O (l)  NH4+1 (aq) + OH-1 (aq)
Kc = 1.8x10-5
+1
-1
If the initial [NH3]o = 0.5 M, [NH4 ] = 0.0025 M, and [OH ] = 0.0025 M, is the system at equilibrium?
If not, which way will the reaction shift, left or right?
Kc =
Q=
[NH4+1][OH-1]
[NH3]
[0.0025]2
[0.5]
Q = 1.25x10-5 < 1.8x10-5 so the reaction goes right
11. For the equation:
CS2 (g) + 3 Cl2 (g)  S2Cl2 (g) + CCl4 (g), Kc = 4.8x10-2. If you start with [CS2] = 0.025
M, [Cl2] = 0.175 M, [S2Cl2]= 0.58 M, and [CCl4] = 0.042 M, is the reaction at equilibrium? If not, which way will the
reaction go to reach equilibrium (left or right)?
QQ= = [S[S
QQ= =(0.58)(0.042)
(0.58)(0.042) = 2.3
sCl
2Cl
2][CCl
2][CCl
4] 4]
CS2 (g) + 3 Cl2 (g) 
S2Cl2 (g)
+ CCl4
3
3
[CS[CS
(0.025)(0.175)
(0.025)(0.175)3 3
2][Cl
2][Cl
2] 2]
[]o 0.025
0.175
0.58
0.042
-2
QQ= >182
Kc so
>>>>
reaction
4.8 x10
goes
left
so rxn goes left
Calculating Equilibrium Conditions
12. For the equation:
PCl5 (g)  PCl3 (g) + Cl2 (g), you start with 0.25 atm of each of the products as well as the
reactants. The Kp value is 0.125. Is the reaction at equilibrium? Prove it. What are the equilibrium pressures of all
species?
Q = PPCl3PCl2 Q = (0.25)2 = 0.25
PCl5(s)
 PCl3 (g) +
Cl2 (g)
PPCl5
(0.25)
Po 0.25
0.25
0.25
Q > Kp 0.25 > 0.125 so there are too many
 x
-x
-x
products so the reaction moves to the left
Peq 0.25 + x
0.25 -x
0.25 - x
Kp = PPCl3PCl2 0.125 = (0.25-x)2
x = 0.055
PPCl5
(0.25+x)
PPCl5 = 0.305
PPCl3 = PCl2 = 0.195
13. For the equation:
H2O (g) + CO (g)  CO2 (g) + H2 (g)
Kc = 0.235
If 2 moles of each of H2O and CO are put into a 10 L container, what is the concentration of all species at
equilibrium?
Kc = [CO2][H2]
0.235 = (x)2
H2O (g)
+
CO (g) 
CO2 (g) +
H2 (g)
[H2O][CO]
(0.2-x)2
Po 0.2
0.2
0
0
 x
Peq 0.2 - x
-x
0.2 -x
+x
x
+x
x
x = 0.065
[CO2] = [H2] = 0.065
[H2O] = [CO] = 0.135
14. For the equation:
SO2Cl2 (g)  SO2 (g) + Cl2 (g)
Kp = 4.8
If enough SO2Cl2 is put into a container so its pressure is 8 atm, what is the equilibrium pressure of all species.
What is the total pressure?
SO2Cl2 (g)
Po 8
 x
Peq 8-x

SO2 (g) +
0
+x
x
Cl2 (g)
0
+x
x
Kp =
Kp =
PSO2PCl2
PSO2Cl2
[x]2
8-x
Kp = 4.25
PSO2 = PCl2 = 4.25
PSO2Cl2 = 3.75
Ptot = 3.75+ 4.25 + 4.25 = 12.25
15. For the equation:
CaCO3 (s)  CaO (s) + CO2 (g)
Kp = 2.4
If 200 g CaCO3 is put into a 20 L container at 500 K, how many grams of it remain at equilibrium?
CaCO3 (s) 
CaO (s) +
Po
 
Peq
CO2 (g)
0
+x
x
Kp = 2.4 = PCO2
PCO2 = 2.4
PV = nRT
(2.4 atm)(20 L) = n(0.0821 Latm/molK)(500 K)
n = 1.17moles reacted away
1.17 moles CaCO3* (100.1 g/mole) = 117 g reacted
200 g to begin with – 117 g reacted = 83 g left
16. For the equation:
2 KClO3 (s)  2 KCl (s) + 3 O2 (g), you start with some KClO3 that decomposes into the
products. At equilibrium, there is some solid remaining and the total pressure in the flask is 0.29 atm. Calculate the
value of Kp.
2 KClO3 (s) 
Po

Peq
2 KCl (s) +
3 O2 (g)
0
+0.29
0.29
KP =
P3O2 = 0.0244
17. For the equation:
NH4Cl (s)  NH3 (g) + HCl (g), you start with some NH4Cl that decomposes into the
products. At equilibrium, there is some solid remaining and the total pressure in the flask is 1.8 atm. Calculate the
value of Kp.
NH4Cl (s) 
Po

Peq
NH3 (g) +
0
+0.9
0.9
HCl (g)
0
+0. 9
0.9
KP =
PNH3PHCl = (0.9)(0.9) = 0.81
18. For the equation:
COBr2 (g) CO (g) + Br2 (g), you start with 4 moles in a 10 L vessel of COBr 2. The
reaction has a Kc = 0.76. What are the equilibrium concentrations of all species?
COBr2 (g) 
[]o
0.4

-x
[]eq
0.4-x
Kc =
CO (g) +
0
+x
x
Br2 (g)
0
+x
x
[CO][Br2]
[COBr2]
0.76 =
x2
x = 0.29
[0.4-x]
[COBr2] = 0.11 [CO] = [Br2] = 0.29
19. For the equation:
H2 (g) + CO2 (g)  H2O (g) + CO (g), you start with 2 atm of each of the reactants and
none of the products. The Kp = 3.4. What are the equilibrium pressures of all species?
H2 (g) +
CO2 (g)

H2O (g) +
CO (g) Kp = PH2OPCO
Po 2
2
0
0
PH2PCO2
 -x
-x
+x
+x
3.4 =
x2
x = 1.3
Peq 2-x
2-x
x
x
[2-x]2
PCO2 = PH2 = 0.7
PCO = PH2O = 1.3
20. For the equation:
2 CH2Cl2 (g)  CH4 (g) + CCl4 (g), you start with 0.25 M of CH2Cl2 and is has a Kc value
of 0.84. What are the equilibrium concentrations of all species?
2 CH2Cl2 (g)  CH4 (g) +
CCl4 (g)
Kc = [CH4][CCl4]
[]o
0.25
0
0
[CH2Cl2] 2
0.84 =
x2
x = 0.0809

-2x
+x
+x
[0.25-2x]2
[]eq
0.25-2x
x
x
[CH2Cl2] = 0.0882
[CH4] = [CCl4] = 0.0809
21. For the equation:
NH4HS (s)  NH3 (g) + H2S (g), you start with 100 grams of NH4HS (s) in a 2.5 L flask at
500 K. The Kp value is 1.45. How many grams of the solid remain at equilibrium?
NH4HS (s)  NH3 (g) +
Po
0

+x
Peq
x
H2S (g)
0
+x
x
KP =
PNH3PH2S x2 = 1.45
x = 1.20 atm
n = PV/RT = (1.2 atm)(2.5 L)/[(0.0821)(500 K)]
n = 0.0731 moles NH3 so there must have been the
same moles of NH4HS that broke up
(0.0731 mol)(51.1 g/mol) = 3.735 g broke up so:
100 g – 3.735 g = 96.265 g remain
LeChatelier’s Principle
1. State LeChatelier’s Principle.A change in any factor that effects the equilibrium of a system will cause the system to
shift in such a way to reduce or counteract that initial change.
2. For the following reaction:
Heat + CaCO3 (s) + CO2 (g) + H2O (l)  Ca+2 (aq) + 2 HCO3-1 (aq)
What will be the effect of doing each of the following actions on the above equilibrium?
a) Adding CaCO3 (s)
Left
Right No Change
b) Removing Ca+2 (aq)
Left
Right No Change
c) Removing CO2 (g)
Left
Right No Change
d) Adding NaHCO3 (s)
Left
Right No Change
e) Adding Ne (g)
Left
Right No Change
f) Adding CO2 (g)
Left
Right No Change
g) Increasing temperature
Left
Right No Change
h) Decreasing volume
Left
Right No Change
3. For the following reaction:
2 SO2 (g) + O2 (g)  2 SO3 (g) + Heat
What will be the effect of doing each of the following actions on the above equilibrium?
a) Decreasing temperature
Left
Right No Change
b) Increasing O2 (g)
Left
Right No Change
c) Decreasing SO2 (g)
Left
Right No Change
d) Increasing volume
Left
Right No Change
e) Increasing SO3 (g)
Left
Right No Change
f) Adding N2 (g)
Left
Right No Change
Ksp
1.
For each of the substances below, write the solubility equation as well as the K sp equation.
a. AgCl
AgCl  Ag+ + Cl-1
Ksp = [Ag+][Cl-1]
+2
-1
b. PbI2
PbI2  Pb + 2 I
Ksp = [Pb+2][I-1]2
+
-2
c. Ag2CO3
Ag2CO3  2 Ag + CO3
Ksp = [Ag+]2 [CO3-2]
+2
-3
d. Ca3(PO4)2
Ca3(PO4)2  3 Ca + 2 PO4
Ksp = [Ca+2]3 [PO4-3]2
Problem Solving with Ksp
1. Calculate the Ksp of CaCrO4 if a saturated solution has [Ca+2] = 4.5 x10-5
CaCrO4  Ca+2 + CrO4-2
Ksp = [Ca+2][CrO4-2] = (4.5 x10-5) (4.5 x10-5) = 2.025 x10-9
2. Calculate the Ksp of Fe(OH)3 if a saturated solution has [Fe+3] = 4.2 x10-6
Fe(OH)3  Fe+3 + 3 OH-1
Ksp = [Fe+3][OH-1]3 = (4.2 x10-6) [(3)(4.2 x10-6)]3 = 8.4 x10-21
3. Calculate the solubility (moles/L) of PbCO3 if Ksp = 7.4 x10-14
PbCO3  Pb+2 + CO3-2
Ksp = [Pb+2][CO3-2] 7.4x10-14 = x2
x = 2.72 x10-7 mole/L
-5
4. Calculate the solubility (moles/L) of Ag2SO4 if Ksp = 1.2 x10
Ag2SO4  2 Ag+1 + SO4-2
Ksp = [Ag+1]2 [SO4-2] 1.2x10-5 = 4x3
x = 0.0144 mole/L
-9
5. Calculate the solubility (mg/L) of FePO4 if Ksp = 9.4 x10
FePO4  Fe+3 + PO4-3
Ksp = [Fe+3][PO4-3] 9.4x10-9 = x2
x = 9.70 x10-5 mole/L
-25
6. Calculate the solubility (mg/L) of Al2(CO3)3 if Ksp = 7.2 x10
Al2(CO3)3  2 Al+3 + 3 CO3-2 Ksp = [Al+3]2 [CO3-2]3 7.2x10-25 = 108x5
x = 5.82 x10-6 mole/L *(234 g/mole) *(1000 mg/g) = 1.36 mg/L
7. How many mg of CuCrO4 will dissolve in 50 mL of water (Ksp = 9.4 x10-10)
CuCrO4  Cu+2 + CrO4-2
Ksp = [Cu+2] [CrO4-2]
9.4x10-10 = x2
x = 3.07 x10-5 mole/L *(179.5 g/mole) *(1000 mg/g)*(0.05 L) = 0.276 mg
8. If 100 mg of CaCO3 (Ksp = 3.4 x10-9) is put in 500 mL of water, how many mg remain undissolved?
CaCO3  Ca+2 + CO3-2
Ksp = [Ca+2] [CO3-2]
3.4x10-9 = x2
-5
x = 5.83 x10 mole/L *(179.5 g/mole) *(1000 mg/g)*(0.5 L) = 2.92 mg
100 mg – 2.92 mg dissolved = 97.1 mg undissolved
9. Put the following substances in order of least soluble to most soluble.
NiCO3 (Ksp = 1.4x10-7)
AuCl (Ksp = 2x10-13)
MnCO3 (Ksp = 2.3x10-11) PbCrO4 (Ksp = 2.8x10-13)
Since they are all the same number of pieces, you can compare them directly by K sp values. Thus:
AuCl (Ksp = 2x10-13) < PbCrO4 (Ksp = 2.8x10-13) < MnCO3 (Ksp = 2.3x10-11) < NiCO3 (Ksp = 1.4x10-7)
10. Put the following substances in order of least soluble to most soluble.
Zn(CN)2 (Ksp = 8x10-12) AgBr (Ksp = 5.x10-13)
Pb(OH)2 (Ksp = 1.4x10-15) BaSO4 (Ksp = 1.1x10-10)
These substances have different number of ionic pieces so you can’t compare K sp values directly. If you do a solubility
ICE table for each, the ultimate results will be as follows where x is the solubility:
Zn(CN)2
4x3 = 8x10-12
x = 1.27x10-4 M
2
-13
AgBr
x = 5x10
x = 7.1x10-7 M
3
-15
Pb(OH)2
4x = 1.4x10
x = 7.1x10-6 M
2
-10
BaSO4
x = 1.1x10
x = 1.1x10-5 M
Thus
AgBr < Pb(OH)2 < BaSO4 < Zn(CN)2
11. Prove which of each of the substances below is the most soluble.
a. AgBr (Ksp = 5.4x10-13) vs. AgI (Ksp = 8.5 x10-17)
AgBr  Ag+1 + Br-1 Ksp = [Ag+1][Br-1] 5.4x10-13 = x2
AgI  Ag+1 + I-1
Ksp = [Ag+1][I-1] 8.5x10-17 = x2
x = 7.3 x10-7 mole/L
x = 9.2 x10-9 mole/L
b. PbCl2 (Ksp = 1.7 x10-5) vs. PbBr2 (Ksp = 6.6 x10-6)
PbCl2  Pb+2 + 2 Cl-1
Ksp = [Pb+2][Cl-1]2 1.7x10-5 = 4x3
PbBr2  Pb+2 + 2 Br-1
Ksp = [Pb+2][Br-1]2 6.6x10-6 = 4x3
x = 0.0162 mole/L
x = 0.0118 mole/L
c. AgCl (Ksp = 1.8 x10-10) vs. Ag2CrO4 (Ksp = 1.1 x10-12)
AgCl  Ag+1 + Cl-1
Ksp = [Ag+1][Cl-1] 1.8x10-10 = x2
+1
-1
Ag2CrO4  2 Ag + CrO4
Ksp = [Ag+1]2 [CrO4-2] 1.1x10-12 = 4x3
x = 1.3 x10-5 mole/L
x = 6.5x10-5 mole/L
d. CaCO3 (Ksp = 3.4 x10-9) vs. Mg(OH)2(Ksp = 5.6 x10-12)
CaCO3  Ca+2 + CO3-2
Ksp = [Ca+2][CO3-2] 3.4x10-9 = x2
+2
-1
Mg(OH)2  Mg + 2 OH
Ksp = [Mg+2][OH-1]2 5.6x10-12 = 4x3
x = 5.8 x10-5 mole/L
x = 1.12 x10-4 mole/L
Common Ion Effect
1. Calculate the solubility (moles/L) of MgS (Ksp = 5.2 x10-16) in:
a. Pure water
MgS  Mg+2 + S-2
Ksp = [Mg+2][S-2] 5.2x10-16 = x2
b. A 0.25 M solution of MgCl2
MgS  Mg+2 + S-2
Ksp = [Mg+2][S-2] 5.2x10-16 = (0.25)(x)
x = 2.3 x10-8 mole/L
2. Calculate the solubility (moles/L) of PbCl2 (Ksp = 1.7 x10-5) in:
a. Pure water
PbCl2  Pb+2 + 2 Cl-1
Ksp = [Pb+2][Cl-1]2 1.7x10-5 = 4x3
b. A 0.55 M solution of NaCl
PbCl2  Pb+2 + 2 Cl-1
Ksp = [Pb+2][Cl-1]2 1.7x10-5 = (x)(0.55)2
x = 0.0162 mole/L
3. Calculate the solubility (moles/L) of PbI2 (Ksp = 9.8 x10-9) in:
a. Pure water
PbI2  Pb+2 + 2 I-1
Ksp = [Pb+2][I-1]2 9.8x10-9 = 4x3
b. A 0.005 M solution of AlI3
PbI2  Pb+2 + 2 I-1
Ksp = [Pb+2][I-1]2 9.8x10-9 = (x)(0.015)2
x = 1.35x10-3 mole/L
x = 2.1 x10-15 mole/L
x = 5.6x10-5 mole/L
x = 0.435 mole/L
4. Calculate what mass of Hg2SO4 (Ksp = 6.5 x10-7) will dissolve per liter in:
a. Pure water
Hg2SO4  2 Hg+1 + 2 SO4-2 Ksp = [Hg+1]2 [SO4-2] 6.5x10-7 = 4x3
x = 5.46x10-3 mole/L(497.3 g/mol) = 2.71 g/L
b. A 0.0075 M solution of Na 2SO4
Hg2SO4  2 Hg+1 + 2 SO4-2 Ksp = [Hg+1]2 [SO4-2] 6.5x10-7 = (4x2)(0.0075)
x = 4.65x10-3 mole/L (497.3 g/mol) = 2.31 g/L
5. Calculate what mass of CaCO3 (Ksp = 3.4 x10-9) will dissolve per liter in:
a. Pure water
CaCO3  Ca+2 + CO3-2
Ksp = [Ca+2][CO3-2] 3.4x10-9 = x2
x = 5.83 x10-5 mole/L (100.1 g/mole) = 5.84 x10-3 g/L
b. A 0.45 M solution of CaCl2
CaCO3  Ca+2 + CO3-2
Ksp = [Ca+2][CO3-2] 3.4x10-9 = (0.45)(x)
x = 7.56 x10-9 mole/L (100.1 g/mole) = 7.56 x10-7 g/L
6. Calculate what mass of PbCl2 (Ksp = 1.7 x10-5) will dissolve per liter in:
a. Pure water
PbCl2  Pb+2 + 2 Cl-1
Ksp = [Pb+2][Cl-1]2 1.7x10-5 = 4x3
x = 0.0162 mole/L(278.2 g/mole) = 4.5 g/L
b. A 0.067 M solution of AlCl3
PbCl2  Pb+2 + 2 Cl-1
Ksp = [Pb+2][Cl-1]2 1.7x10-5 = (x)(0.201)2
x = 4.2x10-4 mole/L(278.2 g/mole) = 0.117 g/L
7. AgBr will be the least soluble in 0.10 M:
a. NaBr
b. CaBr2
c. AlBr3
d. Ag2SO4
e. AgNO3
AlBr3 will have the most concentrated [Br-1] or [Ag+1] since there are 3 Br-1 for every 1 AlBr3. Thus, AgBr will be least
soluble in it.
8. MgF2 will be least soluble in 0.25 M:
a. KF b. Mg(NO3)2
c. MgC2O4
d. BaF2
e. LiF
-1
+2
-1
BaF2 will have the most concentrated [F ] or [Mg ] since there are 2 F for every 1 BaF2. Thus, MgF2 will be least
soluble in it.
Precipitation
1. Will a ppt of CaCO3 (Ksp= 3.4 x10-9) form if [Ca+2] = 4 x10-6 M and [CO3-2] = 4 x10-3?
Q = [4 x10-6][4x10-3] = 1.6x10-8 >> 3.4x10-9 so yes, ppt
2. Will a ppt of Ag2CrO4 (Ksp = 1.1 x10-12 ) form if [Ag+] = 3x10-4 and [CrO4-2] = 2x10-4?
Q = [3 x10-4]2 [2x10-4] = 1.8x10-11 >> 1.1x10-12 so yes, ppt
3. Will a ppt of BaCO3 (Ksp = 2.6 x10-9) form if 50 mL of 4x10-5 M Ba+2 is mixed with 50 mL of 8x10-5 M CO3-2?
Q = [(50/100)4 x10-5][(50/100)8x10-5] = 8x10-10 << 2.6x10-9 so no ppt
4. Will a ppt of PbBr2 (Ksp = 6.6 x10-6) form if 150 mL of 2x10-2 M Pb(NO3)2 is mixed with 50 mL of 3x10-2 M AlBr3?
Q = [(150/200)2 x10-2][(50/200)(3) 3x10-2] = 7.6x10-6 >> 6.6x10-6 so yes, ppt
5. What concentration of [OH-1] will cause a ppt of Fe(OH)2 (Ksp = 4.9x10-17) of a 2x10-5 M Fe+2 solution ?
4.9x10-17 = (2x10-5)(OH-1)2
[OH-1] = 1.56x10-6 M
-2
6. What concentration of [C2O4 ] will cause a ppt of Ag2C2O4 (Ksp = 5.4x10-12) of a 4x10-4 M Ag+1 solution?
5.4x10-12 = (4x10-4)2(C2O4-2)
[C2O4-2] = 3.4x10-5 M
7. What mass of Ni(NO3)2*7 H2O (s) will cause a ppt of NiCO3 (Ksp = 1.4x10-7) in a 2 L solution of 3.5x10-4 M CO3-2?
1.4x10-7 = [Ni+2](3.5x10-4)
[Ni+2] = 4x10-4 M
Molar mass of Ni(NO3)2*7H2O = 308.7 g/mole
[Ni+2] = 4x10-4 moles/L * 2 L * 308.7 g/mole = 0.247 grams Ni(NO 3)2*7H2O
8. What mass of Pb(NO3)2 will cause a ppt of PbCl2 (Ksp = 1.7x10-5) in 100 mL of a 2.4x10-2 M solution of AlCl3?
Since there are 3 Cl-1 ions for every AlCl3 then [Cl-1] = 7.2x10-2 M so:
1.7x10-5 = [Pb+2](0.072)2 [Pb+2] = 3.28x10-3 M
Molar mass of Pb(NO3)2 is 331.2 g/mole
Thus: 3.28x10-3 moles/L * 1 L/1000 mL * 100 mL * 331.2 g/mole = 0.109 grams Pb(NO 3)2
Separations by Ksp
1. A solution is made so that [Ca+2] = 0.00050 and [Mg+2] = 0.00050 as well. If Cr2O7-2 is added, answer the following
questions. Know that Ksp CaCr2O7 = 4.8 x10-7 and Ksp MgCr2O7 = 7.6 x10-8
a. Which will ppt 1st? At what [Cr2O7-2] will it begin to ppt?
CaCr2O7  Ca+2 + Cr2O7-2
Ksp = [Ca+2][ Cr2O7-2] 4.8x10-7 = x(0.0005)
x = 9.6 x10-4 mole/L
+2
-2
+2
-2
-8
MgCr2O7  Mg + Cr2O7
Ksp = [Mg ][ Cr2O7 ] 7.6x10 = x(0.0005)
x = 1.5 x10-4 mole/L
MgCr2O7 ppts 1st
b. What is the maximum [Cr2O7-2] that can be made to ppt almost all of one and none of the other?
9.6x10-4 moles/L
c. What is the concentration of the less soluble ion under the conditions specified in “b”?
MgCr2O7  Mg+2 + Cr2O7-2 Ksp = [Mg+2][ Cr2O7-2] 7.6x10-8 = (9.6x10-4)x
x = 7.92 x10-5 mole/L
d. What % of the less soluble ion remains in solution under the conditions specified in “b”?
(7.92x10-5/0.0005)*100% = 15.8%
2. A solution is made so that [Pb+2] = 0.00250 and [Hg+2] = 0.0075. If SO4-2 is added, answer the following questions.
Know that Ksp PbSO4 = 2.5 x10-8 and Ksp HgSO4 = 5.4 x10-7
a. Which will ppt 1st? At what [SO4-2] will it begin to ppt?
PbSO4  Pb+2 + SO4-2
Ksp = [Pb+2][ SO4-2] 2.5x10-8 = x(0.0025)
x = 1 x10-5 mole/L
+2
-2
+2
-2
-7
HgSO4  Hg + SO4
Ksp = [Hg ][ SO4 ] 5.4x10 = x(0.0075)
x = 7.2 x10-5 mole/L
st
PbSO4 ppts 1
b. What is the maximum [SO4-2] that can be made to ppt almost all of one and none of the other?
HgSO4  Hg+2 + SO4-2
Ksp = [Hg+2][ SO4-2] 5.4x10-7 = x(0.0075)
x = 7.2 x10-5 mole/L
c. What is the concentration of the less soluble ion under the conditions specified in “b”?
PbSO4  Pb+2 + SO4-2
Ksp = [Pb+2][ SO4-2] 2.5x10-8 = x(7.2x10-5)
x = 3.5 x10-4 mole/L
d. What % of the less soluble ion remains in solution under the conditions specified in “b”?
(3.5x10-4/0.0025)*100% = 13.9%
3. A solution is made so that [Zn+2] = 0.00250 and [Ag+1] = 0.057. If F-1 is added, answer the following questions. Know
that Ksp ZnF2 = 4.8 x10-7 and Ksp AgF = 7.6 x10-8
a. Which will ppt 1st? At what [F-1] will it begin to ppt?
ZnF2  Zn+2 + 2 F-1
Ksp = [Zn+2][ F-1]2 4.8x10-7 = x2(0.0025)
x = 1.39 x10-2 mole/L
+1
-1
+1
-1
-8
AgF  Ag + F
Ksp = [Ag ][ F ] 7.6x10 = x(0.057)
x = 1.3 x10-6 mole/L
st
AgF ppts 1
b. What is the maximum [F-1] that can be made to ppt almost all of one and none of the other?
1.39x10-2 mole/L
c. What is the concentration of the less soluble ion under the conditions specified in “b”?
AgF  Ag+1 + F-1
Ksp = [Ag+1][ F-1] 7.6x10-8 = 1.39x10-2(x)
x = 5.47 x10-6 mole/L
d. What % of the less soluble ion remains in solution under the conditions specified in “b”?
(5.47x10-6/0.057)*100% = 0.0096%
Combination Problems
1. For the reaction at 2000 K
H2 (g) + CO2 (g)  H2O (g) + CO (g)
For an experiment, the equilibrium values of each substance are as follows:
[H2] = 0.20 M
[CO2] = 0.30 M
[H2O] = [CO] = 0.55 M
a. What is the mole fraction of CO in the equilibrium mixture?
XCO = (0.55)/(0.20+0.30+0.55+0.55) = 0.355
b. Calculate the value of Kc, the equilibrium constant for the reaction above.
Kc = (0.55)2/(0.20)(0.30) = 5.042
c. Determine Kp in terms of Kc for this system. Kp = Kc because ngas = 0
d. When the system is cooled from 2000 K to a lower temperature, 30% of the CO is converted
back to CO2. Calculate the value of Kc at this lower temperature.
Since 30% of CO is converted: 0.55*0.30 = 0.165 changes for each
H2 (g) + CO2 (g) 
H2O (g) +
CO (g)
[]0 0.20
0.30
0.55
0.55
  



[]eq 0.365
0.465
0.385
0.385
Kc = (0.385)2/(0.365)(0.465) = 0.873
e. In a different experiment, 0.50 mole of H2 is mixed with 0.50 mole of CO2 in a 3.0 L reaction
vessel at 2000 K. Calculate the equilibrium concentration, in M, of CO at this temperature.
H2 (g) + CO2 (g) 
H2O (g) +
CO (g)
[]0 0.167
0.167
0
0
 -x 
-x


+x

+x
[]eq 0.167-x 0.167-x
x
x
5.042 = x2 /(0.167-x)2
2.245 = x/(0.167-x)
x = [CO] = 0.1156
2. For the reaction:
PCl5 (g)  PCl3 (g) + Cl2 (g)
It is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased.
a) What is the sign of So for the reaction? Explain.
S should be positive. You are going from one mole of gas on the left to 2 moles on the rights.
This shows an increase in disorder and a positive entropy change
b) What change, if any, will occur in Go for the reaction as the temperature is increased?
Explain. Since S is positive, G should decrease as temperature is increased. This is due to
the equation G = H –TS. Since T is in K and therefore positive, you will be subtracting a
positive value from G.
c) If He gas is added to the original mixture at constant volume and temperature, what will happen
to the partial pressure of Cl2? Explain. Dalton’s law of partial pressures says that the total
pressure is equal to the sum of the individual pressures. Adding more He gas will increase the
total pressure but not change the pressure the Cl2 exerts.
d) If the volume of the reaction mixture is decreased at constant temperature to half the original
volume, what will happen to the number of moles of Cl2 in the reaction vessel? Explain.
By decreasing the volume, you are increasing the pressures of each of the above species. Since
there are unequal amounts of gas on the reactant and product sides of the equation above, the
system will try to reduce the pressure increase it experienced. It can do this by going to the left
side which will reduce the number of moles of Cl2
3. For the reaction:
C(s) + CO2(g)  2 CO(g)
Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the
reaction represented above occurred. As the reaction proceeded, the total pressure in the container was
monitored. When equilibrium was reached, there was still some C(s) remaining in the container.
Results are recorded in the table below.
Time (hours) Total Pressure of gases (atm)
0.0
5.00
2.0
6.26
4.0
7.09
6.0
7.75
8.0
8.37
10.0
8.37
(a) Write the expression for the equilibrium constant, Kp , for the reaction.
Kp = P2CO/PCO2
(b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the volume
of the solid carbon is negligible.)
n = PV/RT = (5 atm)(2 L)/(0.0821)(1160 K) = 0.105 moles
(c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO2(g) is 1.63 atm.
Calculate:
(i) the partial pressure of CO(g) , and
CO2: 5 atm – 1.63 atm = 3.37 atm CO2 reacted *(2 mole CO/1 mole CO2 )= 6.74 atm CO
(ii) the value of the equilibrium constant, Kp .
Kp = P2CO/PCO2 = (6.74)2/1.63 = 27.87
(d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the
gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at
equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst
is negligible.)
Nothing would change. It would just get to equilibrium faster.
(e) In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g
of C(s) , plus CO(g) and CO2(g) , each at a partial pressure of 2.00 atm at 1,160 K. Predict whether
the partial pressure of CO2(g) will increase, decrease, or remain the same as this system approaches
equilibrium. Justify your prediction with a calculation.
Q = (2)2/2 = 2
Q <<< Kp
so reaction will go to the right
This will cause the partial pressure of CO2 to decrease
4. Answer the following questions:
a. A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of
solution at 25 oC. The concentration of Pb+2 (aq) in the saturated solution is found to be 1.3x10-3
M. The chemical equation for the dissolution of PbI2(s) in water is shown below:
PbI2 (s)  Pb+2 (aq) + 2 I-1 (aq)
i) Write the equilibrium-constant expression for the equation.
Ksp = [Pb+2][I-1]2
ii) Calculate the molar concentration of I-1 (aq) in the solution.
[Pb+2] = 1.3x10-3 M *(2 moles I-1/1 mole Pb+2) = 2.6x10-3 M
iii) Calculate the value of the equilibrium constant, Ksp.
Ksp = (1.3x10-3)(2.6x10-3)2 = 8.79x10-9
b. A saturated solution is prepared by PbI2 (s) to distilled water to form 2.0 L of solution at 25 oC.
What are the molar concentrations of Pb+2 (aq) and I-1 (aq) in the solution? Justify your answer.
The concentrations should be the same as above; [Pb+2] = 1.3x10-3 and [I-1] = 2.6x10-3. Changing
the volume does not change the concentration if the solution is still saturated.
c. Soild NaI is added to a saturated solution of PbI2 at 25 oC. Assuming that the volume of the
solution does not change, does the molar concentration of Pb+2 (aq) in the solution increase,
decrease, or stay the same? Justify your answer.
By adding NaI, you are adding [I-1] and thus, by LeChatelier’s Principle, the reaction should go to
the left and form a precipitate. To do this, you must use up some of the Pb+2 so thus [Pb+2]
should decrease
5. The value of Ksp for the salt BaCrO4 is 1.2x10-10. When a 500. mL sample of 8.2x10-6 M Ba(NO3)2
is added to 500. mL of 8.2x10-6 M Na2CrO4, no precipitate is observed.
i) Assuming the volumes are additive, calculate the molar concentrations of Ba+2 (aq) and CrO4-2
(aq) in the 1.00 L of solution.
[Ba+2] = 8.2x10-6 M Ba(NO3)2 *(500/1000) = 4.1x10-6 M
[CrO4-2] = 8.2x10-6 M Na2CrO4 *(500/1000) = 4.1x10-6 M
ii) Use the molar concentrations of Ba+2 (aq) and CrO4-2 (aq) ions as determined above to show
why a precipitate does not form. You must include a calculation as part of your answer.
Q = [Ba+2][CrO4-2] = (4.1x10-6)(4.1x10-6) = 1.681x10-11
Q < Ksp 1.681x10-1 < 1.2x10-10
so no ppt forms
6. Silver chromate dissociates in water according the equation below:
Ag2CrO4 (s)  2 Ag+1 (aq) + CrO4-2 (aq) Ksp = 2.6x10-12 at 25 oC
a) Write the equilibrium-constant expression for the dissolving of Ag2CrO4 (s)
Ksp = [Ag+1]2[CrO4-2]
b) Calculate the concentration, in M, of Ag+1 (aq) in a saturated solution of Ag2CrO4 at 25oC
2.6x10-12 = 4x3
x = 8.66x10-5 M
Since there are 2 Ag+1 ions present, [Ag+1] = 1.732x10-4
c) Calculate the maximum mass, in grams, of Ag2CrO4 that can dissolve in 100. mL of water at 25
o
C.
x = 8.6x10-5 moles/L *(0.10 L) *(331.8 g/mole) = 0.002853 g
d) A 0.100 mol sample of solid AgNO3 is added to a 1.00 L saturated solution of Ag2CrO4.
Assuming no volume change, does [CrO4-2] increase, decrease, or stay the same? Justify.
Since AgNO3 will dissolve and make more Ag+1 ions, the reaction will go left due to Le Chatelier’s
principle. This will cause CrO4-2 to react and ppt so [CrO4-2] will decrease.
7. In a saturated solution of Ag3PO4 at 25 oC, the concentration of Ag+1 (aq) is 5.3x10-5 M. The
equilibrium constant expression for the dissolving of Ag3PO4 (s) in water is shown below:
Ksp = [Ag+1]3[PO4-3]
a) Write the balanced equation for the dissolving of Ag3PO4 in water.
Ag3PO4  3 Ag+1 + PO4-3
b) Calculate the value of Ksp for Ag3PO4 at 25 oC.
[Ag+1] = 5.3x10-5 but there is only 1 PO4-3 for every 3 Ag+1 so:
5.3x10-5 *(1 mole PO4-3/3 mole Ag+1) = 1.77 x10-5 M PO4-3
Ksp = [Ag+1]3[PO4-3] = (5.3x10-5)3(1.77x10-5) = 2.635x10-18
c) A 1.00 L sample of saturated Ag3PO4 solution is allowed to evaporate at 25 oC to a final volume
of 500. mL. What is the [Ag+1] in the solution? Justify your answer.
The concentration does not change in a saturated solution. As 500 mL evaporates, solid Ag3PO4
will ppt out at the bottom of the beaker. The solution will still be saturated, though, and
therefore not change its concentration.