Chapter 17
Solubility and
Simultaneous
Equilibria
CHE1102, Chapter 17
Learn, 1
Solubility Products
Consider the equilibrium that exists in a
saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42(aq)
Saturated solution – no more solute will
dissolve
CHE1102, Chapter 17
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Solubility Product Constant
– equilibrium constant for ionic compounds
that are only slightly soluble
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42]
CHE1102, Chapter 17
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Solubility Products
• Ksp is not the same as solubility.
• Solubility generally expressed as (g/L),
(g/mL), or in mol/L (M).
– Ksp - Only one value for given solid at a given
temperature
• Temperature dependence
Solubility and thus Ksp changes with T
CHE1102, Chapter 17
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Solubility Equilibria
When ionic salt dissolves in water
It dissociates into separate hydrated ions
Initially, no ions in solution
CaF2(s)  Ca2+(aq) + 2F–(aq)
As dissolution occurs, ions build up and collide
Ca2+(aq) + 2F–(aq)  CaF2(s)
At equilibrium
CaF2(s)
Ca2+(aq) + 2F–(aq)
We now have a saturated solution
Ksp = [Ca+][F–]2
CHE1102, Chapter 17
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Writing Ksp Equilibrium Laws
AgCl(s)
Ag+(aq) + Cl–(aq)
PbI2(s)
Pb2+(aq) + 2I–(aq)
Ag2CrO4(s)
2Ag+(aq) + CrO42–(aq)
AuCl3(s)
Au3+(aq) + 3Cl–(aq)
Ksp = [Ag+][Cl–]
Ksp = [Pb2+][I–]2
Ksp = [Ag+]2[CrO42–]
Ksp = [Au3+][Cl–]3
CHE1102, Chapter 17
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Molar Solubility
– # moles of ionic solid that dissolves per 1 L
DI water to form a saturated solution at 25 C
Determine the final equilibrium
concentration of Ca2+ (aq) and CO32- (aq) in a
saturated solution
6.7 x 10-5 moles of CaCO3 (s) will dissolve
per 1 L DI water producing 6.7 x 10-5 M Ca2+
(aq) and 6.7 x 10-5 M CO32- (aq)
CHE1102, Chapter 17
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Given Solubilities, Calculate Ksp
At 25 °C, the solubility of AgCl is 1.34 × 10–5 M.
Calculate the solubility product for AgCl.
AgCl(s)
Ag+(aq) + Cl–(aq)
Ksp = [Ag+][Cl–]
AgCl(s)
I
C
E
Ag+(aq) +
Cl–(aq)
0.00
0. 00
1.34 × 10–5 M
1.34 × 10–5 M
1.34 × 10–5 M
1.34 × 10–5 M
Ksp = (1.34 × 10–5)(1.34 × 10–5)
Ksp = 1.80 × 10–10
CHE1102, Chapter 17
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Learning Check
The solubility of calcium fluoride, CaF2, in pure
water is 2.15 × 10-4 M. What is the value of Ksp?
A. 1.99 × 10–11
B. 3.98 × 10–11
C. 9.94 × 10–12
D. 1.85 × 10-7
CaF2
Ca2+ + 2F–
[Ca2+] = (2.15 × 10–4) [F-] = 2(2.15 × 10–4)
Ksp = [Ca][F]2 = (2.15 × 10–4) (4.3 x 10-4)2
Ksp = 3.98 × 10–11
CHE1102, Chapter 17
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Given Ksp, Calculate Solubility
What is the molar solubility of CuI in water? Determine
the equilibrium concentrations of Cu+ and I–
CuI(s)
Cu+(aq) + I–(aq)
Ksp = [Cu+][I–]
Ksp = 1.3  10–12
CHE1102, Chapter 17
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Molar Solubilities from Ksp
Conc (M)
Initial Conc.
Change
Equilibrium Conc.
CuI(s)
No
entries
No
entries
No
entries
Cu+(aq) + I–(aq)
0.00
+x
0.00
+x
x
x
Solve Ksp expression
• Ksp = 1.3 × 10–12 = (x)(x)
• x2 = 1.3 × 10–12
• x = 1.1 × 10–6 M = calculated molar solubility of CuI
= [Cu+] = [I– ]
CHE1102, Chapter 17
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Given Ksp, Calculate Solubilities
Calculate the solubility of CaF2 in water at 25 °C, if
Ksp = 3.4 × 10–11.
CaF2(s)
Ca2+(aq) + 2F– (aq)
1. Write equilibrium law
Ksp = [Ca2+][F–]2
2. Construct concentration table
Conc (M)
CaF2(s)
Initial Conc.
Change
Equilibrium Conc.
(No entries 0.00
+x
in this
x
column)
Ca2+(aq) 2F–(aq)
0.00
+2x
2CHE1102,
x Chapter 17
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Molar Solubilities from Ksp
3. Solve the Ksp expression
Ksp = [Ca2+][F–]2 = (x) (2x)2
3.4 × 10–11 = 4x3
3.4 ´10
x =
4
-11
3
x = 8.5´10
3
-12
x = 2.0 × 10–4 M = molar solubility of CaF2
[Ca2+] = x = 2.0 × 10–4 M
[F–] = 2x = 2(2.0 × 10–4 M) = 4.0 × 10–4 M
CHE1102, Chapter 17
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Factors Affecting Solubility
• The Common-Ion Effect
– If one of the ions in a solution equilibrium
is already dissolved in the solution, the
equilibrium will shift to the left and the
solubility of the salt will decrease:
BaSO4(s)
Ba2+(aq) + SO42(aq)
CHE1102, Chapter 17
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Factors Affecting Solubility
• pH
– If a substance has a basic anion, it will be more
soluble in an acidic solution.
– Substances with acidic cations are more soluble
in basic solutions.
CHE1102, Chapter 17
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Common Ion Effect
• What happens if another salt, containing one
of the ions in our insoluble salt, is added to a
solution?
Consider PbI2(s)
Pb2+(aq) + 2I–(aq)
– Saturated solution of PbI2 in water
• Add KI
• PbI2 (yellow solid) precipitates out
– Why?
• Le Chatelier’s Principle
• Add product I–
• Equilibrium moves to left and solid
PbI2 forms
CHE1102, Chapter 17
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Learning Check
What effect would adding copper(II) nitrate have on
the solubility of CuS?
A. The solubility would increase
B. The solubility would decrease
C. The solubility would not change
CHE1102, Chapter 17
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Common Ion Effect
Consider three cases
A. What is the molar solubility of
Ag2CrO4 in pure water?
B. What is the molar solubility of Ag2CrO4 in 0.10 M
AgNO3?
C. What is the molar solubility of Ag2CrO4 in 0.10 M
Na2CrO4?
•
•
Ag2CrO4(s)
2Ag+(aq) + CrO42–(aq)
Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12
CHE1102, Chapter 17
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Common Ion Effect
A. What is the solubility of Ag2CrO4 in pure water?
I
C
E
Ag2CrO4(s)
2Ag+(aq) +
(No entries
in this
column)
0.00 M
+2x
CrO42–(aq)
2x
0.00 M
+x
x
Ksp = [Ag+]2[CrO42–] = (2x)2(x) = 1.1  10–12 = 4x3
-12
1.1´10
x= 3
= 3 2.7 ´10-13
4
x = Solubility of Ag2CrO4 = 6.5 x 10-5 M
[CrO42–] = x = 6.5 x 10-5 M
[Ag+] = 2x = 1.3 × 10–4 M
CHE1102, Chapter 17
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Common Ion Effect
B. What is the molar solubility of Ag2CrO4 in 0.10 M
AgNO3 solution? Ksp = 1.1 × 10–12
Ag2CrO4(s)
I (No entries
C in this
E column)
Ksp = 1.1 ×
•
•
•
10–12
2Ag+(aq)
+
CrO42–(aq)
0.10 M
+2x
0.00
+x
≈0.10
x
= (0.10
M)2[x]
-12
1.1 ´ 10
x=
(0.010)
x = Solubility of Ag2CrO4 = 1.1 × 10–10 M
[Ag+] = 0.10 M
[CrO42–] = 1.1 × 10–10 M
CHE1102, Chapter 17
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Common Ion Effect
C. What is the solubility of Ag2CrO4 in 0.100 M Na2CrO4?
Ag2CrO4(s)
I (No entries
C in this
E column)
2Ag+(aq) +
CrO42–(aq)
0.00 M
+2x
0.10 M
+x
2x
≈0.10
Ksp = (2x)2(0.10) = 1.1  10–12 = 4x2(0.10)
1.1´10-12
x=
= 2.7 ´10-12
0.4
x = Solubility of Ag2CrO4 = 1.66 × 10–6 M
[CrO42–] = 0.10 M
[Ag+] = 2x = 3.3 × 10–6 M
CHE1102, Chapter 17
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Common Ion Effect
What have we learned about the solubility of silver
chromate?
A. Dissolving it in pure water the solubility was
3.0 × 10–4 M
B. Dissolving it in AgNO3 solution solubility was
1.1 × 10–10 M
C. Dissolving it in Na2CrO4 solution solubility was
3.2 × 10–5 M
Common ion appearing the most in the formula of the
precipitate decreases the solubility the most
CHE1102, Chapter 17
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Will a Precipitate Form?
• In a solution,
– If Q = Ksp, the system is at equilibrium
and the solution is saturated.
– If Q < Ksp, more solid can dissolve
until Q = Ksp.
– If Q > Ksp, the salt will precipitate until
Q = Ksp.
CHE1102, Chapter 17
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Predicting Precipitation
Will a precipitate of PbI2 form if 100.0 mL of
0.0500 M Pb(NO3)2 are mixed with 200.0 mL
of 0.100 M NaI?
PbI2(s)
Pb2+(aq) + 2I–(aq)
Ksp = [Pb2+][I–]2 = 9.8 × 10–9
Strategy for solving
1. Calculate concentrations in mixture prepared
2. Calculate Qsp = [Pb2+][I–]2
3. Compare Qsp to Ksp
CHE1102, Chapter 17
Learn, 24
Predicting Precipitation
Step 1. Calculate concentrations
– Vtotal = 100.0 mL + 200.0 mL = 300.0 mL
2+
mmol
of
Pb
(100.0 mL)(0.0500 mmol/mL)
[Pb2+ ] =
=
mL of solution
300.0 mL
– [Pb2+] = 1.67 × 10–2 M
mmol
of
I
(200.0 mL)(0.100 mmol/mL)
[I- ] =
=
mL of solution
300.0 mL
– [I–] = 6.67 × 10–2 M
CHE1102, Chapter 17
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Predicting Precipitation
Step 2. Calculate Qsp
– Qsp = [Pb2+][I–]2 = (1.67 × 10–2 M)(6.67 × 10–2 M)2
– Qsp =7.43 × 10–5
Step 3. Compare Qsp and Ksp
– Qsp = 7.43 × 10–5
– Ksp = 9.8 × 109
– Qsp > Ksp so precipitation will occur
CHE1102, Chapter 17
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pH and Solubility
• Mg(OH)2(s)
Mg2+(aq) + 2OH–(aq)
– Increase OH– shift equilibrium to left
– Add H+
shift equilibrium to right
– Le Châtelier’s Principle
• Ag3PO4(s)
3Ag+(aq) + PO43–(aq)
– Add H+
increase solubility
– H+(aq) + PO43–(aq)  HPO42–(aq)
• AgCl(s)
Ag+(aq) + Cl–(aq)
– Adding H+ has no effect on solubility Why?
– Cl– is very, very weak base, so neutral anion
– So adding H+ doesn’t effect Cl– concentration
CHE1102, Chapter 17
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