1. CM4106 Chemical Equilibria & Thermodynamics
Lesson 4
Solubility Equilibria
A Chemistry Education Blog by Mr Tan
http://chemistry-mr-tan-yong-yao.blogspot.sg/

2. Solubility of Common Salts
Courtesy of Chemistry, 10th Edition by Raymond Chang

3. Fundamentals:
AxBy (s) ⇌ xAy+(aq) + yBx(aq)
• The equilibrium constant for this heterogeneous
equilibrium is called the solubility product, Ksp, is written
as:
Ksp = [Ay+(aq)]xeqm [Bx-(aq)]yeqm
• The solubility product, Ksp, of a sparingly soluble salt is
defined as the product of the concentration of the ions
(in M) in a saturated solution at a given temperature
raised to the power of its coefficient in the equilibrium
equation.
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4. Calculating Ksp value from solubility
• The solubility of AgCl at 18°C is 1.46 x 10-3 g/L, what is the
solubility product of AgCl at 18°C?
Since the solubility of AgCl = 1.46 x 10-3 g / L;
[Ag+] = (1.46 x 10-3 g/L) / (143.5 g/mol)
= 1.017 x 10-5 mol/L
AgCl (s) ⇌ Ag+ (aq) + Cl (aq)
Initial / M - 0 0
Change / M -s +s +s
Eqm / M - s s
Ksp = [Ag+] [Cl]
= (1.017 x 10-5)2
= 1.03 x 10-10
The solubility product of AgCl at 18°C is 1.03 x 10-10

5. Calculating Solubility from Ksp
• The Ksp for Ag2CO3 is 8.0 x 10-12 M3, calculate its solubility at this
temperature.
Let s be the solubility of Ag2CO3 in mol / L
Ag2CO3(s) ⇌ 2Ag+(aq) + CO32(aq)
Initial / M - 0 0
Change / M -s +2s +s
Eqm / M - 2s s
Ksp = [Ag+]2[CO32]
8.0 x 10-12 = [2s]2[s]
8.0 x 10-12 = 4 s3
s = 1.26 x 10-4 M
= 1.3 x 10-4 M (2 s.f.)
Solubility of Ag2CO3 = 1.3 x 10-4 M

6. Quantitative Problems
Calculate Ksp given solubility
Calculate solubility given Ksp

7. Relationship between Ksp and Solubility (M)

8. Predicting Precipitation - Will a PPT
form?
Knowing the value of Ksp allows us to predict if a ppt will be
formed when two solutions containing ions to form an insoluble
salt are mixed.
Step 1: Calculate reaction quotient Qsp
Step 2: Compare Qsp with Ksp
Inference
Q>K Precipitation occurs till Q = Ksp
No precipitation is seen because the solution is not saturated
Q<K
with the ions hence they remain dissolved in the solution
Q=K A saturated solution is obtained
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9. Predicting Precipitation (Ksp vs Q)
Will a precipitate form when 0.10 L of 8.0 x 10-3 M Pb(NO3)2 is
added to 0.40 L of 5.0 x 10-3 M Na2SO4? (Ksp = 6.3 x 10-7)
REMEMBER: TAKE INTO ACCOUNT OF CONCENTRATION OF IONS IN
MIXTURE (DILUTION UPON MIXING)
PbSO4 (s) ⇌ Pb2+ (aq) + SO42- (aq)
[Pb2+] = (0.10/0.50) x 8.0 x 10-3 [SO42-] = (0.40/0.50) x 5.0 x 10-3
= 0.0016 M = 0.0040 M
Q = [Pb2+][SO42-]
= (0.0016) (0.0040)
= 6.4 x 10-6 > Ksp = 6.3 x 10-7
Q > Ksp  ppt will form

10. Factors affecting solubility
1. Common Ion Effect
2. pH of solution
3. Formation of Complexes

11. 1. Common Ion Effect
Solubility of a substance is affected by the presence of other
solutes, especially if there is a common ion present
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
What happens to the solubility of CaF2 if the solution already
contains Ca2+ (aq) or F- (aq)?
1. Equilibrium position shifts to the left
2. Solubility of CaF2 decreases

12. 1. Common Ion Effect
Calculate the molar solubility of CaF2 at 25 oC in 0.010 M
Ca(NO3)2 solution
CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq)
Initial / M - 0.010 0
Change / M - +s + 2s
Eqm / M - 0.010 + s 2s
Ksp = [Ca2+][F-]2 = (0.010 + s)(2s)2
Since s is small, assume 0.010 + s ≈ 0.010
(CaF2 is a sparingly soluble salt and the solubility is further suppressed by the
presence of common ion effect)
3.9 x 10-11 = (0.010)(2s)2 Assumption is valid, s << 0.010
s = solubility = 3.1 x 10-5 mol/L

13. 2. pH of Solution
 If a substance has a basic anion, it will be more soluble in
an acidic solution.
 If a substance has an acidic cation, it will be more soluble
in an basic solution.

14. 2. pH of Solution
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq) Ksp = 1.8 x 10-11 (25 oC)
Calculate the molar solubility of Mg(OH)2 in pure water.
What is the pH of the resulting solution?
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)
Initial / M - 0 0
Ksp = [Mg2+][OH-]2
= (x)(2x)2 Change / M - +x +2x
1.8 x 10-11 = 4x3 Eqm / M - x 2x
x = Solubility = 1.651 x 10-4 mol/L = 1.7 x10-4 M (2 s.f.)
[OH-] = 2x = 3.302 x 10-4 M
pOH = - lg(3.302 10-4) = 3.48
pH = 10.5 (1 d.p.)
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15. 2. pH of Solution
What is the solubility of Mg(OH)2 in a less alkaline solution buffered
at pH 9?
pOH = 5 Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2OH- (aq)
[OH-] = 1.0 x 10-5 M Initial / M - 0 1.0 x 10-5
Change / M - +x -
Eqm / M - x 1.0 x 10-5
Ksp = [Mg2+] [OH-]2 Ksp = 1.8 x 10-11 (25 oC)
= [Mg2+](1.0 x 10-5)2
[Mg2+] = 0.18 M (2 s.f.)
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16. With the
addition of
acid

17. 3. Complexation
When aqueous ammonia is added dropwise to a Cu2+ solution, a light blue precipitate is
observed. As aqueous ammonia is added in excess, the precipitate dissolves to give a deep
blue solution.
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18. 3. Complexation
1. When aqueous ammonia is added dropwise to a Cu2+ solution, a light blue precipitate is
observed. As aqueous ammonia is added in excess, the precipitate dissolves to give a
deep blue solution.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH (aq)
When added to Cu2+(aq) solution, Cu2+(aq) + 2OH (aq) ⇌ Cu(OH)2(s) – (1)
Hence a blue precipitate is observed.
As more NH3(aq) is added, Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq)
[Cu2+] drops as Cu(NH3)42+(aq) is formed, thus position of equilibrium in equation
1 shifts left to produce more Cu2+(aq), hence Cu(OH)2(s) dissolves.
[Cu(H2O)6]2+ + 2 OH- (aq) → [Cu(H2O)4(OH)2] (s)
[Cu(H2O)6]2+ + 4 NH3 (aq) → [Cu(NH3)4(H2O)2]2+ (aq)
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19. Ksp values