The document discusses solubility equilibria and solubility product constants (Ksp). It defines Ksp as the product of ion concentrations raised to their coefficients in the solubility reaction equation. It provides examples of using Ksp to calculate solubility from solubility data and vice versa. Additionally, it discusses factors that affect solubility such as common ion effect, pH, and complexation.

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