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Question:

The electrochemical cell shown below is a concentration cell. M|M²⁺(saturated solution of a sparingly soluble salt, MX₂)||M²⁺(0.001 moldm⁻³)|M The emf of the cell depends on the difference in concentrations of M²⁺ ions at the two electrodes. The emf of the cell at 298K is 0.059 V. The solubility product (Ksp; mol³ dm⁻⁹) of MX₂ at 298K based on the information available for the given concentration cell is: [Take 2.303 × R × 298/F = 0.059 V]

1×10⁻⁵

4×10⁻⁵

4×10⁻²

1×10⁻²

Solution:

E.M.F of concentration cell = 0.059/n log₁₀[M²⁺ₐ/M²⁺c]
Given that emf = 0.059 V
0.059 = 0.059/2 log₁₀[M²⁺ₐ/0.001]
1 = 1/2 log₁₀[M²⁺ₐ/0.001]
2 = log₁₀[M²⁺ₐ/0.001]
10² = M²⁺ₐ/0.001
100 = M²⁺ₐ/0.001
M²⁺ₐ = 100 × 0.001 = 0.1
Let s be the solubility of MX₂
MX₂ <=> M²⁺ + 2X⁻
Ksp = [M²⁺][X⁻]² = (s)(2s)² = 4s³
Since [M²⁺] = 0.1
s = 0.1
Ksp = 4s³ = 4(0.1)³ = 4 × 10⁻³ = 4 × 10⁻² mol³ dm⁻⁹