Q. 23 The solubility product (Ksp) of Mg(OH)2 at 25°C is 5.6 × 10−11. Its solubility in water is S × 10−2 g/L, where the value of S is ____ (up to two decimal places). (Given: Molecular weight of Mg(OH)2 = 58.3 g mol−1)
Step-by-Step Calculation
Mg(OH)2 dissociates as Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH–(aq), so Ksp = [Mg2+][OH–]2 = S × (2S)2 = 4S3.
Solve 4S3 = 5.6 × 10-11: S3 = 1.4 × 10-11, S = (1.4 × 10-11)1/3 ≈ 2.41 × 10-4 mol/L.
Convert to g/L: (2.41 × 10-4 mol/L) × 58.3 g/mol ≈ 0.01405 g/L = 1.405 × 10-2 g/L.
Exact calculation: Smolar = [5.6×10-11 / 4]1/3 = [1.4×10-11]1/3.
S = 2.408 × 10-4 mol/L, mass solubility = 2.408×10-4 × 58.3 ≈ 0.01403664 g/L = 1.40 × 10-2 g/L, thus S = 1.40.
Common Options Explained
- 1.10: Underestimates S3 factor, common if using Ksp=1.8×10-11 instead.
- 1.40 (Correct): Matches given Ksp=5.6×10-11 and MW=58.3 precisely.
- 1.75: Assumes 2S3 error (for 1:1 electrolyte).
- 2.10: From older Ksp≈1×10-10 values.
