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Question:

At 25°C, the solubility product of Mg(OH)₂ is 1.0 × 10⁻¹¹. At which pH will Mg²⁺ ions start precipitating in the form of Mg(OH)₂ from a solution of 0.001 M Mg²⁺ ions?

9

11

10

8

Solution:

Ksp = [Mg²⁺][OH⁻]²
[OH⁻] = √(Ksp/[Mg²⁺]) = √(10⁻¹¹/10⁻³)= 10⁻⁴
pOH = 4
and pH = 10