What is the minimum \(pH\) of a solution \(0.10\, \ M\) in \(Mg^{2+}\) from which \(Mg(OH)_2\) will not precipitate?

Given: \(K_{sp}\) of \(Mg(OH)_2\) \(= 1.2 \times 10^{-11} M^3\)

9.04

The correct answer is option 1. 9.04.

Given, \(K_{sp}\) of \(Mg(OH)_2\) \(= 1.2 \times 10^{-11} M^3\)

\(K_{sp}\, \ Mg(OH)_2\, \ = \, \ [Mg^{2+}][OH^-]^2\)

\(⇒ 1.2 \times 10^{-11} =  [0.1][OH^-]^2\)

\(⇒ [OH^-]^2 = 1.2 \times 10^{-10}\)

\(⇒ [OH^-] = 1.0954 \times 10^{-5}M \)

Thus,

\(pOH = -log_{10}(1.0954 \times 10^{-5}) = 4.96\)

Hence, \(pH =  14 - 4.96 = 9.04\)