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 \(Mg(OH)_2\) \(= 1.2 \times 10^{-11} M^3\)
Answer & explanation
Correct answer: option 1
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, \(pH = 14 - 4.96 = 9.04\)