Question

If $K_{sp}$ of $Mg(OH)_2$ is $1.2 ? 10^{-11}$. Then the highest pH of the $0.1\, M$ solution of $Mg^{2+}$ ion from which $Mg(OH)_2$ is not precipitated is

• A. $4.96$
• B. $6.96$
• C. $7.54$
• D. $9.04$

Answer 1

Answer: (D)

$9.04$

Answer 2

As $K_{sp}$ of $Mg(OH)_2 = [Mg^{2+}] [OH^-]^2$ $\therefore \left[OH^{-}\right]^{2}=\frac{1.2\times10^{-11}}{0.1}=1.2\times10^{-10}$ $\therefore \left[OH^{-}\right]^{2}=1.1\times10^{-5}$ $\therefore \left[H^{+}\right]=10^{-14}/1.1\times10^{-5}=0.91\times10^{-9}$ $pH=-log\left[H^{+}\right]=log\,0.91\times10^{-9}=9.04$