Question

The $K_{sp} $ of $Mg(OH)_{2}$ is $ 1\times 10^{-12}.\, 0.01\, M\, Mg(OH)_{2}$ will precipitate at the limited $pH$ of

• A. 3
• B. 9
• C. 5
• D. 8

Answer 1

Answer: (B)

$Mg ( OH )_{2} \rightleftharpoons Mg ^{2+}+2 OH ^{-}$
The solubility product,
$K_{ sp } \text { of } Mg ( OH )_{2} =[ Mg ]^{2+}\left[ OH ^{-}\right]^{2}$
$1 \times 10^{-12} =[0.01]\left[ OH ^{-}\right]^{2}$
$\left[ OH ^{-}\right]^{2} =\frac{1 \times 10^{-12}}{0.01}=1 \times 10^{-10}$
$\left[ OH ^{-}\right] =10^{-5}$
$\because \left[ H ^{+}\right]\left[ OH ^{-}\right]=10^{-14}$
$\left[ H ^{+}\right]\left[10^{-5}\right]=10^{-14}$
$\left[ H ^{+}\right]=\frac{10^{-14}}{10^{-5}}=10^{-9}$
$pH = -\log \left[ H ^{+}\right]$
$=-\log 10^{-9}=9$