Question

What will be the $ pH $ at which precipitation of $ Zn(OH)_2 $ will take place from a solution containing $ 0.3 \,M \,Zn^{2+} $ ion? $ K_{sp} $ of $ Zn(OH)_2 = 1.2 \times 10^{-12} $

• A. $ 5.7 $
• B. $ 8.3 $
• C. $ 11.4 $
• D. $ 6.2 $

Answer 1

Answer: (B)

$K_{sp}$ of $Zn\left(OH\right)_{2}$
$=\left[Zn^{2+}\right] \left[OH^{-}\right]^{2}$
$1.2\times10^{-12}=0.3\times\left[OH^{-}\right]^{2}$
$\left[OH^{-}\right]=2\times10^{-6}$
$pOH=-log \left[2\times10^{-6}\right]=5.699$
$\therefore pH=14-5.699$
$=8.301$