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)

$8.3$

Answer 2

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