The molar solubility of Zn(OH)_ 2 in 0.1 ,M ,NaOH solution is x 1… - Vidyadip

MCQ

The molar solubility of $\mathrm{Zn}(\mathrm{OH})_{2}$ in $0.1 \,\mathrm{M} \,\mathrm{NaOH}$ solution is $\mathrm{x} \times 10^{-18} \,\mathrm{M}$. The value of $\mathrm{x}$ is ...... . (Nearest integer)

(Given : The solubility product of $\mathrm{Zn}(\mathrm{OH})_{2}$ is $\left.2 \times 10^{-20}\right)$

A
$1$
B
$3$
✓
$2$
D
$4$

✓

Answer

Correct option: C.

$2$

c
$\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Zn}^{+2}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$

$\quad\quad\quad\quad\quad\quad\quad S\quad\quad\quad\quad (0.1+2S) \simeq\,0.1$

$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}(0.1)^{2}$

$2 \times 10^{-20}=\mathrm{s} \times 10^{-2} \Rightarrow \mathrm{S}=2 \times 10^{-18}$

$=\mathrm{x} \times 10^{-18}$

$\mathrm{x}=2$

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