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6-2.Equilibrium-II (Ionic Equilibrium)
hard
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
$3$
C
$2$
D
$4$
(JEE MAIN-2021)
Solution
$\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Zn}^{+2}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$
$\quad\quad\quad\quad\quad\quad\quad S\quad\quad\quad\quad (0.1+2S) \simeq\,0.1$
$\mathrm{K}_{\mathrm{sp}}=\mathrm{S}(0.1)^{2}$
$2 \times 10^{-20}=\mathrm{s} \times 10^{-2} \Rightarrow \mathrm{S}=2 \times 10^{-18}$
$=\mathrm{x} \times 10^{-18}$
$\mathrm{x}=2$
Standard 11
Chemistry
