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At $298 \mathrm{~K}$ the molar solubility of $\mathrm{Cd}(\mathrm{OH})_2$ in $0.1 \mathrm{M} \mathrm{KOH}$ solution is $x \times 10^{-\mathrm{y}}$. The values of $x$ and $y$ are respectively (at $298 \mathrm{~K}, \mathrm{~K}_{\mathrm{sp}}$ of $\mathrm{Cd}(\mathrm{OH})_2=2.5 \times 10^{-14}$ )
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The correct answer is:
25,13
$\mathrm{KOH} \rightarrow \mathrm{K}^{+}+\mathrm{OH}^{-}$
$\mathrm{Cd}(\mathrm{OH})_2 \rightleftharpoons \begin{array}{cc}0.1 & 0.1 \\ \mathrm{Cd}^{2+} & +2 \mathrm{OH}^{-} \\ \mathrm{S} & 2 \mathrm{~S}\end{array}$
Solubility product $\mathrm{k}_{\mathrm{sp}}=[\mathrm{S}]^1[2 \mathrm{~S}+0.1]^2$
$\begin{aligned} & \mathrm{k}_{\mathrm{sp}}=[\mathrm{S}]^1[0.1]^2(\because 0.1>>>2 \mathrm{~S}) \\ & 2.5 \times 10^{-14}=[\mathrm{S}] \times 0.01 \\ & {[\mathrm{~S}]=2.5 \times 10^{-12}} \\ & {[\mathrm{~S}]=25 \times 10^{-13}=\mathrm{x} \times 10^{-\mathrm{y}}}\end{aligned}$
So, $x=25, y=13$.
$\mathrm{Cd}(\mathrm{OH})_2 \rightleftharpoons \begin{array}{cc}0.1 & 0.1 \\ \mathrm{Cd}^{2+} & +2 \mathrm{OH}^{-} \\ \mathrm{S} & 2 \mathrm{~S}\end{array}$
Solubility product $\mathrm{k}_{\mathrm{sp}}=[\mathrm{S}]^1[2 \mathrm{~S}+0.1]^2$
$\begin{aligned} & \mathrm{k}_{\mathrm{sp}}=[\mathrm{S}]^1[0.1]^2(\because 0.1>>>2 \mathrm{~S}) \\ & 2.5 \times 10^{-14}=[\mathrm{S}] \times 0.01 \\ & {[\mathrm{~S}]=2.5 \times 10^{-12}} \\ & {[\mathrm{~S}]=25 \times 10^{-13}=\mathrm{x} \times 10^{-\mathrm{y}}}\end{aligned}$
So, $x=25, y=13$.
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