(B) $KOH \rightarrow K^{+} + OH^{-}$$Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^{-}$Solubility product $K_{sp} = [Cd^{2+}][OH^{-}]^2$Given $[OH^{-}] =

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(B) $KOH \rightarrow K^{+} + OH^{-}$$Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^{-}$Solubility product $K_{sp} = [Cd^{2+}][OH^{-}]^2$Given $[OH^{-}] = 0.1 \ M$ due to common ion effect.$K_{sp} = [S] \times [0.1]^2$$2.5 \times 10^{-14} = [S] \times 0.01$$[S] = \frac{2.5 \times 10^{-14}}{10^{-2}} = 2.5 \times 10^{-12}$$[S] = 25 \times 10^{-13} = x \times 10^{-y}$Thus,$x = 25$ and $y = 13$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Common ion effect

Medium

At $298 \ K$ the molar solubility of $Cd(OH)_2$ in $0.1 \ M \ KOH$ solution is $x \times 10^{-y}$. The values of $x$ and $y$ are respectively (at $298 \ K, \ K_{sp}$ of $Cd(OH)_2 = 2.5 \times 10^{-14}$)

TS EAMCET 2023 All TS EAMCET

A
$2.5, 14$
B
$25, 13$
C
$25, 14$
D
$2.5, 16$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Common ion effect

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At $298 \ K$ the molar solubility of $Cd(OH)_2$ in $0.1 \ M \ KOH$ solution is $x \times 10^{-y}$. The values of $x$ and $y$ are respectively (at $298 \ K, \ K_{sp}$ of $Cd(OH)_2 = 2.5 \times 10^{-14}$)

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