pH of a saturated solution of is 9. The solubility product of is [2019]

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Solution

Q.
pH of a saturated solution of $Ca {(OH)}_{2}$ is 9. The solubility product $(K_{s p})$ of $Ca {(OH)}_{2}$ is [2019]

1 $0.5 \times 10^{- 10}$

2 $0.5 \times 10^{- 15}$

3 $0.25 \times 10^{- 10}$

4 $0.125 \times 10^{- 15}$

Ans.
(2)

pH of the saturated solution of $Ca {(OH)}_{2}$ = 9

$∴$ ${pOH of the saturated solution of Ca(OH)}_{2} = 14 - 9 = 5$

$\Rightarrow [{OH}^{-}] = 10^{- 5}$ $(∵ pH + pOH = 14)$

$\begin{matrix} Ca {(OH)}_{2} & ⇌ & {Ca}^{2 +} & + & 2 {OH}^{-} \\ s & 2 s \\ \frac{1}{2} \times 10^{- 5} & 10^{- 5} \end{matrix}$

$K_{s p} = [{Ca}^{2 +}] {[{OH}^{-}]}^{2} = [\frac{1}{2} \times 10^{- 5}] {[10^{- 5}]}^{2}$

$= 0.5 \times 10^{- 15}$

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