6-2.Equilibrium-II (Ionic Equilibrium)
medium

$\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_{2}$ ts $9 .$ The solublity product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ is

A

 $0.5 \times 10^{-15}$

B

 $0.25 \times 10^{-10}$

C

 $0.125 \times 10^{-15}$

D

 $0.5 \times 10^{-10}$

(NEET-2019)

Solution

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

$\mathrm{pH}=9 ; \mathrm{pOH}=5 ; \mathrm{[OH]}^{-}=10^{-5}=2 \mathrm{S}$

$\mathrm{S}=\frac{10^{-5}}{2}$

$\mathrm{K}_{sp}-\left|\mathrm{Ca}^{+2}\right|\left|\mathrm{OH}^{-}\right|^{2}$

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

$\mathrm{K}_{\mathrm{sp}}=4 \mathrm{S}^{3}$

$\mathrm{K}_{\mathrm{sp}}=4 \times\left(\frac{10^{-5}}{2}\right)^{3}$

$\mathrm{K}_{\mathrm{sp}}=0.5 \times 10^{-15}$

Standard 11
Chemistry

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