Equilibrium - Result Question 66
####69. $pH$ of a saturated solution of $Ca(OH)_2$ is 9 . The solubility product $(K _{s p})$ of $Ca(OH)_2$ is: [2019]
(a) $0.510^{-15}$
(b) $0.2510^{-10}$
(c) $0.12510^{-15}$
(d) $0.5 \quad 10^{-10}$
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Solution:
- (a) $Ca(OH)_2 \rightarrow Ca^{2+}+2 OH^{-}$ $pH=9, pOH=14-9=5$
$[OH]=10^{-5}$
$[Ca^{2+}]=\frac{10^{-5}}{2}$
$K _{s p}=[Ca^{2+}][OH^{-}]_2=(\frac{10^{-5}}{2}) \times(10^{-5})^{2}$
$ =0.5 \times 10^{-15} $
