Equilibrium Ques 67
67. $pH$ of a saturated solution of $Ca(OH)_2$ is $9$ . The solubility product $(K _{s p})$ of $Ca(OH)_2$ is:
[2019]
(a) $0.5\times 10^{-15}$
(b) $0.25\times 10^{-10}$
(c) $0.125\times 10^{-15}$
(d) $0.5 \times 10^{-10}$
Show Answer
Answer:
Correct Answer: 67.(a)
Solution:
- (a) $Ca(OH)_2 \rightleftharpoons Ca^{2+}+2 OH^{-}$ $pH=9,
pOH=14-9=5$
$[OH]=10^{-5}$
$[Ca^{2+}]=\frac{10^{-5}}{2}$
$K _{s p}=[Ca^{2+}][OH^{-}]_2=(\frac{10^{-5}}{2}) \times(10^{-5})^{2}$
$ =0.5 \times 10^{-15} $
BETA
