Equilibrium - Result Question 76
####80. $pH$ of a saturated solution of $Ba(OH)_2$ is 12 . The value of solubility product $(K _{s p})$ of $Ba(OH)_2$ is :
(a) $3.3 \times 10^{-7}$
(b) $5.0 \times 10^{-7}$
(c) $4.0 \times 10^{-6}$
(d) $5.0 \times 10^{-6}$
[2012]
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Solution:
- (b) Given $pH=12$
or $[H^{+}]=10^{-12}$
Since, $\quad[H^{+}][OH^{-}]=10^{-14}$
$\therefore \quad[OH^{-}]=\frac{10^{-14}}{10^{-12}}=10^{-2}$
$ \begin{aligned} & Ba(OH)_2 \rightarrow Ba^{2+}+2 OH^{-} \\ & {[OH^{-}]=10^{-2}} \\ & 2 s=10^{-2} \\ & s=\frac{10^{-2}}{2}=5 \times 10^{-3} m, K _{s p}=s \cdot(2 s)^{2} \\ & \Rightarrow \quad K _{s p}=5 \times 10^{-7} \end{aligned} $
