- Home
- Standard 11
- Chemistry
6-2.Equilibrium-II (Ionic Equilibrium)
medium
$pH$ of a saturated solution of $Ba(OH)_2$ is $12.$ The value of solubility product $(K_{sp})$ of $Ba(OH)_2$ is
A
$3.3 \times 10^{-7}$
B
$5.0 \times 10^{-7}$
C
$4.0 \times 10^{-6}$
D
$5.0 \times 10^{-6}$
(AIPMT-2012)
Solution
$\mathrm{pH}$ of solution $=12$
$\left[\mathrm{H}^{+}\right]=10^{-12}$
$\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-12}}=10^{-2}$
$\mathrm{Ba}(\mathrm{OH})_{2} \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}$
$\quad \quad \quad \quad \quad \quad s \quad \quad \quad 2 s$
$2 s=10^{-2} \Rightarrow s=\frac{10^{-2}}{2}$
$K_{s p}=(s)(2 s)^{2}=4 s^{3}$
$=4 \times\left(\frac{10^{-2}}{2}\right)^{3}=\frac{4}{8} \times 10^{-6}$
$=5 \times 10^{-7}$
Standard 11
Chemistry
