Question

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} $

Answer 1

Answer: (B)

$Ba ( OH )_{2} \rightleftarrows Ba ^{2+}+2 OH ^{-}$
$pH =12 \Rightarrow p ( OH )=14- pH$
$\rightarrow p ( OH )=14-12=2$
$\left[ OH ^{-}\right]=10^{- POH }=10^{-2} $ or $ 1 \times 10^{-2}$
as conc. of $Ba ^{2+}$ is half of $OH ^{-}$
$\rightarrow Ba ^{2+}=0.5 \times 10^{-2}$
$K _{ sp }=\left(0.5 \times 10^{-2}\right)\left(1 \times 10^{-2}\right)^{2}$
$K _{ sp }=0.5 \times 10^{-6}$
$=5 \times 10^{-7}$