First slide

Solubility product(KSP)

Question

pH of saturated solution of $Ba (OH)_{2}$ is 12.The value of solubility product of $K_{sp} of Ba (OH)_{2}$ is

Difficult

A

$3.3 \times 10^{- 7}$
B

$5.0 \times 10^{- 7}$
C

$4.0 \times 10^{- 4}$
D

$5.0 \times 10^{6}$

Solution

Given $pH of Ba (OH)_{2} = 12$
$pOH = 14 - pH = 14 - 12 = 2$
We know that, $pOH = - \log [{OH}^{-}]$
$2 = - \log [{OH}^{-}]$
$[{OH}^{-}] = 10^{- 2}$
$Ba (OH)_{2} dissolve in water as$
$Ba (OH)_{2} (s) ⇌ {Ba}^{2 +} + 2 {OH}^{-}$
$S mol L^{- 1} S 2 S$
$[{OH}^{-}] = 2 s = 10^{- 2}$
$K_{sp} = [{Ba}^{2 +}] {[{OH}^{-}]}^{2} = \frac{(10^{- 2})}{2} \times {(10^{- 2})}^{2}$
$= 0.5 \times 10^{- 6} = 5 \times 10^{- 7}$

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