First slide

Solubility product(KSP)

Question

pH of $Ba {(OH)}_{2}$ solution is 12. Its solubility product is

Easy

A

$10^{- 6} M^{3}$
B

$4 \times 10^{- 6} M^{3}$
C

$0.5 \times 10^{- 7} M^{3}$
D

$5 \times 10^{- 7} M^{3}$

Solution

Since pH = 12 $∴$ pOH = 14 - 12 = 2
$∴ [{OH}^{-}] = 10^{- 2} M$
We know $Ba {(OH)}_{2}$ $⇌ {Ba}^{+ +} + 2 {OH}^{-}$
$∴$ $[{Ba}^{+ +}] = \frac{[{OH}^{-}]}{2}$
$∴$ $K_{SP} = [{Ba}^{+ +}] {[{OH}^{-}]}^{2} = (\frac{10^{- 2}}{2}) \times {(10^{- 2})}^{2}$
= $5 \times 10^{- 7} M^{3}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App