First slide

Solubility product(KSP)

Question

The $K_{SP}$ for $Cr {(OH)}_{3}$ is $1.6 \times 10^{- 30} .$ The molar solubility of this compound in water is

Easy

A

$2 \sqrt{1.6 \times 10^{- 30}}$
B

$4 \sqrt{1.6 \times 10^{- 30}}$
C

$\sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}$
D

$1.6 \times 10^{- 30} / 27$

Solution

Let molar solubility of $Cr {(OH)}_{3}$ = $s m o l L^{- 1}$
$Cr {(OH)}_{3} (s) ⇌ {Cr}^{3 +} (aq) + \underset{3 s}{3 {OH}^{-} (aq)} K_{SP} = 1.6 \times 10^{- 30} = [{Cr}^{3 +}] {[{OH}^{-}]}^{3} = (s) {(3 s)}^{3} = 27 s^{4} ∴ s^{4} = \frac{1.6 \times 10^{- 30}}{27} ∴ s = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}$

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