First slide

Solubility product(KSP)

Question

Calculate the solubility of $A_{2} X_{3}$ in pure water, assuming that neither kind of ion reacts with water. The solubility product of $A_{2} X_{3}, K_{sp} = 1.1 \times 10^{- 23}$

Moderate

A

$2.0 \times 10^{- 5} mol L^{- 1}$
B

$5.0 \times 10^{- 5} mol L^{- 1}$
C

$1.0 \times 10^{- 5} mol L^{- 1}$
D

$4.0 \times 10^{- 3} mol L^{- 1}$

Solution

$A_{2} X_{3} ⟶ 2 A^{3 +} + 3 X^{2 -}$
$K_{sp} = {[A^{3 +}]}^{2} {[X^{2 -}]}^{3} = 1.1 \times 10^{- 23}$
$If S = solubility of A_{2} X_{3}, then$
$[A^{3 +}] = 2 S; [X^{2 -}] = 3 S$
therefore, $K_{sp} = (2 S)^{2} (3 S)^{3}$
$= (108) S^{5} = 1.1 \times 10^{- 23}$
Thus, $S^{5} = 1 \times 10^{- 25}$
$S = 1.0 \times 10^{- 5} mol L$

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