Question

Solubility of $M{X}_2$ type electrolytes is $0.5\times {10}^{-4}$ mole/lit. Then find out $K_{sp}$ of electrolytes.

• A. $5\times {10}^{12}$
• B. $25\times {10}^{-10}$
• C. $1.25\times {10}^{-13}$
• D. $5\times {10}^{-13}$

Answer 1

The correct option is D $5\times {10}^{-13}$

An electrolyte $M{X}_2$ undergoes dissociation as follows :-

$M{X}_2\rightleftharpoons M^{+2}+2X^{-}$

Concentration	$M{X}_2$	$M^{+2}$	$X^{-}$
Initial concentration	1	0	0
Concentration at Equilibrium	1-s	s	2s

Thus from the above condition we can say that,

$K_{sp}=s\times (2s{)}^2=4\times (s{)}^3$

Here, s (the solubility ) is $0.5\times {10}^{-4}mole/lit.$

$\therefore$ $K_{sp}=4\times (0.5\times {10}^{-4}{)}^3$

$\therefore$ $K_{sp}=5\times {10}^{-13}$

Hence the correct answer is (D).