Question

Solubility of $M{X}_2$ type electrolytes is $0.5\times 10^{-4}mol/L,$ then find out $K_{sp}$ of electrolytes:-

• A. $5\times 10^{-12}$
• B. $25\times 10^{-10}$
• C. $1\times 10^{-13}$
• D. $5\times 10^{-13}$

Answer 1

Answer: (D)

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.
$\therefore K_{sp}=4\times {\left(0.5\times 10^{-4}\right)}^3$
$\therefore K_{sp}=5\times 10^{-13}$