Q. Solubility of $ MX_2 $ type electrolytes is $ 0.5 \times 10^{-4} \,mol/L , $ then find out $ K_{sp} $ of electrolytes:-
AIPMTAIPMT 2002Equilibrium
Solution:
An electrolyte $MX _{2}$ undergoes dissociation as follows :-
$MX _{2} \rightleftharpoons M ^{+2}+2 X ^{-}$
Concentration
$MX _{2}$
$M ^{+2}$
$X^-$
Initial concentration
1
0
0
Concentration at Equilibrium
1-s
s
2s
Thus from the above condition we can say that,
$K _{ sp }= s \times(2 s )^{2}=4 \times( s )^{3}$
Here, $s$ (the solubility) is $0.5 \times 10^{-4}$ mole / lit.
$\therefore K_{s p}=4 \times\left(0.5 \times 10^{-4}\right)^{3}$
$\therefore K _{ sp }=5 \times 10^{-13}$
| Concentration | $MX _{2}$ | $M ^{+2}$ | $X^-$ |
| Initial concentration | 1 | 0 | 0 |
| Concentration at Equilibrium | 1-s | s | 2s |