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6-2.Equilibrium-II (Ionic Equilibrium)
medium
Solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4}\,mol/litre$ , then $K_{sp}$ of electrolyte is
A
$5\times 10^{-12}$
B
$2.5\times 10^{-10}$
C
$1\times 10^{-13}$
D
$5\times 10^{-13}$
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 $0.5 \times 10^{-4}\, mole/lit.$
$\therefore K_{\text {sp }}=4 \times\left(0.5 \times 10^{-4}\right)^3$
$\therefore K _{ sp }=5 \times 10^{-13}$
Standard 11
Chemistry
