Equilibrium - Result Question 89
####93. Solubility of $M X_2$-type eletrolytes is $0.5 \times 10^{-4}$ $mol /$ lit, then find out $K _{s p}$ of electrolytes [2002]
(a) $5 \times 10^{-12}$
(b) $25 \times 10^{-10}$
(c) $1 \times 10^{-13}$
(d) $5 \times 10^{-13}$
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Solution:
- (d) Given $s=0.5 \times 10^{-4} moles / L$
$[M X_2 \rightarrow M^{2+}+2 X]$
$\because$ For $M X_2, K _{s p}=s \times(2 s)^{2}=4 s^{3}$
$K _{s p}=4 \times(0.5 \times 10^{-4})^{3}=4 \times 0.125 \times 10^{-12}$
$=0.5 \times 10^{-12}=5 \times 10^{-13}$
