Equilibrium - Result Question 93
####97. The solubility product of a sparingly soluble salt $B A_2$ is $4 \times 10^{-12}$. The solubility of $B A_2$ is
[1999]
(a) $4 \times 10^{-4}$
(b) $4 \times 10^{-12}$
(c) $4 \times 10^{-3}$
(d) $1 \times 10^{-4}$
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Solution:
(d) $B A_2 \to \underset{s}{B+2 A}$
Solubility product $=[s][2 s]^{2}=4 s^{3}$
$4 \times 10^{-12}=4 s^{3}$ or $s=\sqrt[3]{\frac{4^{\prime} 10^{-12}}{4}}$
$\therefore \quad s=10^{-4}$
