Equilibrium - Result Question 87
####91. The solubility product of a sparingly soluble salt $A X_2$ is $3.2 \times 10^{-11}$. Its solubility ( in moles/ litre) is
(a) $5.6 \times 10^{-6}$
(b) $3.1 \times 10^{-4}$
(c) $2 \times 10^{-4}$
(d) $4 \times 10^{-4}$
[2004]
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Solution:
- (c) For $A X_2 ; K _{s p}=4 s^{3} \therefore 3.2 \times 10^{-11}=4 s^{3}$
or $s=3 \sqrt{\frac{3.2 \times 10^{-11}}{4}}=2 \times 10^{-4}$
