Question-15 The solubility product of PbI_2 is 1.08\times10^{-7}. Calculate its solubility in mol\ dm^{-3} at 298\ K.
(A) 2.018\times10^{-3}
(B) 2.011\times10^{-9}
(C) 1.259\times10^{-9}
(D) 3.0\times10^{-3}
Answer : (D)
Explanation:
For PbI₂,
PbI_2(s) \rightleftharpoons Pb^{2+}{(aq)} + 2I^-{(aq)}∴ x = 1, y = 2
K_{sp} = x^x y^y S^{x+y} = (1)^1 (2)^2 S^{1+2} = 4S^3∴ S = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{1.08 \times 10^{-7}}{4}} = \sqrt[3]{\frac{108 \times 10^{-9}}{4}}
= \sqrt[3]{27 \times 10^{-9}} = 3 \times 10^{-3} mol dm^{-3}