For a sparingly soluble salt AB2, the equilibrium concentrations of A^2+ ions and B^- ions are 1.2 x 10^-4 M

Question

For a sparingly soluble salt \(AB_2,\) the equilibrium concentrations of \(A^{2+}\) ions and \(B^-\) ions are \(1.2 \times 10^{-4}\) M and \(0.24 \times 10^{-3}\) M, respectively. The solubility product of \(AB_2\) is : 

(1) \(0.069 \times 10^{-12 }\)

(2) \(6.91 \times 10^{-12 }\)

(3) \(0.276 \times 10 ^{-12}\)

(4) \(27.65 \times 10 ^{-12}\)

Answer 1

Correct option is (2) \(6.91 \times 10^{-12 }\)

\(AB_2 \rightarrow A^{2+}+ 2B^-\)

\(K_{sp} = [A^{+2}] [B^-]^2\)

\(=[1.2 \times 10 ^{-4}] [0.24 \times 10 ^{-3}] ^2\)

\(=0.0691 \times 10 ^{-10}\)

\(= 6.91 \times 10 ^{-12}\)