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Chemistry Question on Equilibrium

At 25∘^{\circ}C, the solubility product of Hg2Cl2_2Cl_2 in water is 3.2×1−17mol3dm−93.2 \times 1^{-17} mol^3 dm^{-9} What is the solubility of Hg2Cl2_2Cl_2 in water at 25∘^{\circ}C?

A

1.2×10−12M1.2 \times 10^{-12} M

B

3.0×10−6M3.0 \times 10^{-6} M

C

2×10−6M2 \times 10^{-6} M

D

1.2×10−16M1.2 \times 10^{-16} M

Answer

2×10−6M2 \times 10^{-6} M

Explanation

Solution

Hg2Cl2→Hg22++2Cl−Hg_2Cl_2 \rightarrow Hg_2^{2+} +2Cl^-
               Ksp=4S3=32×10−18\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, K_{sp} =4S^3 =32 \times 10^{-18}
                  S=2×10−6M\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, S=2 \times 10^{-6} M