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Chemistry Question on Solutions

The solubility product of AgCl is 4.0×10−104.0 \times 10^{-10} at 298 K. The solubility of AgCl in 0.04 M CaCl2CaCl_2 will be

A

2.0×10−5M2.0 \times 10^{-5}M

B

1.0×10−4M1.0 \times 10^{-4}M

C

5.0×10−9M5.0 \times 10^{-9}M

D

2.2×10−4M2.2 \times 10^{-4}M

Answer

5.0×10−9M5.0 \times 10^{-9}M

Explanation

Solution

In CaCl2⋅[Cl−]=2×0.04=0.08 mol L−1CaCl_{2}\cdot [Cl^{-}]=2 \times 0.04=0.08\,mol\,L^{-1} [Cl−][Cl^{-}] from AgClAgCl is too small and is neglected Ksp=[Ag+][Cl−]K_{sp}=[Ag^{+}] [Cl^{-}] 4×10−10=[Ag+]×0.084 \times 10^{-10}=[Ag^{+}]\times 0.08 4×10−10=[Ag+]×0.084 \times 10^{-10}=[Ag^{+}]\times 0.08 [Ag+]=4×10−100.08=48×10−8\left[Ag^{+}\right]=\frac{4\times10^{-10}}{0.08}=\frac{4}{8}\times10^{-8} =5.0×10−9 M=5.0 \times 10^{-9}\,M