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Chemistry Question on Equilibrium

Given that, the solubility product KspK_{sp} of AgClAgCl is 1.8×10−101.8 \times 10^{-10}, the concentration of Cl−Cl^- ions that must be exceeded before AgClAgCl will precipitate from a solution containing 4×10−3 M Ag+4 \times 10^{-3}\, M \,Ag^+ ions is

A

4.5×10−8 M4.5 \times 10^{-8}\, M

B

4×10−8 M4 \times 10^{-8}\, M

C

1.8×10−8 M1.8 \times 10^{-8}\, M

D

1×10−8 M1 \times 10^{-8}\, M

Answer

4.5×10−8 M4.5 \times 10^{-8}\, M

Explanation

Solution

Ksp=[Ag+][Cl−]K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right] [Cl−]=1.8×10−104×10−3\left[Cl^{-}\right]=\frac{1.8\times10^{-10}}{4\times10^{-3}} =4.5×10−8 M=4.5\times10^{-8}\,M If concentration of Cl−Cl^{-} exceeds 4.5×10−8 M4.5\times10^{-8}\,M then ionic product becomes greater than KspK_{sp} and precipitation takes place.