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Chemistry Question on Solubility Equilibria Of Sparingly Soluble Salts

The solubility of AgClAgCl is 1×10−5 mol/L1 \times 10^{-5}\, mol/L . Its solubility in 0.10.1 molar sodium chloride solution is

A

1×10−101 \times 10^{-10}

B

1×10−51 \times 10^{-5}

C

1×10−91 \times 10^{-9}

D

1×10−41 \times 10^{-4}

Answer

1×10−91 \times 10^{-9}

Explanation

Solution

KspK_{s p} of AgCl=( solubility of AgCl)2AgCl =(\text { solubility of } AgCl )^{2}

=(1×10−5)2=1×10−10=\left(1 \times 10^{-5}\right)^{2}=1 \times 10^{-10}

Suppose its solubility in 0.1MNaCl0.1 M NaCl is xmol/Lx mol / L

AgCl⇌Ag+x+Cl−xAgCl \rightleftharpoons \underset{x}{Ag ^{+}}+ \underset{x}{Cl ^{-}}

NaCl⇌Na+0.1M+Cl−0.1MNaCl \rightleftharpoons \underset{0.1M}{Na ^{+}}+\underset{0.1 M }{ Cl ^{-}}

[Cl−]=(x+0.1)M\left[ Cl ^{-}\right]=(x +0.1) M

KspK_{s p} of AgCl=[Ag+][Cl−]AgCl =\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]

=x×(x+0.1)=x \times(x+0.1)

1×10−10=x2+0.1x1 \times 10^{-10}=x^{2}+0.1 x

Higher power of xx are neglected

1×10−10=0.1x1 \times 10^{-10} =0.1 x

x=1×10−9Mx =1 \times 10^{-9} M