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Chemistry Question on Solubility Equilibria Of Sparingly Soluble Salts

The concentration of Ag+A{{g}^{+}} ion in a given saturated solution of AgClAgCl at 25∘C25{}^\circ C is 1.06×10−5g−1.06\times {{10}^{-5}}g- ion per litre. Thus, the solubility product of AgClAgCl is

A

0.353×10−100.353\times {{10}^{-10}}

B

0.530×10−100.530\times {{10}^{-10}}

C

1.12×10−101.12\times {{10}^{-10}}

D

2.12×10−102.12\times {{10}^{-10}}

Answer

1.12×10−101.12\times {{10}^{-10}}

Explanation

Solution

[Ag+]=1.06×10−5 g−ion/L[Ag^+] = 1.06 \times 10^{-5} \,g - ion/L =1.06×10−5 mol/L=[Cl−] = 1.06 \times 10^{-5} \,mol/L = [Cl^-] AgCl(s)<=>Ag++Cl−AgCl(s) {<=>} Ag^+ + Cl^- Ksp=[Ag+][Cl−]K_{sp} = [Ag^+][Cl^-] =(1.06×10−5)2= (1.06 \times 10^{-5})^2 =1.12×10−10 = 1.12 \times 10^{-10}