question_answer

The solubility of \[AgCl(s)\] with solubility product \[1.6\times {{10}^{-10}}\]in 0.1 M \[NaCl\] solution would be

A)  \[1.26\times {{10}^{-5}}M\]    

B)  \[1.6\times {{10}^{-11}}M\]

C)  \[1.6\times {{10}^{-9}}M\]                  

D)  Zero

Correct Answer: B

Solution :

\[\therefore {{K}_{sp}}(AgCl)=S(S+0.1)\] \[\because S\ll 0.1\] \[\therefore S+0.1<<0.1\] \[\therefore 1.6\times {{10}^{-10}}=S=0.1\] \[\therefore S=1.6\times {{10}^{-9}}M\]