The pH of a solution of AgCI(s) with solubility product `1.6 xx 10^(-10)` in 0.1 M Nacl solution would be :

Options

A. `1.26xx10^(-5)M`
B. `1.6xx10^(-9)M`
C. `1.6xx10^(-11)M`
D. zero

Answer

B

Explanation

`{:(,underset((s))(AgCl),hArr,Ag^(+),+,Cl^(-)),("Initial",-,,0,0),(Eq^(m),-,,S^(1),S^(1)):}`
`{:(,NaCl,to,Na^(+),+,Cl^(-)),("Initial",C,,0,,0),("Final",0,,C,,C):}`
So Ionic product `Q=K_(sp)=[Ag^(+)][Cl^(-)]`
`" "K_(sp)=(S^(1))" "(C)`
`becalseCl^(-)` ion due to common ion obtained from NaCl
`S^(1)=(K_(sp))/(C)=(1.6xx10^(-10))/(0.1)=1.6xx10^(-9)`