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. `1.6xx10^(-11)` M
B. Zero
C. `1.26xx10^(-7)` M
D. `1.6xx10^(-8)` M

Answer

D

Explanation

Follow the steps
`" "NaCl(aq) harr Na^(+)" "Cl^(-)`
`{:("Initially", " "0.1 M, " "0 ," "0), ("At equilibrium"," " 0, " "0.1 ," "0.1+S):}`
`" "AgCl harr Ag^(+)+Cl^(-)`
`{:("Initially", " "a, " "0 ," "0), ("At equilibrium"," " 0, " "S ," "0.1+S):}`
`K_(sp)=1.6xx10^(-10)=[Ag^(+)][Cl^(-)]`
= S(0.1+S)
`1.6xx10^(-10)=S xx 0.1`
`S=1.6xx10^(-9)` M