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. Zero
B. `1.26 xx 10^(-5)M`
C. `1.6 xx 10^(-9) M`
D. `1.6 xx 10^(-11) M`

Answer

C

Explanation

`AgCl hArr Ag^(+) + Cl^(-)`
`S hArr S + 0.1`
`K_(sp) = [Ag^(+)][Cl^(-)]`
`1.6 xx 10^(-10) = S xx (S + 0.1) = S xx 0.1`
`1.6 xx 10^(-9) = S`