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 xx10^(-9) M `
D. `1.6 xx 10^(-11)`M

Answer

C

Explanation

`{:("Ag"Cl,hArr,"Ag"^(+),+,Cl^(-)),(,,S,,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 `