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

Answer

B

Explanation

Let S be the solubility of AgCI in moles per litre.
`underset(S)(AgCI (aq)) hArr underset(S)(Ag^(+))(aq) + underset((S+0.1))(CI^(-)(aq))`
(0.1 NaCI solution also provides 0.1 M `CI^(-)` ions)
`Ks_(p) =[Ag^(+)] [CI^(-)]`
`1.6 xx 10^(-10) =S [S +0.1] = S(0.1)`
`("Because " S lt lt lt 0.1)`
`S=(1.6xx 10^(-10))/(0.1) =1.6 xx 10^(9) M`