The solubility of AgCl(s) with solubility product `1.6 xx 10^(-10)` in 0.1 M NaCl solution would be

Options

A. `1.6xx10^(-11)` M
B. 0
C. `1.26xx10^(-5)` M
D. `1.6xx10^(-9)` M

Answer

D

Explanation

`{:(NaCl to , Na^(+) + , Cl^-),(0.1M, 0,0),(0,0.1 M, 0.1 +S):}`
`{:(AgCl hArr , Ag^(+)+, Cl^-),(0.1 M, 0,0),(x-S,S,S+0.1):}`
`K_(sp)=1.6xx10^(-10) = [Ag^+][Ci]=[S]` [S+0.1]
`because K_(sp)` is small , S is neglected with respect to 0.1 M
`1.6xx10^(-10) =Sxx0.1`
`S=1.6xx10^(-9)` M