The solubility of AgCl(s) having solubility product `1.6×10^(-10)`, in `0.2 M` NaCl solution will be

Options

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

Answer

B

Explanation

`{:(AgCl(s),overset(+aq)hArr ,Ag^(+)(aq),+,Cl^(-)(aq)),(S,,S,,S+0.M"(from NaCl)"),(,,,,~=0.1M (" as " S lt lt 0.1)):}`
`K_(sp)=[Ag^(+)][Cl^(-)]`
`1.6xx10^(-10)=S(0.1) "or" S=1.6xx10^(-9)M`