The solubility product of `AgCl` is `1.8xx10^(-10)` at `18^(@)C`. The solubility of `AgCl` in `0.1 M` solution of sodium chloride would be

Options

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

Answer

B

Explanation

Key Idea - As solubility of AgCl(s) is asked in 0.1 M NaCl solution, so in the calculation, solublity of `Cl^(-)` (from NaCl) must be added to the solubility of `Cl^(-)` (from AgCl).
Let s be the solubility of `Ag^(+)` and `Cl^(-)` in AgCl before the addition of NaCl.
`{:(NaCl(aq)rarrNa^(+)(aq)+Cl^(-)(aq)),(" 0.1 M 0 0"),(" 0 0.1 M 0.1+s"),(" "AgCl(s)hArrAg^(+)(aq)+Cl^(-)(aq)),(" s s+0.1"):}`
Given, `K_(sp)=1.6xx10^(-10)=[Ag^(+)][Cl^(-)]`
or `1.6xx10^(-10)=s(0.1+s)=0.1 s+s^(2)`
`because K_(sp)` is small, so s is very less in comparison with 0.1. Hence, `s^(2)` can be neglected.
Thus, `1.6xx10^(-10)=0.1s`
or `s=1.6xx10^(-9)M`