The solubility product of AgCl in water is `1.5xx10^(-10)` .Calculate its solubility in 0.01 M NaCl aqueous solution.

Explanation

As NaCl dissociates completely, therefore, in 0.01 M NaCl solution, `[Cl^(-)] = 0.01M`
If solubility of AgCl in 0.01 M NaCl solution is s mol `L^(-1)`, then from AgCl that dissolves,
`[Ag^(+)]=[Cl^(-)]=s ` mol `L^(-1) :. ` Total `[Cl^(-)]=0.01 + s ~= 0.01M`
`K_(sp) ` for AgCl `=[Ag^(+)] [ Cl^(-)] = s xx 0.01 = 0.001 s :. 0.01 s = 1.5 xx 10^(-5) xx 10^(-10) or s= 1.5 xx 10^(-8) M`