The solubility product of AgCI in water is`1.5xx10^(-10)`. Calculate its solubility in `0.01M NaCI`.

Explanation

Let the solubility of AgCI in water be S mol `L^(-1)`.
`AgCI(s) hArr underset(S)(Ag^(+)(aq)) + underset(S)(CI^(-)(aq))`
`NaCI overset((aq))(to) underset(0.01M)(Na^(+) (aq)) +underset(0.01M)(CI^(-) (aq))`
`"Now"" "[Ag^(+)]=S , [CI^(-)] =[CI^(-)] " from " AgCI + [CI^(-)] " from " NaCI`
`=S + 0.01 ~~ 0.01 ("as" S lt lt 0.01)`
`"Now"" "K_(sp) =[Ag^(+)][CI^(-)] " or " 1.5 xx 10^(-10) =S xx 0.01`
`S=(1.5 xx 10^(-10))/(0.01) =1.5 xx 10^(-8) "mol "L^(-1)`