A sample of AgCI was treated with `5.00mL` of `1.5M` `Na_(2)CO_(3)` solubility to give `Ag_(2)CO_(3)` . The remaining solution contained `0.0026g of CI^(-)` per litre. Calculate the solubility product of AgCI. `(K_(SP)for Ag_(2)CO_(3)=8.2xx10^(-12))`

Explanation

(I). Calculation of `Ag^(+)` ion concentration
`Ag_(2)CO_(3)(s) overset((aq))(hArr) 2Ag^(+) (aq) +CO_(3)^(2-) (aq)`
`K_(sp) =[Ag^(+) (aq)]^(2) [CO_(3)^(2-)(aq)]`
`K_(sp) =8.2 xx 10^(-12), [CO_(3)^(2-)] =1.5 M" "[:. Na_(2)CO_(3) " is " 1.5 M]`
`[Ag^(+) (aq)] =[(K_(sp))/(CO_(3)^(2-) (aq))]^(1//2) =((8.2xx10^(-12))/(1.5))^(1//2) =2.34 xx 10^(-6)M`
(II). Calculation of `K_(sp)` of AgCI
`[Ag^(+) (aq)] =2.34l xx 10^(-6) M`
`[CI^(-) (aq)] =((0.0026 gL^(-1)))/((35.5 g " mol"^(-2)))=7.32 xx 10^(-5) M`
`K_(sp) =[Ag^(+) (aq)] xx[CI^(-)(aq)]`
`=(2.34 xx 10^(-6) M) xx (7.32 xx 10^(-5)M) =1.71 xx 10^(-10)`