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

`1.5 M Na_(2)CO_(3) ` gives `[CO_(3)^(2-)]=1.5M`
`K_(sp)` for `Ag_(2)CO_(3)=[Ag^(+)]^(2)[CO_(3)^(2-)]`
`:. [Ag^(+)]=sqrt((K_(sp) "for" Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2xx10^(-12))/(1.5))=2.34xx106(-6)M`
`K_(sp) "for" AgCl=[Ag^(+)][Cl^(-)]=(2.34xx10^(-6))((0.0026)/(35.5))=1.71xx10^(-10)`