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

`Ag_2CO_3 hArr 2Ag^(+) + CO_3^(2-)`
`K_(sp) (Ag_2CO_3) =[Ag^+]^2 [CO_3^(2-)]`
1.5 M `Na_2CO_3` gives , `[CO_3^(2-)]` =1.5 M
Concentration of `Ag^+` ion in solution ,
`[Ag^+]=(((K_(sp) )(Ag_2CO_3))/([CO_3^(2-)]))^(1/2)`
`=((8.2xx10^(-12))/1.5)^(1/2)=2.34xx10^(-6)` M
`[CI^-]=(0.0026 gL^(-1))/(35.5 g mol^(-1))=7.32xx10^(-5)` M
`K_(sp)=[Ag^+][Cl^-]`
`=(2.34xx10^(-6))(7.32xx10^(-5))`
`=1.71xx10^(-10)`