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

Initial `[Na_(2)CO_(3)]=[CO_(3)^(2-)]=1.5 M`
Equilibrium `[Cl^(-)]=[NaCl]=(0.0026)/(35.5)=0.0000734M`
`2AgCl(s)+underset(1.5M)overset(1.5M)(Na_(2)CO_(3)) Leftrightarrow Ag_(2)(CO)_(3) (s)+underset(0.0000732M"-eqb. Concn.")overset("0-initial concn.")(2NaCl)`
`[Ag^(+)]=sqrt(K_(sp)(Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2 xx 10^(-12))/(1.5))=2.338 xx 10^(-6)M`
`K_(sp)=[Ag^(+)] [Cl^(-)]=(2.338 xx 10^(-6)) (0.0000732) =1.71 xx 10^(-10)`.