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))`

Answer

A

Explanation

`Ag_(2)CO_(3) hArr 2Ag^(o+) +CO_(3)^(2-)`
`K_(sp) = [Ag^(o+)] [CO_(3)^(2-)]`
or `[Ag^(o+)]^(2) = (K_(sp))/([CO_(3)^(2-)]`
or `[Ag^(o+)] = sqrt((K_(sp))/([CO_(3)^(2-)]))`
`[C^(Theta)] = 0.0026 g L^(-1) = (0.0026)/(35.5) mol L^(-1)`
`AgCI hArr Ag^(o+) + CI^(Theta)`
`K_(sp)(AgCI)=sqrt((K_(sp))/([CO_(3)^(2-)]))[CI^(Theta)]`
`=sqrt((8.2xx10^(-12))/(1.5))xx(0.0026)/(35.5) = 1.71 xx 10^(-10)`