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))`

Options

A. `1.1 xx 10^(-2)`
B. `1.71 xx 10^(-11)`
C. `1.71 xx 10^(-10)`
D. `1.32 xx 10^(-9)`

Answer

C

Explanation

`[Ag^(+)] = sqrt((K_(sp))/([CO_(3)^(2-)])) = 2.34 xx 10^(-6)m`
`K_(spAgCl) = [Ag^(+)][Al^(-)] = 1.71 xx 10^(-10)`