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