InterviewSolution
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InterviewSolution
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A sample of AgCl was treated with 5.00 ml of 1.5 M `Na_(2)CO_(3)` solution to give `Ag_(2)CO_(3)`. The remaining solution contained 0.0026 g of `Cl^(-)` per litre. Calculate the solubility product of AgCl `(K_(sp) "for" Ag_(2)CO_(3)=8.2xx10^(-2))` |
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Answer» `1.5 M Na_(2)CO_(3) ` gives `[CO_(3)^(2-)]=1.5M` `K_(sp)` for `Ag_(2)CO_(3)=[Ag^(+)]^(2)[CO_(3)^(2-)]` `:. [Ag^(+)]=sqrt((K_(sp) "for" Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2xx10^(-12))/(1.5))=2.34xx106(-6)M` `K_(sp) "for" AgCl=[Ag^(+)][Cl^(-)]=(2.34xx10^(-6))((0.0026)/(35.5))=1.71xx10^(-10)` |
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