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The molar solubility of silver sulphate is `1.5xx10^(-2)mol L^(-1)`. The solubility product of the salt will beA. `2.25xx10^(-4)`B. `1.4xx10^(-5)`C. `1.7xx10^(-6)`D. `3.0xx10^(-3)` |
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Answer» Correct Answer - B Solubility equilibrium equation: `Ag_(2)SO_(4)(s)hArr2Ag^(+)(aq.)+SO_(4)^(2-)(aq.)` 1 mol of `Ag_(2)SO_(4)` produces 2 mol of `Ag^(+)` and 1 mol of `SO_(4)^(2-)` in solution. Therefore, when `1.5xx10^(-2)` mol `Ag_(2)SO_(4)` is dissolved in 1 L of solution, the concentrations are `C_(Ag^(+))= 2(1.5xx10^(-2)M)` `= 3.0xx10^(-2)M` `C_(SO_(4)^(2-))= 1.5xx10^(-2)M` Solubility product: `K_(sp)=C_(Ag^(+))^(2)C_(SO_(4)^(2-))` `= (3.0xx10^(-2))^(2)(1.5xx10^(-2))` `= 1.4xx10^(-5)` |
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