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At `20^(@)C`, the `Ag^(+)` ion concentration in a saturated solution `Ag_(2)CrO_(4)` is `1.5 x10^(-4)` mol `//` litre. At `20^(@)C`, the solubility product of `Ag_(2)CrO_(4)` would beA. `3.3750 xx 10^(-12)`B. `1.6875 xx 10^(-10)`C. `1.6875 xx 10^(-12)`D. `1.6875 xx 10^(-11)` |
Answer» Correct Answer - C `Ag_(2)CrO_(4) hArr 2Ag^(+)+CrO_(4)^(-2)` Given `[2Ag^(+)]=1.5 xx 10^(-4) mol L^(-1)` Thus, `[Ag^(+)]=0.75xx 10^(-4) mol L^(-1)` Thus `[CrO_(4)^(-2)]=0.75 xx 10^(-4) mol L^(-1)` `K_(sp)=[Ag^(+)]^(2) [CrO_(4)^(-2)]^(2)` `=(1.5 xx 10^(-4))^(2)(0.75 xx 10^(-4))` `=1.6875xx 10^(-12)mol^(3)L^(-3)` |
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