InterviewSolution
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InterviewSolution
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The relation between `K_(p)` and `K_(c)` is `K_(p)=K_(c)(RT)^(Deltan)` unit of `K_(p)=(atm)^(Deltan)`, unit of `K_(c)=(mol L^(-1))^(Deltan)` `H_(3)ClO_(4)` is a tribasic acid, it undergoes ionisation as `H_(3)ClO_(4) hArr H^(o+)+H_(2)ClO_(4)^(-), K_(1)` `H_(2)ClO_(4)^(-) hArr H^(o+)+HClO_(4)^(2-), K_(2)` `HClO_(4)^(2-) hArr H^(o+)+ClO_(4)^(3-), K_(3)` Then, equilibrium constant for the following reaction will be: `H_(3)ClO_(4) hArr 3H^(o+)+ClO_(4)^(3-)`A. `K_(1)K_(2)K_(3)`B. `((K_(1)K_(3))^(2))/(K_(2))`C. `K_(1)/K_(2)`D. `(K_(1)K_(2))/(K_(3)^(2))` |
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Answer» Correct Answer - A `H_(3)PO_(4) overset(K_(1))(rarr)H^(o+)+H_(2)PO_(4)^(ɵ)` `H_(2)PO_(4)^(ɵ) overset(K_(2))(rarr)H^(o+)+HPO_(4)^(2-)` `HPO_(4)^(2-)overset(K_(3))(rarr)H^(o+)+PO_(4)^(3-)` `:.` For `H_(3)PO_(4) overset(K)(rarr) 3H^(o+)+PO_(4)^(3-)` `K=K_(1)xxK_(2)xxK_(3)` |
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