InterviewSolution
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InterviewSolution
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If the dissociation constants of two weak acids `HA_(1)` and `HA_(2)` are `K_(1)` and `K_(2)`, then the relative strengths of `HA_(1)` and `HA_(2)` are given byA. `(K_(1))/(K_(2))`B. `(K_(1)+sqrt(K_(2)))/(K_(2)+sqrt(K_(1)))`C. `((K_(1))/(K_(2)))^(1//2)`D. `((K_(1))/(K_(2)))^(3//2)` |
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Answer» Correct Answer - C The dissociation of a weak acid is related to its `K_(a)` value as `{:(,HA,hArr ,H^(+),+,A^(-)),("Initial conc. (M)",1.0,,0,,0),("Eqm. conc.",C(1-alpha),,Calpha,,Calpha):}` or `K_(a)=(C alpha.C alpha)/(C(1-alpha))=(C alpha^(2))/((1-alpha))` If `a lt lt 1` and `C=1` mol then `K_(a)=alpha^(2)` or `alpha=(K_(a))^(1//2)` Ratio of relative strengths of two acids depend on depends of dissociation i.e., `alpha` `:. (alpha_(1))/(alpha_(2))=((K_(1))/(K_(2)))^(1//2)` |
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