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A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :A. `10 xx 10^(-4)`B. `10 xx 10^(-10)`C. `10 xx 10^(10)`D. `1.0 xx 10^(14)` |
Answer» Correct Answer - C For the weak acid `HA`, `HA hArr H^(o+) + A^(Theta)` `K_(a) = ([H^(o+)]xx[A^(Theta)])/([HA])` Reaction of weak acid with strong base. `HA + OH^(Theta) hArr A^(Theta) + H_(2)O` `K = ([A^(Theta)])/([HA]xx[overset(Theta)OH])` Divident (1) by (2), we get `(K_(a))/(K) = [H^(o+)] [overset(Theta)OH] = K_(w)` or `K = (K_(a))/(K_(w)) = (10^(-14))/(10^(-14)) = 10^(10)` |
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