Given:(i) HCN_{(aq)} + H_2O_{(l)} rightlefth | vedclass

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Question

(B) The strength of a base is determined by its ability to accept a proton $(H^{+})$.In reaction $(ii)$,$CN^{-}$ reacts with $H_2O$ to form $OH^{-}$.

Vedclass · Accepted Answer

$OH^{-} > CN^{-} > H_2O$

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(B) The strength of a base is determined by its ability to accept a proton $(H^{+})$.In reaction $(ii)$,$CN^{-}$ reacts with $H_2O$ to form $OH^{-}$. The equilibrium constant $K_b = 1.6 	imes 10^{-5}$ indicates the extent of this reaction.Since $OH^{-}$ is the product of the base dissociation of $CN^{-}$,and $K_b$ is relatively large,$OH^{-}$ is a stronger base than $CN^{-}$.Comparing the conjugate bases: $OH^{-}$ is the conjugate base of $H_2O$,and $CN^{-}$ is the conjugate base of $HCN$.Since $HCN$ is a weak acid $(K_a = 6.2 	imes 10^{-10})$ and $H_2O$ is an extremely weak acid,the conjugate base of the weaker acid is stronger.Thus,the order of relative base strength is $OH^{-} > CN^{-} > H_2O$.

Given:
$(i) \ HCN_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + CN^{-}_{(aq)}$
$K_a = 6.2 \times 10^{-10}$
$(ii) \ CN^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons HCN_{(aq)} + OH^{-}_{(aq)}$
$K_b = 1.6 \times 10^{-5}$
These equilibria show the following order of the relative base strength:

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Given:$(i) \ HCN_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + CN^{-}_{(aq)}$$K_a = 6.2 \times 10^{-10}$$(ii) \ CN^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons HCN_{(aq)} + OH^{-}_{(aq)}$$K_b = 1.6 \times 10^{-5}$These equilibria show the following order of the relative base strength: