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Given
$(i)$ $\begin{gathered}
HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
{K_a} = 6.2 \times {10^{ - 10}} \hfill \\
\end{gathered} $
$(ii)$ $\begin{gathered}
C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
{K_b} = 1.6 \times {10^{ - 5}} \hfill \\
\end{gathered} $
These equilibria show the following order of the relative base strength
$O{H^ - } > {H_2}O > C{N^ - }$
$O{H^ - } > C{N^ - } > {H_2}O$
${H_2}O > C{N^ - } > O{H^ - }$
$C{N^ - } > {H_2}O > O{H^ - }$
Solution
The more is the value of equilibrium constant, the more is the completion of reaction or more is the concentration of products i.e . the order of relative strength would be
$O{H^ – } > C{N^ – } > {H_2}O$
