Given (i) $\mathrm{HCN}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ightleftharpoons \mathrm{H}_{3}…

Given
(i) $\mathrm{HCN}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{CN}^{-}(\mathrm{aq})$
$K_{\mathrm{a}}=6.2 \times 10^{-10}$
(ii) $\mathrm{CN}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ightleftharpoons \mathrm{HCN}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$
$K_{\mathrm{b}}=1.6 \times 10^{-5}$
These equilibria show the following order of the relative base strength,
  1. $\mathrm{OH}^{-}>\mathrm{H}_{2} \mathrm{O}>\mathrm{CN}^{-}$
  2. $\mathrm{OH}^{-}>\mathrm{CN}^{-}>\mathrm{H}_{2} \mathrm{O}$
  3. $\mathrm{H}_{2} \mathrm{O}>\mathrm{CN}^{-}>\mathrm{OH}^{-}$
  4. $\mathrm{CN}^{-}>\mathrm{H}_{2} \mathrm{O}>\mathrm{OH}^{-}$

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
$$
\mathrm{OH}^{-}>\mathrm{CN}^{-}>\mathrm{H}_{2} \mathrm{O}
$$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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