Question

Given
$\left(i\right) HCN\left(aq\right) + H_{2}O\left( l\right) {\rightleftharpoons} H_{3}O^{+}\left(aq\right) + CN^{-} \left(aq\right)K_{a} =6.2 \times 10^{-10}$
$\left(ii\right) CN \left(aq\right) + H_{2}O\left(l\right) {\rightleftharpoons} HCN \left(aq\right) + OH^{-} \left(aq\right) K_{b}= 1.6 \times 10^{-5}.$
These equilibria show the following order of the relative base strength,

• A. $OH^{-} > H_{2}O > CN^{-}$
• B. $OH^{-} > CN^{-} > H_{2}O$
• C. $ H_{2}O > CN^{-} > OH^{-}$
• D. $ CN^{-} > H_{2}O > OH^{-}$

Answer 1

Answer: (B)

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
$OH^{-} > CN^{-} > H_{2}O$