Question

What is the percentage hydrolysis of $ NaCN $ in $ \dfrac{N}{80} $ solution when the dissociation constant for HCN in the solution is $ 1\cdot 3\times {{10}^{-9}} $ and $ {{K}_{w}} $ is $ 1\cdot 0\times {{10}^{-14}} $ ?
(A) $ 2\cdot 48% $
(B) $ 5\cdot 26% $
(C) $ 8\cdot 20% $
(D) $ 9\cdot 60% $

Answer 1

Hint: To find the percentage hydrolysis we need to find equilibrium hydrolysis constant i.e. ... Given in the question is dissociation constant for acid and dissociation constant for water i.e. $ {{K}_{w}} $ . $ {{K}_{h}} $ is given as the ratio of dissociation constant for water to the dissociation constant for acid. After finding $ {{K}_{h}} $ we will find the degree of hydrolysis which will give us the percentage hydrolysis for $ NaCN $ .

Formula Used: $ {{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{a}}} $ where $ {{K}_{h}} $ hydrolysis equilibrium constant
 $ {{K}_{w}} $ dissociation constant for water
 $ {{K}_{a}} $ dissociation constant for acid.

Complete Step By Step Solution
Given, $ {{K}_{a}} $ = $ 1\cdot 3\times {{10}^{-9}} $ and $ {{K}_{w}} $ = $ 1\cdot 0\times {{10}^{-14}} $
The reaction will be given as- $ NaCN+{{H}_{2}}O\rightleftharpoons NaOH+HCN $ where $ NaCN $ is salt $ NaOH $ is base and $ HCN $ is acid.
   $ {{K}_{h}} $ is given as- $ \dfrac{[HCN][O{{H}^{-}}]}{[C{{N}^{-}}]} $ and $ {{K}_{w}} $ is given as- $ [{{H}^{+}}][O{{H}^{-}}] $
Dissociation of HCN will be- $ HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}} $
So, dissociation constant for an acid will be, $ {{K}_{a}} $ $ =\dfrac{[{{H}^{+}}][C{{N}^{-}}]}{[HCN]} $
So, $ \dfrac{{{K}_{w}}}{{{K}_{a}}}=\dfrac{[{{H}^{+}}][O{{H}^{-}}][HCN]}{[{{H}^{+}}][C{{N}^{-}}]}=\dfrac{[O{{H}^{-}}][HCN]}{[C{{N}^{-}}]}={{K}_{h}} $
Therefore, $ {{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{a}}} $
 $ \begin{align}
  & \Rightarrow {{K}_{h}}=\dfrac{1\cdot 0\times {{10}^{-14}}}{1\cdot 3\times {{10}^{-9}}} \\
 & \Rightarrow 0\cdot 769\times {{10}^{-14+9}} \\
 & \Rightarrow 0\cdot 769\times {{10}^{-5}} \\
\end{align} $
Let ‘h’ be the degree of hydrolysis and ‘c’ be the concentration of given salt which is $ \dfrac{N}{80} $ .
So, concentration for $ O{{H}^{-}}\,and\,HCN=hc $
And $ [C{{N}^{-}}]=(1-h)c $
Therefore, we get, $ {{K}_{h}} $ = $ \dfrac{[HCN][O{{H}^{-}}]}{[C{{N}^{-}}]} $ $ =\dfrac{hc\times hc}{(1-h)c} $ (‘h’ is very small because HCN is a weak acid so degree of hydrolysis will be less for its salt so we can ignore h in denominator)
 $ \Rightarrow {{K}_{h}}=\dfrac{{{h}^{2}}{{c}^{2}}}{1\times c}={{h}^{2}}c $
 $ \Rightarrow h=\sqrt{\dfrac{{{K}_{h}}}{c}} $
 $ \Rightarrow \sqrt{\dfrac{0\cdot 769\times {{10}^{-5}}\times 80}{1}} $
 $ \Rightarrow \sqrt{61\cdot 52\times {{10}^{-5}}} $
 $ \Rightarrow \sqrt{615\cdot 2\times {{10}^{-6}}} $
 $ \Rightarrow 24\cdot 8\times {{10}^{-3}} $
 $ \Rightarrow 2\cdot 48% $
So, the correct choice is (A).

Note
Hydrolysis is defined as dissolving a salt of weak acid or weak base in water. It helps in breaking down proteins and fats. The higher the $ {{K}_{a}} $ value, the greater the no. of hydrogen ions liberated per mole of acid in the solution and hence stronger is the acid. Low values of $ {{K}_{a}} $ means that the acid does not dissociate well and that it is a weak acid. The more easily the acid dissociates, and the stronger it is i.e. the weaker the base it is. Oftentimes, the $ {{K}_{a}} $ value is expressed by using the $ p{{K}_{a}} $ . The larger the value of $ p{{K}_{a}} $ , the smaller the extent of dissociation i.e. ability to donate a proton in aqueous solution is less.