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Question: What is the percentage hydrolysis of \( NaCN \) in \( \dfrac{N}{80} \) solution when the dissociatio...

What is the percentage hydrolysis of NaCNNaCN in N80\dfrac{N}{80} solution when the dissociation constant for HCN in the solution is 1⋅3×10−91\cdot 3\times {{10}^{-9}} and Kw{{K}_{w}} is 1⋅0×10−141\cdot 0\times {{10}^{-14}} ?
(A) 2⋅482\cdot 48%
(B) 5⋅265\cdot 26%
(C) 8⋅208\cdot 20%
(D) 9⋅609\cdot 60%

Explanation

Solution

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. Kw{{K}_{w}} . Kh{{K}_{h}} is given as the ratio of dissociation constant for water to the dissociation constant for acid. After finding Kh{{K}_{h}} we will find the degree of hydrolysis which will give us the percentage hydrolysis for NaCNNaCN .

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

Complete Step By Step Solution
Given, Ka{{K}_{a}} = 1⋅3×10−91\cdot 3\times {{10}^{-9}} and Kw{{K}_{w}} = 1⋅0×10−141\cdot 0\times {{10}^{-14}}
The reaction will be given as- NaCN+H2O⇌NaOH+HCNNaCN+{{H}_{2}}O\rightleftharpoons NaOH+HCN where NaCNNaCN is salt NaOHNaOH is base and HCNHCN is acid.
Kh{{K}_{h}} is given as- [HCN][OH−][CN−]\dfrac{[HCN][O{{H}^{-}}]}{[C{{N}^{-}}]} and Kw{{K}_{w}} is given as- [H+][OH−][{{H}^{+}}][O{{H}^{-}}]
Dissociation of HCN will be- HCN⇌H++CN−HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}}
So, dissociation constant for an acid will be, Ka{{K}_{a}} =[H+][CN−][HCN]=\dfrac{[{{H}^{+}}][C{{N}^{-}}]}{[HCN]}
So, KwKa=[H+][OH−][HCN][H+][CN−]=[OH−][HCN][CN−]=Kh\dfrac{{{K}_{w}}}{{{K}_{a}}}=\dfrac{[{{H}^{+}}][O{{H}^{-}}][HCN]}{[{{H}^{+}}][C{{N}^{-}}]}=\dfrac{[O{{H}^{-}}][HCN]}{[C{{N}^{-}}]}={{K}_{h}}
Therefore, Kh=KwKa{{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}
⇒Kh=1⋅0×10−141⋅3×10−9 ⇒0⋅769×10−14+9 ⇒0⋅769×10−5 \begin{aligned} & \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{aligned}
Let ‘h’ be the degree of hydrolysis and ‘c’ be the concentration of given salt which is N80\dfrac{N}{80} .
So, concentration for OH− and HCN=hcO{{H}^{-}}\,and\,HCN=hc
And [CN−]=(1−h)c[C{{N}^{-}}]=(1-h)c
Therefore, we get, Kh{{K}_{h}} = [HCN][OH−][CN−]\dfrac{[HCN][O{{H}^{-}}]}{[C{{N}^{-}}]} =hc×hc(1−h)c=\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)
⇒Kh=h2c21×c=h2c\Rightarrow {{K}_{h}}=\dfrac{{{h}^{2}}{{c}^{2}}}{1\times c}={{h}^{2}}c
⇒h=Khc\Rightarrow h=\sqrt{\dfrac{{{K}_{h}}}{c}}
⇒0⋅769×10−5×801\Rightarrow \sqrt{\dfrac{0\cdot 769\times {{10}^{-5}}\times 80}{1}}
⇒61⋅52×10−5\Rightarrow \sqrt{61\cdot 52\times {{10}^{-5}}}
⇒615⋅2×10−6\Rightarrow \sqrt{615\cdot 2\times {{10}^{-6}}}
⇒24⋅8×10−3\Rightarrow 24\cdot 8\times {{10}^{-3}}
⇒2⋅48\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 Ka{{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 Ka{{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 Ka{{K}_{a}} value is expressed by using the pKap{{K}_{a}} . The larger the value of pKap{{K}_{a}} , the smaller the extent of dissociation i.e. ability to donate a proton in aqueous solution is less.