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What is the pH of a \[1.0 \times {10^{ - 2}}\] molar solution of HCN? (for HCN, ${K_a} = 4.0 \times {10^{ - 10}}$)
A.$10$
B.Between $7$ and $10$
C.$7$
D.Between $4$ and $7$

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Last updated date: 13th Jun 2024
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Answer
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Hint:HCN is a weak acid. Hence it will not dissociate completely. In order to find pH we need to calculate ${H^ + }$ ion concentration. For that first we need to calculate the degree of dissociation of HCN.

Complete step by step answer:HCN dissociates as,
$HCN \rightleftharpoons {H^ + } + C{N^ - }$
Since HCN is a weak acid, it will not dissociate completely in equilibrium. A considerable amount of HCN , ${H^ + }$ and $C{N^ - }$ will be present in equilibrium. HCN is the un-dissociated form and ${H^ + }$ and $C{N^ - }$ are ionised forms. For a strong acid, the concentration of the un-dissociated acid will be negligible at equilibrium.
Let c be the initial concentration of HCN and $\alpha $ be the degree of dissociation of HCN. Then at equilibrium,
Concentration of ${H^ + }$, $[{H^ + }]$ = $c\alpha $
For a weak acid, the acidity constant, ${K_a} = c{\alpha ^2}$
${K_a}$ and $c$ are given in the question.
${K_a} = 4.0 \times {10^{ - 10}}$
$c = 1.0 \times {10^{ - 2}}M$
Now, we can find out the value of degree of dissociation, $\alpha $ using the equation,
$\alpha = \sqrt[{}]{{\dfrac{{{K_a}}}{c}}}$
Substituting the values,
$\alpha = \sqrt {\dfrac{{4.0 \times {{10}^{ - 10}}}}{{1.0 \times {{10}^{ - 2}}}}} = \sqrt {4 \times {{10}^{ - 8}}} = 2 \times {10^{ - 4}}$
Using this we can find out the concentration of ${H^ + }$ , $[{H^ + }]$.
$[{H^ + }]$ = $c\alpha $
Substituting the values,
$[{H^ + }] = 1.0 \times {10^{ - 2}} \times 2 \times {10^{ - 4}} = 2 \times {10^{ - 6}}M$
pH can be find out using the formula,
$pH = - \log [{H^ + }]$
Substituting the value of $[{H^ + }]$, we get,
$pH = - \log [2 \times {10^{ - 6}}] = 5.699$
i.e. pH of a \[1.0 \times {10^{ - 2}}\] molar solution of HCN is $5.699$ . This value is between $4$ and $7$.
Therefore, the correct option is D.

Note:
There is a simple way to solve this problem.
We have,
$[{H^ + }]$ = $c\alpha $ and ${K_a} = c{\alpha ^2}$.
Multiply the equation, ${K_a} = c{\alpha ^2}$ with c , we get,
${K_a}c = {c^2}{\alpha ^2}$
$c\alpha = \sqrt {{K_a}c} $
So we get,
$[{H^ + }] = \sqrt {{K_a}c} $
In this way, we can find out the concentration of ${H^ + }$ ion directly. Then there is no need to find the degree of dissociation $\alpha $. The difference between two methods is, if we know the concept we don’t need to remember the equations in the first method. In the second method, we need to remember the equations. Anyway, one can easily derive the second formula from the first method.