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Calculate the $pH$ of $0.10M$ solution of ${{NaCN}}$(aq). ${K_a}$ for ${{HCN}}$ is $4.9\times{10^{ - 10}}$ at ${{2}}{{{5}}^{{0}}}{{C}}$.
(A) $11.15$
(B) $2.85$
(C) $8.75$
(D) $7$

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Last updated date: 22nd Feb 2024
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Hint: Hydrolysis reactions are inverse condensation reactions in which two molecules join together into larger one and eject a water molecule. In hydrolysis reactions water molecules are added to the compound to break it down into ions.
Salt given in the question is a salt of weak acid and strong base so we will apply the formula for calculating the pH for hydrolysis of salt of weak acid and strong base.

Formulas used: $pH = 7 + \dfrac{1}{2}[p{K_a} + \log c]$
Where $p{K_a}$ denotes the negative logarithm of the dissociation constant (${K_a}$) and $c$ is the concentration

Complete step by step answer:
The hydrolysis of salt of weak acid and strong base gives alkaline solutions due to the presence of free ${{O}}{{{H}}^{{ - }}}$ ions. Here the salt given in the question is a salt of weak acid and strong base. Since salt is a strong electrolyte it is completely ionised into ions. The dissociation of ${{NaCN}}$ is given as follows:

$NaCN + {H_2}O \to N{a^ + } + C{N^ - }$

Since cation is from a strong base it remains the same in solution. But the anion ${{C}}{{{N}}^{{ - }}}$reacts with water to form ${{HCN}}$ . This process is known as hydrolysis and is represented by the equation:

$C{N^ - } + {H_2}O \rightleftharpoons HCN + O{H^ - }$ (aq)

This is also called as anion hydrolysis. The ${{pH}}$ of the hydrolysed salt solution is:

$pH = 7 + \dfrac{1}{2}[p{K_a} + \log c]$ ---------------------(1)

Where $p{K_a}$ denotes the negative logarithm of the dissociation constant (${K_a}$) and $c$ is the concentration.
Let us calculate the values of $p{K_a} = - \log {K_a} = - \log 4.9\times{10^{ - 10}} = 9.309$

Now let us find the value of $\log c$

$\log c = \log 0.1 = - 1$. Putting the values in equation 1 we get the value of ${{pH}}$ :

$pH = 7 + \dfrac{1}{2}[9.309 - 1] = 11.15$ .

So the $pH$ of solution of $0.10M$ ${{NaCN}}$ is $11.15$ .

So, the correct answer is Option A.

Note: The salts of strong acids and strong bases do not undergo hydrolysis and the resulting solution is neutral. The solution of salt of weak acid and strong base is alkaline and the solution of salt of weak base and strong acid is acidic. Thus, the $pH$ of a solution is dependent upon the relative strengths of the acid and base from which it was produced.
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