Question

The \[p{K_a}\] of $HCN$is $9.30$. The$pH$of a solution by mixing $2.5\,moles$ of $KCN$and $2.5\,moles$ of $HCN$ in water and making up the total volume of $500\,mL$ is:
(i) $9.30$
(ii) $7.30$
(iii) $10.30$
(iv) $8.30$

Answer 1

Hint:As seen from the \[p{K_a}\] value of $HCN$ it can be seen that it is comparatively a weak acid hence a solution of $HCN$ with its salt $KCN$ forms an acidic buffer. An acidic buffer is the one which maintains the$pH$of the solution on slight dilution or upon addition of a slight amount of acid or base. Applying Henderson’s equation you can calculate the $pH$of the solution.

Complete step-by-step answer:$HCN$ is a weak acid. Hence when it forms a solution with $KCN$, one of its salt it forms an acidic buffer which maintains the $pH$of the solution on slight dilution or upon addition of a slight amount of acid or base.
The$pH$of the buffer solution can be estimated using Henderson Hasselbach equation which is given as:
$pH\, = \,p{K_a}\, + \,{\log _{10}}\left( {\dfrac{{concentration\,of\,salt}}{{concentration\,of\,an\,acid}}} \right)..............\left( 1 \right)$.
$HCN$ being a weak acid, it dissociates as: . The ionization can be neglected as the concentration of ${H^ + }$ and $C{N^ - }$ formed is negligibly small.
Whereas, $KCN$being a salt it completely dissociates in water to give, $KCN \rightleftharpoons {K^ + } + C{N^ - }$.
Therefore, concentration of an acid $ = \,\left[ {HCN} \right]$ and concentration of salt $ = \,\left[ {C{N^ - }} \right]$ which is equivalent to$\left[ {KCN} \right]$.
Now, it is given the solution is formed by mixing $2.5\,moles$ of $KCN$and $2.5\,moles$ of $HCN$ in water and making up the total volume of $500\,mL$.
Therefore, $\left[ {HCN} \right]\, = \,\dfrac{{2.5\,moles}}{{500\,mL}}\, = \,\dfrac{{2.5\,moles}}{{500 \times {{10}^{ - 3}}\,L}}\, = \,5\,M$.
Also, $\left[ {C{N^ - }} \right]\, = \,\dfrac{{2.5\,moles}}{{500\,mL}}\, = \,\dfrac{{2.5\,moles}}{{500 \times {{10}^{ - 3}}\,L}}\, = \,5\,M$.
$p{K_a}\, = \,9.30$
Putting all the values in equation $\left( 1 \right)$ we get,
$pH\, = \,9.30\, + \,{\log _{10}}\left( {\dfrac{{2.5}}{{2.5}}} \right)\, = \,9.30 + {\log _{10}}\left( 1 \right)$
Now, ${\log _{10}}\left( 1 \right)\, = \,0$.
Therefore, $pH\, = \,9.30 + 0\, = \,9.30$.

Hence the correct answer is (i) $9.30$.

Note:For this question you must have a basic idea about buffers and its types. If the buffer given is a basic buffer then from the Henderson Hasselbach equation you can calculate the$pOH$ of the solution. Then using $pH + pOH\, = \,14$, you can then calculate the$pH$ of the given solution.