Question

${K_a}$ for $HCN$ is $5 \times {10^{ - 10}}$ at $25^\circ C$. For maintaining a constant $pH = 9$, the volume $5M$ $KCN$ solution required to be added to $10ml$ of $2M$ $HCN$ solution is :
A. $4ml$
B. $2.5ml$
C. $2ml$
D. $6.4ml$

Answer 1

Hint: Here in the question we can observe that we are given a mixture of weak acid $HCN$ and a salt of it that is $KCN$ . So it is an acidic buffer. In an acidic buffer the $pH$remains constant. $KCN$ is formed when $HCN$ reacts with $KOH$.

Complete step by step answer: As in the question we are given the ${K_a}$ for $HCN$ at $25^\circ C$ that is $5 \times {10^{ - 10}}$. The $pH = 9$ is constant. For an acidic buffer we can write $ \Rightarrow pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}}$.
Here $[salt]$ i s the concentration of salt , that is $KCN$. $[acid]$ is the concentration of acid that is $HCN$. Let the volume of $KCN$ added be $v$$ml$. The total volume of the solution will be $V = (v + 10)ml$. Total moles of $KCN$ will be $5v$$mmoles$, as $moles = concentration \times volume$. Concentration of $KCN$ will be $ \Rightarrow [salt] = [KCN] = \dfrac{{moles}}{V} = \dfrac{{5v}}{{v + 10}}$.
 Similarly the moles of $HCN$ will be $20mmoles$, the total volume will be $V = (v + 10)ml$. So the concentration of $HCN$ will be $ \Rightarrow [acid] = [HCN] = \dfrac{{moles}}{V} = \dfrac{{20}}{{v + 10}}$.
Now we can directly apply the formula of acidic buffer:
 $
   \Rightarrow pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}} \\
   \Rightarrow 9 = - \log (5 \times {10^{ - 10}}) + \log \dfrac{{[\dfrac{{5v}}{{v + 10}}]}}{{[\dfrac{{20}}{{v + 10}}]}} \\
   \Rightarrow 9 = 9.3 + \log \dfrac{{5v}}{{20}} \\
   \Rightarrow - 0.3 = \log (v) - \log 4 \\
   \Rightarrow \log (v) = 0.6 - 0.3 = 0.3 \\
   \Rightarrow v = 2 \\
 $
We will have to add $2ml$ $KCN$in order to maintain a constant $pH$. So from the above explanation and calculation it is clear to us that

The correct answer of the given question is option: C. $2ml$

Additional information:
$HCN$ or hydrogen cyanide is a very useful compound . It is used commercially for mining, electroplating , fumigation and chemical synthesis. It is also used in the production of plastics , dyes, pesticides and synthetic fibres.

Note:
Always remember that we use the formula $ \Rightarrow pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}}$ for an acidic buffer. In a buffer solution the $pH$ always remains constant. Always try to avoid calculation errors and silly mistakes while solving the numerical. SO solve the question carefully.