Question

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

• A. 4 mL
• B. 2.5 mL
• C. 2 mL
• D. 6.4 mL

Answer 1

Answer: (C)

2 mL

Answer 2

$pH = pK_{a}+ log \frac{\left[Salt\right]}{\left[Acid\right]}$ $\quad\quad=pK_{a}+ log \frac{\left[KCN\right]}{\left[HCN\right]}\quad\quad\quad\quad\quad...\left(i\right)$ Let the volume of KCN solution required be V mL $\therefore\quad\left[KCN\right]=\frac{5\times V}{V+10} and \left[HCN\right] = \frac{10\times2}{V +10}$ Now from eqn. $\left(i\right),$ $pH=- log \left(5\times10^{-10}\right)+log \left[\frac{5\times V}{V+10} /\frac{10\times2}{V+10}\right]$ $9 = -log\left(5\times10^{-10}\right)+log \frac{V}{4}$ On solving, $V = 1.99 \approx 2 mL$