Q. $K_{a}$forHCN is $5 \times 10^{- 10}$ at $25℃$. 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) 2 mL
B) 4 mL
C) 8.2 mL
D) 6.4 mL
Explanation
(a)
$pH = - \log{\left( 5 \times 10^{- 10} \right) + \log\left\lbrack \frac{5 \times V}{V + 10}/\frac{10 \times 2}{V + 10} \right\rbrack}$
$9 = - \log{\left( 5 \times 10^{- 10} \right) + \log\frac{V}{4}}$
On solving, $V = 2\ mL$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.