Question

A buffer solution that has a pH of $4.72$ is prepared by mixing of $1\text{M NaOH}$ and $1M\text{HCN}$. What is the ratio of volumes of the acid to the base? $(p{K}_b{\text{of CN}}^{-}=9.28)$

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 3 : 1

Answer 1

The correct option is C 2 : 1

Let the volume of NaOH be $V_1$ and volume of $HCN$ be $V_2$
M eq of $HCN=V_2-V_1$
M eq of $NaCN=V_1$ Now this will form acidic buffer whose pH is given by
$pH=p{K}_a+log\frac{\left[NaCN\right]}{\left[HCN\right]}$ Now given
$p{K}_b{\text{of CN}}^{-}=9.28$
$\therefore p{K}_{a}ofHCN=4.72$, putting these values in above equation
$4.72=4.72+log\frac{V_1}{V_2-V_1}$
so $\frac{V_1}{V_2-V_1}=1$
Hence $V_2=2V_1$