# Question #a4e04

May 26, 2017

The $p {K}_{a}$ of $H N {O}_{2} = 3.25$

Let $x$ L of 0.100M solution of.$H N {O}_{2}$ be mixed with $\left(1 - x\right)$ L of.$N a N {O}_{2}$ to make $1$ L of buffer solution.

Now $x$ L of 0.100M solution of $H N {O}_{2}$ contains $0.1 x$ mol $H N {O}_{2}$ in 1 L mixture

and $\left(1 - x\right)$ L of.$N a N {O}_{2}$ solution contains $0.2 \left(1 - x\right)$ mol $N a N {O}_{2}$ in 1 L mixture.

Now Using the Henderson-Hasselbalch equation

$p H = p {K}_{a} + {\log}_{10} \left(\frac{\left[N a N {O}_{2}\right]}{\left[H N {O}_{2}\right]}\right)$

$\implies 3.35 = 3.25 + {\log}_{10} \left(\frac{0.2 \left(1 - x\right)}{0.1 x}\right)$

$\implies {\log}_{10} \left(\frac{0.2 \left(1 - x\right)}{0.1 x}\right) = 0.01$

$\implies \frac{2 \left(1 - x\right)}{x} = {10}^{0.01} \approx 1.02$

$\implies \frac{1}{x} - \frac{x}{x} = 0.51$

$\implies \frac{1}{x} - 1 = 0.51$

$\implies \frac{1}{x} = 1.51$

$\implies x = \frac{1}{1.51} \approx \frac{2}{3}$

So to prepare $1$ L buffer $\frac{2}{3}$ L of 0.100M solution of $H N {O}_{2}$ should be mixed with $\frac{1}{3}$ L of 0.200M solution of $N a N {O}_{2}$.