Question

What volumes of 0.50 M HNO2 and 0.50 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.62?

Answer 1

Question

What volumes of $0.50\,\mathrm{M}\,\mathrm{HNO}_2$ and $0.50\,\mathrm{M}\,\mathrm{NaNO}_2$ must be mixed to prepare $1.00\,\mathrm{L}$ of a solution buffered at $\mathrm{pH} = 3.62$?

Solution

pH of given buffer solution is calculated as

$$pH = pK_a + \lg \frac{[\mathrm{NO}_2^-]}{[\mathrm{HNO}_2]}$$

For nitrous acid $pK_a = 3.39$

Thus, we have

$$3.62 = 3.39 + \lg \frac{[\mathrm{NO}_2^-]}{[\mathrm{HNO}_2]},$$

whence

$$\lg \frac{[\mathrm{NO}_2^-]}{[\mathrm{HNO}_2]} = 0.23$$

and

$$\frac{[\mathrm{NO}_2^-]}{[\mathrm{HNO}_2]} = 10^{0.23} = 1.698$$

Assuming $\mathrm{NaNO}_2$ dissosiates completely, we can state $[\mathrm{NO}_2^-] = [\mathrm{NaNO}_2]$

The concentrations of $\mathrm{NaNO}_2$ and $\mathrm{HNO}_2$ in the buffer solution are related to the concentrations of initial solutions $([\mathrm{NO}_2^-]_0$ and $[\mathrm{HNO}_2]_0)$ as follows

$$[\mathrm{NO}_2^-] = \frac{[\mathrm{NO}_2^-]_0 \cdot V_{(\mathrm{NaNO}_2)}}{V_{bs}} = \frac{0.50 \cdot V_{(\mathrm{NaNO}_2)}}{1.00} = 0.50 \cdot V_{(\mathrm{NaNO}_2)}$$ $$[\mathrm{HNO}_2] = \frac{[\mathrm{HNO}_2]_0 \cdot V_{(\mathrm{HNO}_2)}}{V_{bs}} = \frac{0.50 \cdot V_{(\mathrm{HNO}_2)}}{1.00} = 0.50 \cdot V_{(\mathrm{HNO}_2)},$$

where $V_{(\mathrm{NaNO}_2)}$ and $V_{(\mathrm{HNO}_2)}$- volumes of the initial solutions, $V_{bs}$ - volume of resulting buffer solution.

We know that

$$V_{(\mathrm{NaNO}_2)} + V_{(\mathrm{HNO}_2)} = V_{bs} = 1.00$$

We have also calculated that

$$\frac{[\mathrm{NO}_2^-]}{[\mathrm{HNO}_2]} = \frac{0.50 \cdot V_{(\mathrm{NaNO}_2)}}{0.50 \cdot V_{(\mathrm{HNO}_2)}} = \frac{V_{(\mathrm{NaNO}_2)}}{V_{(\mathrm{HNO}_2)}} = 1.698$$

Let us assign $V_{(\mathrm{NaNO}_2)} = x$ and $V_{(\mathrm{HNO}_2)} = y$

We have the set of two equations with two unknown values:

$$\left\{ \begin{array}{l} x + y = 1.00 \\ \frac{x}{y} = 1.698 \end{array} \right.$$ $$\begin{array}{l} x = 1.698 \cdot y \\ 1.698 \cdot y + y = 1.00 \\ 2.698 \cdot y = 1.00 \\ y = \frac{1.00}{2.698} = 0.37 \\ x = 1.00 - y = 1.00 - 0.37 = 0.63 \end{array}$$

Answer: $V_{(\mathrm{NaNO}_2)} = 0.63\,\mathrm{L}$

$$V_{(\mathrm{HNO}_2)} = 0.37\,\mathrm{L}$$