Question

Calculate the pH of the solution obtained by adding 0.02 lit of 1 mol./lit nitrous acid
(pKa=3.34) to 0.04 lit of 0.5 mol/lit sodium nitrite solution.

Answer 1

Calculate the pH of the solution obtained by adding 0.02 lit of 1 mol/lit nitrous acid (pKa=3.34) to 0.04 lit of 0.5 mol/lit sodium nitrite solution.

Solution:

1. Determine the concentration of nitrous acid in solution, if quantity of nitrous acid in solution is $v=1 \, \text{mol/l} \cdot 0.02L = 0.02 \, \text{mole}$:

$$C(\mathrm{HNO_2}) = \frac{0.02}{0.02 + 0.04} = 0.33 \, \text{mol/L}$$

2. Determine the concentration of sodium nitrite in solution, if quantity of sodium nitrite in solution is $v=0.5 \, \text{mol/l} \cdot 0.04L = 0.02 \, \text{mole}$:

$$C(\mathrm{NaNO_2}) = \frac{0.02}{0.02 + 0.04} = 0.33 \, \text{mol/L}$$

3. Calculate the pH of the solution according to equation:

$$\mathrm{pH} = \mathrm{pK_a} + \lg \frac{C(\mathrm{NaNO_2})}{C(\mathrm{HNO_2})},$$

pH of the solution is: $\mathrm{pH} = 3.34 + \lg \frac{0.33}{0.33} = 3.34$

**Answer**: pH of the solution is 3.34.