A mixture of methanoic acid and sodium methanoate in aqueous solution acts as an acidic buffer solution - AQA - A-Level Chemistry - Question 5 - 2021 - Paper 3

The Henderson-Hasselbalch equation for an acidic buffer is:

$\text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right)$

Using the values:

• pH = 4.05
• pKₐ = 3.75

We can rearrange this to find the ratio of the concentrations:

$4.05 = 3.75 + \log \left( \frac{[HCOO^-]}{[HCOOH]} \right)$

$\log \left( \frac{[HCOO^-]}{[HCOOH]} \right) = 4.05 - 3.75 = 0.30$

From the logarithmic identity, we can express the ratio:

$\frac{[HCOO^-]}{[HCOOH]} = 10^{0.30} \approx 2.00$

This means,

$[HCOO^-] = 2.00 \times [HCOOH]$

Given that the initial concentration of HCOOH is 0.100 mol dm⁻³:

• Volume of solution = 25.0 cm³ = 0.025 dm³

Thus,

$[HCOO^-] = 2.00 \times [HCOOH] = 2.00 \times 0.100 \approx 0.200 \text{ mol dm}^{-3}$

To find the amount of sodium methanoate (HCOONa), we use:

$\text{Amount} = [HCOO^-] \times \text{Volume} = 0.200 \text{ mol dm}^{-3} \times 0.025 \text{ dm}^{3} = 0.00500 ext{ mol}$

Now, using the molar mass of HCOONa (approximately 82.03 g/mol), the mass is:

$\text{Mass} = 0.00500 ext{ mol} \times 82.03 ext{ g/mol} \approx 0.410 ext{ g}$