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

Using the Henderson-Hasselbalch equation:

$pH = pK_a + ext{log} \left( \frac{[A^-]}{[HA]} \right)$

We substitute the values:

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

Rearranging gives:

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

This implies:

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

Let [HCOOH] be the concentration of methanoic acid:

For the original solution of methanoic acid:

$[HCOOH] = 0.100 \, \text{mol dm}^{-3}$

Then using the ratio:

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

To find the amount of sodium methanoate needed, we calculate:

$\text{Amount} = [HCOO^-] \times \text{volume} = 0.200 \, \text{mol dm}^{-3} \times 0.0250 \, \text{dm}^{3} = 0.00500 \, \text{mol}$

Now, to find the mass of sodium methanoate (M = 82.03 g/mol):

$\text{Mass} = \text{Amount} \times M = 0.00500 \, \text{mol} \times 82.03 \, \text{g/mol} = 0.41015 \, \text{g} \approx 0.41 \, \text{g}$