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 is expressed as: [ ext{pH} = ext{pK}_a + \log\left( \frac{[A^-]}{[HA]} \right) ]
where [A⁻] is the concentration of the salt and [HA] is the concentration of the acid. Plugging in the values: [ 4.05 = 3.75 + \log\left( \frac{[HCOO^-]}{0.100} \right) ]
Rearranging gives: [ \log\left( \frac{[HCOO^-]}{0.100} \right) = 4.05 - 3.75 = 0.30 ]
Taking the antilog of both sides: [ \frac{[HCOO^-]}{0.100} = 10^{0.30} ] [ [HCOO^-] = 0.100 \times 10^{0.30} = 0.200 \text{ mol dm}^{-3} ]

Using the formula for molarity: [ \text{amount (mol)} = [HCOO^-] \times \text{volume (dm}^3) ]
Substituting the known values: [ \text{amount} = 0.200 \text{ mol dm}^{-3} \times 0.0250 ext{ dm}^3 = 0.00500 \text{ mol} ]

The mass can be calculated using the molar mass of sodium methanoate (HCOONa), which is approximately 82.03 g/mol:
[ ext{mass} = ext{amount} \times M ]
[ ext{mass} = 0.00500 ext{ mol} \times 82.03 ext{ g/mol} = 0.41015 ext{ g} ]
Thus, the mass of sodium methanoate required is approximately 0.41 g.