Which amount of sodium hydroxide would react exactly with 7.5 g of a diprotic acid, H₂A (Mₐ = 150)? A 50 cm³ of 0.05 mol dm⁻³ NaOH(aq) B 100 cm³ of 0.50 mol dm⁻³ NaOH(aq) C 100 cm³ of 1.0 mol dm⁻³ NaOH(aq) D 100 cm³ of 2.0 mol dm⁻³ NaOH(aq) - AQA - A-Level Chemistry - Question 6 - 2019 - Paper 3

Since the acid is diprotic, it can donate 2 moles of H⁺ ions for every mole of acid. Thus, for 0.05 moles of the acid, the required moles of NaOH will be:

$n_{NaOH} = 2 \times n_{H₂A} = 2 \times 0.05 \, \text{mol} = 0.1 \, \text{mol}$

We need to find out which option provides exactly 0.1 moles of NaOH. Using the concentration formula:

$C = \frac{n}{V}$

rewriting for volume gives:

$V = \frac{n}{C}$

For:

• Option C: 0.1 mol in 1.0 mol/dm³:

$V = \frac{0.1 \, \text{mol}}{1.0 \, \text{mol/dm}^3} = 0.1 \, \text{dm}^3 = 100 \, \text{cm}^3$

This shows that 100 cm³ of 1.0 mol dm⁻³ NaOH(aq) is the correct answer.