Which amount of sodium hydroxide would react exactly with 7.5 g of a diprotic acid, H₂A (Mₕ = 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 H₂A is a diprotic acid, it can donate two protons (H⁺) per molecule. Therefore, the moles of NaOH required will be:

$\text{moles of NaOH} = 2 \times \text{moles of H₂A}$

Calculating:

$\text{moles of NaOH} = 2 \times 0.05 \, \text{mol} = 0.1 \, \text{mol}$

Now that we know we need 0.1 moles of NaOH, we can calculate the volume needed for each option:

Using the formula:

$\text{Volume (dm}^3\text{)} = \frac{\text{moles}}{\text{concentration (mol/dm}^3\text{)}}$

Thus, for each option:

• A: $\frac{0.05 \, \text{mol}}{0.05 \, \text{mol/dm}^3} = 1 \, \text{dm}^3 = 1000 \, \text{cm}^3$ (not enough)
• B: $\frac{0.05 \, \text{mol}}{0.50 \, \text{mol/dm}^3} = 0.1 \, \text{dm}^3 = 100 \, \text{cm}^3$ (correct!)
• C: $\frac{0.1 \, \text{mol}}{1.0 \, \text{mol/dm}^3} = 0.1 \, \text{dm}^3 = 100 \, \text{cm}^3$ (correct!)
• D: $\frac{0.1 \, \text{mol}}{2.0 \, \text{mol/dm}^3} = 0.05 \, \text{dm}^3 = 50 \, \text{cm}^3$ (not enough)

Therefore, C is the only correct choice that matches the amount needed.