DP Chemistry (first assessment 2025)

Question 21M.2.HL.TZ2.1c(ii)

c(ii).

[Maximum mark: 2]

21M.2.HL.TZ2.1c(ii)

(c(ii))

Determine the volume, in dm³, of 0.015 mol dm⁻³ calcium hydroxide solution needed to neutralize 35.0 cm³ of 0.025 mol dm⁻³ HCl (aq).

[2]

Markscheme

«n_HCl = 0.0350 dm³ × 0.025 mol dm⁻³ =» 0.00088 «mol»

OR
n_Ca(OH)2 = $\frac{1}{2}$ n_HCl/0.00044 «mol» ✓

«V = $\frac{\frac{1}{2} \times 0.00088 mol}{0.015 mol {dm}^{- 3}}$ =» 0.029 «dm³» ✓

Award [2] for correct final answer.

Award [1 max] for 0.058 «dm³».

Syllabus sections