| Date | May 2018 | Marks available | 1 | Reference code | 18M.2.sl.TZ2.1 |
| Level | SL | Paper | 2 | Time zone | TZ2 |
| Command term | Formulate | Question number | 1 | Adapted from | N/A |
Question
A student determined the percentage of the active ingredient magnesium hydroxide, Mg(OH)2, in a 1.24 g antacid tablet.
The antacid tablet was added to 50.00 cm3 of 0.100 mol dm−3 sulfuric acid, which was in excess.
Calculate the amount, in mol, of H2SO4.
Formulate the equation for the reaction of H2SO4 with Mg(OH)2.
The excess sulfuric acid required 20.80 cm3 of 0.1133 mol dm−3 NaOH for neutralization.
Calculate the amount of excess acid present.
Calculate the amount of H2SO4 that reacted with Mg(OH)2.
Determine the mass of Mg(OH)2 in the antacid tablet.
Calculate the percentage by mass of magnesium hydroxide in the 1.24 g antacid tablet to three significant figures.
Markscheme
n(H2SO4) «= 0.0500 dm3 × 0.100 mol dm–3» = 0.00500/5.00 × 10–3«mol»
[1 mark]
H2SO4(aq) + Mg(OH)2(s) → MgSO4(aq) + 2H2O(l)
Accept an ionic equation.
[1 mark]
«n(H2SO4) =\(\frac{1}{2}\) × n(NaOH) = \(\frac{1}{2}\) (0.02080 dm3 × 0.1133 mol dm–3)»
0.001178/1.178 × 10–3 «mol»
[1 mark]
n(H2SO4) reacted «= 0.00500 – 0.001178» = 0.00382/3.82 × 10–3 «mol»
[1 mark]
n(Mg(OH)2) «= n(H2SO4) =» = 0.00382/3.82 × 10–3 «mol»
m(Mg(OH)2) «= 0.00382 mol × 58.33 g mol–1» = 0.223 «g»
Award [2] for correct final answer.
[2 marks]
% Mg(OH)2 «= \(\frac{{0.223{\text{ g}}}}{{1.24{\text{ g}}}}\) × 100» = 18.0 «%»
Answer must show three significant figures.
[1 mark]
