Question
Iron (II) sulfide reacts with hydrochloric acid to form hydrogen sulfide, H2S.
(a) (i) Draw the Lewis (electron dot) structure of hydrogen sulfide. [1]
(ii) Predict the shape of the hydrogen sulfide molecule.[1]
(b) In aqueous solution, hydrogen sulfide acts as an acid.
(i) State the formula of its conjugate base. [1]
(ii) Saturated aqueous hydrogen sulfide has a concentration of 0.10 mol dm–3 and a pH of 4.0. Demonstrate whether it is a strong or weak acid.[1]
(iii) Calculate the hydroxide ion concentration in saturated aqueous hydrogen sulfide. [1]
(c) A gaseous sample of nitrogen, contaminated only with hydrogen sulfide, was reacted with excess sodium hydroxide solution at constant temperature. The volume of the gas changed from 550 cm3 to 525 cm3.
Determine the mole percentage of hydrogen sulfide in the sample, stating one assumption you made. [3]
Mole percentage H2S:
Assumption:
Answer/Explanation
Ans:
2. a i 
2. a ii bent/non-linear/angular/v-shaped
2. b i HS−
2. b ii weak AND strong acid of this concentration/[H+ ] = 0.1 mol dm−3 would have pH = 1 OR weak AND [H+ ] = 10−4 < 0.1 «therefore only fraction of acid dissociated»
2. b iii 10−10 «mol dm−3»
2. c Mole percentage H2S: volume of H2S = «550 − 525 = » 25 «cm3» mol % H2S = « × 100 = » 4.5 «%»
Assumption: «both» gases behave as ideal gases Award [2] for correct final answer of 4.5 Accept “volume of gas α mol of gas”. Accept “reaction goes to completion”. Accept “nitrogen is insoluble/does not react with NaOH/only H2S reacts with NaOH”.
Question
Limescale, CaCO3(s), can be removed from water kettles by using vinegar, a dilute solution of ethanoic acid, CH3COOH(aq).
Predict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.[2]
Dissolved carbon dioxide causes unpolluted rain to have a pH of approximately 5, but other dissolved gases can result in a much lower pH. State one environmental effect of acid rain.[1]
Write an equation to show ammonia, NH3, acting as a Brønsted–Lowry base and a different equation to show it acting as a Lewis base.
[2]
Determine the pH of 0.010 mol dm−3 2,2-dimethylpropanoic acid solution.
Ka (2,2-dimethylpropanoic acid) = 9.333 × 10−6[2]
Explain, using appropriate equations, how a suitably concentrated solution formed by the partial neutralization of 2,2-dimethylpropanoic acid with sodium hydroxide acts as a buffer solution.[2]
Answer/Explanation
Markscheme
slower rate with ethanoic acid
OR
smaller temperature rise with ethanoic acid
[H+] lower
OR
ethanoic acid is weak
OR
ethanoic acid is partially dissociated
Accept experimental observations such as “slower bubbling” or “feels less warm”.
[2 marks]
Any one of:
corrosion of materials/metals/carbonate materials
destruction of plant/aquatic life
«indirect» effect on human health
Accept “lowering pH of oceans/lakes/waterways”.
[1 mark]
Brønsted–Lowry base:
NH3 + H+ → NH4+
Lewis base:
NH3 + BF3 → H3NBF3
Accept “AlCl3 as an example of Lewis acid”.
Accept other valid equations such as Cu2+ + 4NH3 → [Cu(NH3)4]2+.
[2 marks]
[H+] «\( = \sqrt {{{\text{K}}_{\text{a}}} \times \left[ {{{\text{C}}_5}{{\text{H}}_{10}}{{\text{O}}_2}} \right]} = \sqrt {9.333 \times {{10}^{ – 6}} \times 0.010} {\text{ }}\)» = 3.055 × 10–4 «mol dm–3»
«pH =» 3.51
Accept “pH = 3.52”.
Award [2] for correct final answer.
Accept other calculation methods.
[2 marks]
(CH3)3CCOOH(aq) + OH–(aq) → (CH3)3CCOO–(aq) + H2O(l)
OR
(CH3)3CCOOH(aq) + OH–(aq) \( \rightleftharpoons \) (CH3)3CCOO–(aq) + H2O(l) AND addition of alkali causes equilibrium to move to right
(CH3)3CCOO–(aq) + H+(aq) → (CH3)3CCOOH(aq)
OR
(CH3)3CCOO–(aq) + H+(aq) \( \rightleftharpoons \) (CH3)3CCOOH(aq) AND addition of acid causes equilibrium to move to right
Accept “HA” for the acid.
Award [1 max] for correct explanations of buffering with addition of acid AND base without equilibrium equations.
[2 marks]