Entropy is a measure of the disorder of a system. The enthalpy change of solution, ΔH°sol, of the Group 2 sulfates becomes more endothermic down the group. An aqueous solution of sodium hydroxide has a pH of 13.25 at 298 K. The Ka for ethanoic acid is 1.75 × 10⁻⁵ mol dm⁻³ at 298 K. Titration curves for two different acid-base reactions, M and N, are shown. [Figure showing two titration curves: reaction M (strong acid-strong base) and reaction N (weak acid-strong base)]
- (a)State and explain the trend in the solubility of the Group 2 sulfates down the group. [3][3]
- (b)(i)Write the expression for Kw, the ionic product of water. Kw = .............................................................. [1][1]
- (b)(ii)The numerical value of Kw increases with increasing temperature. Place a tick (✔) in the appropriate column in each row to show the effect of increasing the temperature of water on the pH and on the ratio [H⁺]: [OH⁻]. | effect of increasing | decrease | stay the same | increase temperature of water | pH | | | ratio [H⁺]: [OH⁻] | | | [2][2]
- (c)Calculate the concentration of this sodium hydroxide solution. concentration = ...................................... mol dm⁻³ [2][2]
- (d)Buffer solutions are used to regulate the pH of a solution to keep its pH value within a narrow range. Write two equations to describe how hydrogencarbonate ions, HCO3⁻, and carbonic acid, H2CO3, control the pH of blood. [2][2]
- (e)(i)When ethanoic acid is dissolved in water, an equilibrium mixture containing two acid-base pairs is formed. Write an equation for this equilibrium. In the boxes label each species acidic or basic to show its behaviour in this equilibrium. CH3CO₂H + [box] ........................ ⇌ ........................ + [box] ........................ [2][2]
- (e)(ii)A buffer solution was prepared by adding 30.0 cm³ of 0.25 mol dm⁻³ ethanoic acid, an excess, to 20.0 cm³ of 0.15 mol dm⁻³ sodium hydroxide. Calculate the pH of the buffer solution formed at 298 K. Give your answer to one decimal place. pH = ...................................... [4][4]
- (f)(i)Use the titration curve for reaction M to deduce the volume of acid added at the end-point for this titration. volume of acid added at the end-point = ...................................... cm³ [1][1]
- (f)(ii)The table shows some acid-base indicators. name of indicator | pH range of colour change ------------------|-------------------------- malachite green | 0.2–1.8 bromocresol green | 3.8–5.4 bromothymol blue | 6.0–7.6 thymolphthalein | 9.3–10.6 Name a suitable indicator for each of the acid-base titrations M and N. Explain your answers. reaction M ..................................................... reaction N ..................................................... explanation ................................................................................................................................................................................................................................................................................................................ [2][2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [1] M1 solubility decreases (down the Group)
- [1] M2 because lattice energy and hydration energy decreases OR lattice energy and hydration energy become less exothermic / more endothermic
- [1] M3 because hydration energy decreases to a greater extent (than does ∆HLatt)
- [1] (Kw = ) [H⁺][OH⁻]
- [1] for each correct tick
- Table entry for pH: decrease (✔)
- Table entry for ratio [H⁺]: [OH⁻]: stay the same (✔)
- [1] [H⁺] = 10⁻¹³·²⁵ = 5.62 × 10⁻¹⁴
- [1] [OH⁻] = Kw/[H⁺] = 1.0 × 10⁻¹⁴/5.62 × 10⁻¹⁴
- [1] [OH⁻] = 0.18 (0.178) (mol dm⁻³) ecf correct answer scores [2]
- [1] HCO3⁻ + H⁺ → H2CO3 OR HCO3⁻ + H⁺ ⇌ CO2 + H2O
- [1] H2CO3 + OH⁻ → HCO3⁻ + H2O
- [1] CH3COOH + H2O ⇌ CH3COO⁻ + H3O⁺
- [1] acid + base ⇌ base + acid
- [1] M1 moles NaOH = 0.15 × 20/1000 = 0.0030 AND initial moles CH3COOH = 0.25 × 30/1000 OR 0.0075
- [1] M2 equilibrium moles CH3COOH = 0.0045 AND equilibrium moles CH3COONa = 0.0030
- [1] M3 [CH3COOH] = 0.0045/0.05 = 0.090 AND [CH3COONa] = 0.003/0.05 = 0.060 ]H⁺] = Ka × [CH3COOH]/[CH3COONa[ = 2.625 × 10⁻⁵
- [1] M4 pH = -log[H⁺] = 4.6 correct answer scores [4]
- [1] end point = 28 cm³
- [1] M1 reaction M bromothymol (blue) / bromocresol (green) AND reaction N bromothymol (blue) / thymolphthalein
- [1] M2 (both indicators have) a pH range / colour change within / in end-point / vertical region / sharp fall of the graph
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