A-LevelChemistryEquilibria9701/43 · Oct/Nov 2020 · Q47 Marks
Water and octan-1-ol form two layers when mixed. Ethanamide is more soluble in water than it is in octan-1-ol. When 1.00g of ethanamide is added to 50.0 cm³ of water and this is then shaken with 50.0 cm³ of octan-1-ol, it is found that the water layer contains 0.935 g of ethanamide at equilibrium.
- (a)(i)Give the mathematical expression for each of the terms pH and Kw. pH = Kw =[2]
- (a)(ii)Calculate the pH of 0.027 moldm³ NaOH(aq).[1]
- (b)The Ka value of chloric(I) acid, HClO, is 3.72 × 10⁸ moldm³. Calculate the pH of 0.010moldm³ HClO(aq).[1]
- (c)(i)Calculate the partition coefficient, Kpc, for ethanamide in water and octan-1-ol. Kpc =[1]
- (c)(ii)The 50.0 cm³ of water containing 0.935 g of ethanamide is then shaken with 100.0 cm³ of pure octan-1-ol under the same conditions. Calculate the mass of ethanamide that is dissolved in the 100.0 cm³ of octan-1-ol at equilibrium. mass of ethanamide = g[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [2] (pH =) -log[H+] OR –lg[H+] [1]
- (Kw =) [H+][OH-] [1]
- [1] [H+] = 1 × 10⁻¹⁴/0.027 = 3.7037 × 10⁻¹³
- pH = -log(3.7037 × 10⁻¹³)= 12.4 [1] min 3sf
- [1] [H+] = √3.72 × 10⁻⁸ × 0.010 = 1.9287 × 10⁻⁵
- pH = -log(1.9287 × 10⁻⁵)= 4.7 [1] min 2sf
- [1] Kpc = (0.935 / 50) / (0.065 / 50)
- Kpc = 14.4 (14.38) [1] min 3sf
- [2] M1: 14.4 = ((0.935 – x) / 50) / (x / 100) [1] ecf from 4(c)(i)
- M2: x = 0.114 g [1] min 2sf ecf from M1
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