A-LevelChemistryGroup 29701/43 · Oct/Nov 2018 · Q419 Marks

(a) The enthalpy change of solution, ΔHsol∘, of the Group 2 sulfates becomes more endothermic down the group. State and explain the trend in the solubility of the Group 2 sulfates down the group.

(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 temperature of water | decrease | stay the same | increase ---|---|---|--- pH | | | ratio [H+]: [OH-] | | | [2][2]
(c)An aqueous solution of sodium hydroxide has a pH of 13.25 at 298 K. 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)The Ka for ethanoic acid is 1.75 × 10⁻⁵ mol dm⁻³ at 298 K.
(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] + [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)Titration curves for two different acid-base reactions, M and N, are shown. [Figure: Two titration curves. Reaction M shows a strong base-weak acid titration. Reaction N shows a strong acid-strong base titration.]
(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]

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] pH: decrease
[1] ratio [H+]: [OH-]: increase
[1] [H+] = 10–13.25 = 5.62 × 10-14
[1] [OH¯] = Kw/[H+] = 1.0 × 10~14/5.62 × 10-14 = 0.18 (0.178) (mol dm³)
[1] HCO3¯ + H+ ⇌ H2CO3 OR HCO3¯ + H+ ⇌ CO2 + H2O
[1] H2CO3 + OH¯ ⇌ HCO3¯ + H2O
[1] CH3COOH + H2O ⇌ CH3COO¯ + H3O+
[1] acid + base ⇌ base + acid (labels)
[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 AND ]H+] = Ka × [CH3COOH]/[CH3COONa[ = 2.625 × 10¯⁵
[1] M4 pH = -log[H+] = 4.6
[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