A-LevelChemistryEquilibria9701/41 · Oct/Nov 2024 · Q116 Marks

Disodium phosphate, (Na+)2(HPO42¯), reacts with an acid to form monosodium phosphate, Na+(H2PO4¯).

(a)(i)Identify the ions that are a conjugate acid-base pair in this reaction, using the formulae of the species involved. conjugate acid conjugate base[1]
(a)(ii)Define buffer solution.[2]
(a)(iii)Write two equations to show how a mixture of (Na+)2(HPO42-) and Na⁺(H2PO4¯) can act as a buffer solution. equation 1 equation 2[2]
(a)(iv)Identify one inorganic ion that acts as a buffer in blood.[1]
(b)Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63g of E is dissolved in water to make 250cm³ of solution F. Solution F has a pH of 13.09 at 298 K.
(b)(i)Show that the concentration of hydroxide ions in solution F is 0.123moldm-3.[2]
(b)(ii)Explain why the concentration of compound E in solution F is 0.0615moldm-3.[1]
(b)(iii)Use the concentration given in (ii) to identify compound E. compound E[1]
(c)Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10–4 moldm-3 at 298 K. Calculate the solubility product, Ksp, of magnesium hydroxide. Include units. KSD = units[3]
(d)Explain why compound E is much more soluble than magnesium hydroxide.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

[1] conjugate acid H2PO4- conjugate base HPO42-
[1] allows only small changes in pH / resists changes in pH
[1] when small amounts of acid or alkali/base/ H+ or OH- are added
[1] Na+2HPO42- + H+ → Na+H2PO4 + Na+ OR HPO42- + H+ → H2PO4-
[1] Na+H2PO4 + NaOH → Na2HPO42- + H2O OR H2PO4 + OH- → HPO42- + H2O
[1] HCO3¯/ hydrogen carbonate
[1] either 10^-13.09 or [H+]= 8.1 × 10^-14 seen
[1] 8.1x10^-14 × [OH-] = 1 × 10^-14 seen OR [OH-] = 1x10^-14 / 8.1×10^-14 seen OR pOH = 0.91 AND [OH-] = 10^-0.91 seen
[1] 2 moles of OH- in one mole of E / X(OH)2 OR there are two hydroxide ions in each formula unit of E / X(OH)2
[1] 0.0615 × 0.25 = 0.0154 moles of E
[1] RFM = 2.63/0.0154 = 170.8/171 OR RAM = 34 – 2.63/0.0154 = 137.1/137 AND barium hydroxide / Ba(OH)2
[1] Ksp = [Mg2+][OH-]² OR Ksp = 1.4 × 10-4 × (2 × 1.4 × 10-4)2
[1] Ksp = 1.10 × 10-11
[1] mol³ dm-9
[1] ∆Hlatt and ∆Hhyd less exothermic (down the group) OR AHlatt/LE and AHhyd more exothermic/more negative for Mg
[1] AHlatt changes more than ∆Hhyd (down the group) OR AHlatt changes more / faster for Mg
[1] AHsol becomes more exothermic (down the group) OR AHSol becomes less exothermic / less negative for Mg

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