O-LevelChemistryAcids, Bases & Salts5070/21 · May/June 2018 · Q311 Marks

Barium chloride is a soluble salt and barium sulfate is an insoluble salt.

(a)(i)Barium sulfate can be prepared by the reaction between aqueous barium chloride and dilute sulfuric acid. Describe the preparation of a pure, dry sample of barium sulfate from aqueous barium chloride and dilute sulfuric acid.[3]
(a)(ii)Write the ionic equation, including state symbols, for this reaction.[2]
(b)Barium chloride can be prepared by reacting barium carbonate with dilute hydrochloric acid. BaCO3 + 2HCl → BaCl2 + H2O + CO2 Excess barium carbonate is reacted with 40.0 cm³ of 1.50 mol/dm³ hydrochloric acid. After purification the percentage yield of barium chloride was 75.0%. Calculate the mass of barium chloride prepared. Give your answer to three significant figures. [Mr: BaCl2, 208] mass of barium chloride ..................................................... g[3]
(c)A barium ion has the formula 138 Ba2+. 56 Complete the table about this ion. subatomic particles number of subatomic particles electrons neutrons protons[3]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

[3] 3(a)(i) mix (solutions) together AND then filter (1) wash the residue (with distilled water or solvent) (1) place residue in warm place (to dry) / place in oven (to dry) / use of filter paper (to dry) / leave (to dry) on the windowsill (1)
[2] 3(a)(ii) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) formulae and balancing correct (1) correct state symbols – dependent on correct formulae (1)
[3] 3(b) moles of hydrochloric acid = 6 × 10⁻²/0.06 (1) (moles of barium chloride = moles of HCl ÷ 2) = 3 × 10⁻²/0.03 (1) (mass of barium chloride = moles × 208 × 0.75) = 4.68 (1)
[3] 3(c) subatomic particles number of subatomic particles electrons 54 (1) neutrons 82 (1) protons 56 (1)

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