O-LevelChemistryAcids, Bases & Salts5070/22 · May/June 2018 · Q310 Marks

Silver chloride, AgCl, is an insoluble salt and silver nitrate is a soluble salt.

(a)Silver chloride can be prepared by the reaction between aqueous silver nitrate and dilute hydrochloric acid.
(a)(i)Describe the preparation of a pure, dry sample of silver chloride from aqueous silver nitrate and dilute hydrochloric acid.[3]
(a)(ii)Write the ionic equation, including state symbols, for this reaction.[2]
(b)Silver chloride decomposes in the presence of light to make silver and chlorine.
(b)(i)What is the name of the type of reaction that requires light?[1]
(b)(ii)Construct the equation for the decomposition of silver chloride.[1]
(c)Silver nitrate can be prepared by reacting silver oxide with dilute nitric acid. Ag2O + 2HNO3 → 2AgNO3 + H2O Excess silver oxide is reacted with 30.0 cm³ of 0.150 mol/dm³ nitric acid. After purification the percentage yield of silver nitrate is 80.0%. Calculate the mass of silver nitrate prepared. Give your answer to three significant figures. [M₁: AgNO3, 170] mass of silver nitrate ..................................................... g[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 / add solutions (to each other) AND then filter / take the mixture AND filter (1) wash the residue (with distilled water / solvent) (1) place residue in warm place (to dry) / place in oven (to dry) / use of filter paper (to dry) / leave (to dry) on windowsill (1)
[2] 3(a)(ii) Ag+(aq) + Cl¯(aq) → AgCl(s) formulae and balance correct (1) correct state symbols dependent on correct formulae (1)
[1] 3(b)(i) photochemical (1)
[1] 3(b)(ii) 2AgCl → 2Ag + Cl2 (1)
[3] 3(c) (moles of nitric acid) = 4.5 × 10⁻³ (1) EITHER moles from first marking point × 170 OR 0.765 g (silver nitrate) (1) (0.765 x 0.8) = 0.612 g (silver nitrate) (1) OR moles from first marking point × 0.8 OR 3.6 × 10⁻³ moles (silver nitrate) (1) (3.6 × 10⁻³ × 170) = 0.612 g silver nitrate (1)

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