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Questions on halide ions, the silver nitrate test, concentrated sulfuric acid, and chlorine in water
The halide ions as reducing agents and the increase in reducing ability down the group, the reactions of solid sodium halides with concentrated sulfuric acid with equations and observations, the silver nitrate test with the colours of the precipitates and their solubility in ammonia, why the test is acidified with nitric acid, identifying a halide from the mass of its silver halide precipitate, the hydrogen halides dissolving in water as acids and giving white smoke with ammonia, chlorine with water as disproportionation, chlorine with water in sunlight, chlorine with cold dilute sodium hydroxide to make bleach, the benefits and risks of chlorinating water, and the method of required practical 4 for identifying anions and cations.
21 original questions · 73 marks · the halide ions, the silver nitrate test, concentrated sulfuric acid, and chlorine in water notes · Group 2 and Group 7 · Flashcards for this unit
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Describe how test-tube reactions could be used to find out whether an unknown solution contains carbonate, sulfate or chloride ions. Give the order of the tests, the reagents and the observations, and explain the order.
Mark scheme
Indicative content
- test for carbonate first by adding dilute nitric acid: effervescence, and the gas turns limewater milky
- test a fresh portion for sulfate by adding dilute hydrochloric acid, then barium chloride solution: a white precipitate
- test a fresh portion for chloride by adding dilute nitric acid, then silver nitrate solution: a white precipitate
- the chloride is confirmed when the white precipitate dissolves in dilute ammonia, which neither bromide nor iodide precipitates do
- carbonate is tested first because carbonate ions give white precipitates with both barium and silver ions
- sulfate is tested before chloride because silver sulfate can precipitate, and the acids are chosen so as not to add the ion being tested for
Levels
- Level 3 (5–6 marks) all three tests are given in the right order with the reagents, the acids and the observations, the ammonia confirmation for chloride is included, and both reasons for the order are explained, carbonate interfering with both later tests and silver sulfate interfering with the chloride test; the answer explains why each acid is chosen
- Level 2 (3–4 marks) the three tests have the right reagents and observations; the order is only partly justified, the ammonia confirmation is left out, or an acid is chosen that adds the ion under test
- Level 1 (1–2 marks) one or two tests are described; the usual slip is to acidify the barium chloride test with sulfuric acid or the silver nitrate test with hydrochloric acid
Describe and explain the reactions of solid sodium chloride, sodium bromide and sodium iodide with concentrated sulfuric acid. Include observations, equations and the trend in the reducing ability of the halide ions.
Mark scheme
Indicative content
- with sodium chloride, misty white fumes of hydrogen chloride form, NaCl + H₂SO₄ → NaHSO₄ + HCl, an acid–base reaction that is not redox
- with sodium bromide, misty fumes of hydrogen bromide form first, and then brown fumes of bromine and colourless, choking sulfur dioxide
- 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O, in which sulfur is reduced from +6 to +4 and bromide is oxidised to bromine
- with sodium iodide there are few misty fumes, but purple vapour or a black solid of iodine, a yellow solid of sulfur and a bad-egg smell of hydrogen sulfide
- iodide reduces sulfur to +4 in SO₂, 0 in S and −2 in H₂S, for example 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O
- reducing ability increases from chloride to iodide because the larger ion's outer electrons are further from the nucleus and more shielded, so lost more easily
Levels
- Level 3 (5–6 marks) the observations for all three halides are given and matched to the products, the chloride reaction is identified as acid–base only, the bromide and iodide reactions are shown with equations and the oxidation states of sulfur in each product, and the trend in reducing power is explained through the size and shielding of the halide ion
- Level 2 (3–4 marks) the observations for the three halides are mostly right and one equation carries oxidation states; the chloride reaction is not called acid–base, the sulfur products from iodide are incomplete, or the trend is unexplained
- Level 1 (1–2 marks) misty or brown fumes are mentioned; the chloride is often said to reduce the acid to sulfur dioxide, and no oxidation states are given
Explain the trend in the reducing ability of the halide ions from fluoride to iodide.
Mark scheme
- B1 reducing ability increases from fluoride to iodide
- B1 an iodide ion is larger, so its outer electrons are further from the nucleus
- B1 its outer electrons are more shielded by inner shells
- B1 they are less strongly attracted, so an iodide ion loses an electron more easily; reject 'iodine loses electrons'
Describe what is seen when concentrated sulfuric acid is added to solid sodium bromide, naming the substance responsible for each observation, and give the equation for the redox step.
Mark scheme
- B1 misty white fumes, due to hydrogen bromide
- B1 a brown gas (orange-brown fumes), due to bromine
- B1 a colourless, choking gas, sulfur dioxide, is also formed; ignore 'steamy fumes' for SO₂
- B1 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O; allow the ionic form 2Br⁻ + H₂SO₄ + 2H⁺ → Br₂ + SO₂ + 2H₂O
Evaluate the use of chlorine to treat public drinking water.
Mark scheme
- B1 chlorine (as HClO and ClO⁻) disinfects the water so bacteria cannot survive, preventing the spread of diseases such as cholera and typhoid
- B1 some chlorine remains in the water, so it continues to disinfect it in the pipes
- B1 chlorine is toxic, and it can react with organic matter to form chlorinated hydrocarbons that may cause cancer
- B1 a conclusion: the benefit to public health outweighs the risk at the low concentrations used; a conclusion with no reason is not credited
Explain how warming a solution with sodium hydroxide solution and testing with red litmus paper identifies ammonium ions, and why the litmus paper must be damp and must not touch the test tube.
Mark scheme
- B1 ammonium ions react with hydroxide ions to give ammonia gas: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
- B1 ammonia is an alkaline gas and turns damp red litmus paper blue; it is the only common alkaline gas
- B1 ammonia must dissolve in water on the paper to form hydroxide ions before the litmus can change colour
- B1 sodium hydroxide on the glass is itself alkaline, so touching it would turn the paper blue even without ammonia
Describe and explain what is seen when hydrogen bromide gas meets moist air, and when a glass rod dipped in concentrated ammonia solution is held in the gas. Give an equation for each change.
Mark scheme
- B1 misty fumes form, because hydrogen bromide is very soluble and dissolves in water vapour to form droplets of acid
- B1 HBr(g) + H₂O(l) → H₃O⁺(aq) + Br⁻(aq)
- B1 dense white smoke forms, which is solid ammonium bromide, made as ammonia takes the proton from the hydrogen bromide
- B1 NH₃(g) + HBr(g) → NH₄Br(s)
Deduce the oxidation state of sulfur in sulfuric acid and in each sulfur-containing product of the reaction of sodium iodide with concentrated sulfuric acid.
Mark scheme
- B1 +6 in sulfuric acid, H₂SO₄
- B1 +4 in sulfur dioxide, SO₂, and 0 in sulfur, S
- B1 −2 in hydrogen sulfide, H₂S
Explain why the reaction of solid sodium chloride with concentrated sulfuric acid is not a redox reaction, and give its equation with state symbols.
Mark scheme
- B1 NaCl(s) + H₂SO₄(l) → NaHSO₄(s) + HCl(g); allow 2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
- B1 no element changes oxidation state: chlorine stays −1 and sulfur stays +6
- B1 chloride ions are too weak a reducing agent to reduce the sulfuric acid, so it acts only as an acid
Describe how acidified silver nitrate and ammonia solution are used to show that a solution contains bromide ions.
Mark scheme
- B1 add dilute nitric acid and then silver nitrate solution
- B1 a cream precipitate of silver bromide forms
- B1 the precipitate does not dissolve in dilute ammonia but dissolves in concentrated ammonia
Explain why dilute nitric acid is added before silver nitrate in the test for halide ions, and why hydrochloric acid cannot be used instead.
Mark scheme
- B1 the acid removes carbonate (and hydroxide) ions
- B1 these would otherwise react with silver ions to form precipitates and give a false positive
- B1 hydrochloric acid adds chloride ions, which would give a white precipitate of silver chloride whatever the sample contained
Suggest how a student could show that a solution containing iodide ions also contains chloride ions, using acidified silver nitrate and ammonia solution.
Mark scheme
- B1 add dilute nitric acid and silver nitrate: a pale yellow mixture of silver chloride and silver iodide precipitates
- B1 add dilute ammonia: silver chloride dissolves while silver iodide remains, and filter
- B1 acidify the filtrate with dilute nitric acid; a white precipitate of silver chloride reappears, showing chloride
Suggest the products formed when concentrated sulfuric acid is added to solid sodium astatide, explain your suggestion, and give an equation for the step that forms hydrogen sulfide.
Mark scheme
- B1 astatine, At₂, and hydrogen sulfide (with sulfur and sulfur dioxide) would form
- B1 astatide is an even stronger reducing agent than iodide, so it reduces sulfur all the way to −2
- B1 8HAt + H₂SO₄ → 4At₂ + H₂S + 4H₂O; allow the ionic form with 8At⁻ and 8H⁺
Give the equation, with state symbols, for the reaction of chlorine with cold dilute sodium hydroxide solution, name the product that acts as a bleach, and give the oxidation state of chlorine in it.
Mark scheme
- B1 Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)
- B1 sodium chlorate(I), NaClO, is the bleach; allow sodium hypochlorite
- B1 chlorine is +1 in NaClO, since oxygen is −2 and the ClO⁻ ion carries a 1− charge
Explain, using oxidation states, why the reaction of chlorine with water is described as disproportionation.
Mark scheme
- B1 Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq)
- B1 chlorine goes from 0 in Cl₂ to −1 in HCl, so it is reduced, and from 0 to +1 in HClO, so it is oxidised
- B1 the same element is oxidised and reduced in the same reaction, which is disproportionation
Calculate the mass of sodium chlorate(I) formed when 6.00 dm³ of chlorine, measured at room temperature and pressure, reacts with excess cold dilute sodium hydroxide. Ar values: O 16.0, Na 23.0, Cl 35.5; molar gas volume = 24.0 dm³ mol⁻¹.
Mark scheme
- M1 amount of Cl₂ = 6.00 ÷ 24.0 = 0.250 mol
- M1 Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l) gives 1 mol NaClO per mol Cl₂, and Mr(NaClO) = 74.5
- A1 mass = 0.250 × 74.5 = 18.6 g, to three significant figures
Calculate the mass of silver chloride precipitated when excess acidified silver nitrate is added to 30.0 cm³ of 0.120 mol dm⁻³ sodium chloride solution. Ar values: Cl 35.5, Ag 107.9.
Mark scheme
- M1 amount of Cl⁻ = 30.0 ÷ 1000 × 0.120 = 3.60 × 10⁻³ mol
- M1 Ag⁺(aq) + Cl⁻(aq) → AgCl(s) is 1 : 1, and Mr(AgCl) = 143.4
- A1 mass = 3.60 × 10⁻³ × 143.4 = 0.516 g, to three significant figures
Compare what is seen when concentrated sulfuric acid is added to solid sodium chloride and to solid sodium iodide.
Mark scheme
- B1 both give misty white fumes of hydrogen halide, though far less with sodium iodide
- B1 sodium iodide also gives purple vapour or a black solid (iodine) and a yellow solid (sulfur)
- B1 sodium iodide gives a smell of bad eggs (hydrogen sulfide); sodium chloride gives nothing further, because chloride does not reduce the acid
Calculate the relative atomic mass of X, and identify X, when 0.250 g of a pure sodium halide NaX gives 0.613 g of dry silver halide AgX with excess acidified silver nitrate. Ar values: Na 23.0, Ag 107.9.
Mark scheme
- M1 amount of NaX = amount of AgX, so 0.250 ÷ (23.0 + x) = 0.613 ÷ (107.9 + x)
- M1 0.250 × (107.9 + x) = 0.613 × (23.0 + x), so 26.98 − 14.10 = 0.363x
- A1 x = 35.5 to three significant figures (Ar has no unit), so X is chlorine and the salt is sodium chloride
Deduce the oxidation state of chlorine in each chlorine-containing species in the reaction 3Cl₂(aq) + 6NaOH(aq) → 5NaCl(aq) + NaClO₃(aq) + 3H₂O(l), which takes place in hot concentrated alkali, and explain why the reaction is a disproportionation.
Mark scheme
- B1 chlorine is 0 in Cl₂ and −1 in NaCl
- B1 chlorine is +5 in NaClO₃, since three oxygens at −2 and the 1− charge of ClO₃⁻ leave +5
- B1 chlorine is both reduced, from 0 to −1, and oxidised, from 0 to +5, in the same reaction, so it disproportionates
Suggest why hydrogen fluoride is a weak acid in water while hydrogen chloride is a strong acid.
Mark scheme
- B1 the H–F bond is much stronger than the H–Cl bond, because the small fluorine atom holds the shared pair close to its nucleus
- B1 so hydrogen fluoride gives its proton to water less readily and ionises only partly, while hydrogen chloride ionises completely