Reactions of acids & making soluble salts (RP1) – GCSE Chemistry exam questions with mark scheme

1Grade 4–55 marks

(a) Name the salt formed when: (i) zinc reacts with sulfuric acid (ii) magnesium oxide reacts with hydrochloric acid. [2]

(b) Calcium carbonate reacts with hydrochloric acid. Write a balanced symbol equation for the reaction. [2]

(a)(i) zinc sulfate allow $\text{ZnSO}_4$(1)
(a)(ii) magnesium chloride allow $\text{MgCl}_2$(1)
(b) all formulae correct: $\text{CaCO}_3 + \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$(1)
(b) balanced: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$ allow multiples; ignore state symbols(1)
(c) bubble the gas through limewater, which turns milky / cloudy (both needed) allow white precipitate forms; ignore 'goes white' without 'milky/cloudy/precipitate'(1)
Answer: (a) zinc sulfate; magnesium chloride (b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (c) limewater turns milky
Examiner tip: The test needs the reagent and the result — “limewater” alone scores nothing.

2Grade 6–76 marks

Copper(II) oxide is an insoluble black solid. It reacts with dilute sulfuric acid to make copper(II) sulfate solution, which is blue.

Describe a method to make pure, dry crystals of copper(II) sulfate from copper(II) oxide and dilute sulfuric acid. [6]

measure a volume of dilute sulfuric acid (e.g. 50 cm³ with a measuring cylinder) into a beaker(1)
warm the acid gently (e.g. with a Bunsen burner or water bath) to speed up the reaction(1)
add copper(II) oxide a spatula at a time, stirring(1)
until copper(II) oxide is in excess / no more dissolves / black solid remains (so all the acid has reacted)(1)
filter to remove the excess copper(II) oxide(1)
heat the filtrate gently / in an evaporating basin to evaporate some of the water (until crystals start to form)(1)
leave to cool and crystallise(1)
remove crystals by filtering and pat dry with filter paper / leave in a warm place to dry(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that would give pure, dry crystals: excess insoluble base added to warm acid, excess removed by filtering, solution concentrated then left to crystallise, and crystals dried.
Level 2	3–4	Most key steps present and broadly in order, but one stage (e.g. removing excess, or crystallising/drying) is missing or unclear, so the product may not be pure or dry.
Level 1	1–2	Some relevant steps given (e.g. mixing the reactants, filtering or heating) but not linked into a workable sequence.
No relevant content	0	Nothing worthy of credit.

Answer: Warm acid, add CuO in excess, filter off excess, evaporate filtrate partly, leave to crystallise, filter and dry crystals.
Examiner tip: Do not boil the solution to dryness — you get a powder, not crystals — and say why the oxide is added in excess.

3Grade 1–3Foundation6 marks

A student adds a piece of magnesium ribbon to some dilute hydrochloric acid in a test tube.

(a) Complete the word equation for the reaction.

magnesium + hydrochloric acid → ____________ + ____________ [2]

(b) Give one thing the student would see during the reaction. [1]

(d) The student wants to make magnesium sulfate instead.

Which acid should the student react with magnesium?

Tick (✓) one box.

☐ citric acid ☐ hydrochloric acid ☐ nitric acid ☐ sulfuric acid [1]

(a) magnesium chloride allow $\text{MgCl}_2$; do not accept magnesium chlorine(1)
(a) hydrogen allow $\text{H}_2$ (either order); do not accept H(1)
(b) any one from: fizzing / bubbles (of gas) / effervescence; magnesium gets smaller / disappears / dissolves ignore 'gas given off' / 'hydrogen produced' (not an observation); ignore 'gets warm' / 'gets hot' (felt, not seen)(1)
(c) (hold a) lighted / burning splint at the mouth of the test tube do not accept glowing splint(1)
(c) (burns with a) squeaky pop allow 'pop' dependent on lighted splint for this mark(1)
(d) sulfuric acid (only box ticked) if more than one box ticked, no mark(1)
Answer: (a) magnesium chloride + hydrogen (b) fizzing (c) lighted splint → squeaky pop (d) sulfuric acid
Examiner tip: A glowing splint is the test for oxygen — writing it here loses the test mark and the result mark.

4Grade 4–55 marks

Salts can be made by reacting acids with different substances.

(a) Complete the table. [3]

Acid	Substance added	Salt made	Other product(s)
nitric acid	potassium hydroxide	(i) ____________	water
(ii) ____________	zinc oxide	zinc chloride	water
sulfuric acid	magnesium carbonate	magnesium sulfate	(iii) ____________

(b) Potassium hydroxide is an alkali. Zinc oxide is a base but is not an alkali.

Give the difference between an alkali and a base. [1]

(c) Give one observation you would make in the reaction of sulfuric acid with magnesium carbonate that you would not make in the reaction of the acid with zinc oxide. [1]

(a)(i) potassium nitrate allow $\text{KNO}_3$; do not accept potassium nitrite / potassium nitride(1)
(a)(ii) hydrochloric acid allow HCl(1)
(a)(iii) water and carbon dioxide (both needed) allow $\text{H}_2\text{O}$ and $\text{CO}_2$(1)
(b) an alkali is a base that dissolves in water / is soluble (in water) allow an alkali is a soluble hydroxide / makes $\text{OH}^-$ ions in solution; allow zinc oxide is insoluble (in water)(1)
(c) fizzing / bubbles / effervescence ignore 'carbon dioxide is given off' (not an observation)(1)
Answer: (a) potassium nitrate; hydrochloric acid; water and carbon dioxide (b) an alkali is a soluble base (c) fizzing
Examiner tip: Carbonates give three products — leaving out water or carbon dioxide in (iii) loses the mark.

5Grade 6–75 marks

The table shows the formulae of some ions.

Positive ions	Negative ions
aluminium $\text{Al}^{3+}$	chloride $\text{Cl}^-$
potassium $\text{K}^+$	nitrate $\text{NO}_3^{\;-}$
	sulfate $\text{SO}_4^{\;2-}$

(a) Use the table to give the formula of potassium sulfate. [1]

(b) Use the table to give the formula of aluminium nitrate. [1]

(c) Aluminium oxide, $\text{Al}_2\text{O}_3$, is an insoluble base. It reacts with dilute sulfuric acid, $\text{H}_2\text{SO}_4$, to make aluminium sulfate and water.

Write a balanced symbol equation for the reaction. [2]

(d) Explain why the formula of aluminium sulfate has two aluminium ions but three sulfate ions. [1]

(a) $\text{K}_2\text{SO}_4$ do not accept $\text{KSO}_4$, $\text{K}_2\text{SO}_4^{\,2-}$ or charges shown incorrectly; ignore correct charges e.g. $(\text{K}^+)_2\text{SO}_4^{\,2-}$(1)
(b) $\text{Al(NO}_3)_3$ (brackets needed) do not accept $\text{AlNO}_{33}$ / $\text{AlNO}_3$(1)
(c) correct formula of aluminium sulfate, $\text{Al}_2(\text{SO}_4)_3$, in an equation with $\text{Al}_2\text{O}_3$ and $\text{H}_2\text{SO}_4$ on the left and $\text{H}_2\text{O}$ on the right(1)
(c) fully balanced: $\text{Al}_2\text{O}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}$ allow multiples; ignore state symbols(1)
(d) the total positive charge must equal the total negative charge / charges must cancel: $2 \times (+3) = +6$ and $3 \times (-2) = -6$ allow 'so the compound is neutral / has no overall charge' with numbers or with reference to the charges 3+ and 2−(1)
Answer: (a) $\text{K}_2\text{SO}_4$ (b) $\text{Al(NO}_3)_3$ (c) $\text{Al}_2\text{O}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2\text{O}$ (d) 2(+3) = +6 balances 3(−2) = −6
Examiner tip: When an ion containing more than one atom appears more than once, it needs brackets: Al(NO₃)₃, not AlNO₃₃.

6Grade 8–9Higher6 marks

Iron reacts with dilute hydrochloric acid to make iron(II) chloride solution and hydrogen.

$$\text{Fe(s)} + 2\text{HCl(aq)} \rightarrow \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

The ionic equation for the reaction is:

$$\text{Fe(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$$

(a) The chloride ions are not shown in the ionic equation. Explain why. [1]

(b) Complete the half equation for the iron.

$$\text{Fe} \rightarrow \text{Fe}^{2+} + \underline{\qquad\qquad}$$ [1]

(c) Explain why the reaction between iron and hydrochloric acid is a redox reaction. Your answer should refer to electrons and name the species oxidised and the species reduced. [3]

(d) Zinc reacts with dilute sulfuric acid in a similar way. Write the ionic equation for this reaction. [1]

(a) they do not change / take part in the reaction / are spectator ions allow they are (dissolved) in solution both before and after the reaction(1)
(b) $2\text{e}^-$ allow 2e; do not accept $2\text{e}^+$ or $\text{e}^{2-}$(1)
(c) iron (atoms) lose (two) electrons, so iron is oxidised(1)
(c) hydrogen ions / $\text{H}^+$ gain electrons, so hydrogen ions are reduced do not accept 'hydrogen is reduced' unless clearly hydrogen ions; do not accept chloride ions reduced(1)
(c) oxidation and reduction happen at the same time / the electrons lost by iron are gained by the hydrogen ions (so it is a redox reaction) allow electrons are transferred from iron to hydrogen ions(1)
(d) $\text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2$ ignore state symbols; do not accept equations including $\text{SO}_4^{\,2-}$ on both sides (not ionic) / $\text{Zn}^{+}$(1)
Answer: (a) spectator ions (unchanged) (b) 2e⁻ (c) Fe loses electrons → oxidised; H⁺ gains electrons → reduced; both at once = redox (d) Zn + 2H⁺ → Zn²⁺ + H₂
Examiner tip: Say which species is reduced — 'hydrogen ions gain electrons', not just 'hydrogen is reduced' or 'the acid is reduced'.

7Grade 8–9Higher7 marks

A student makes copper(II) nitrate by reacting copper(II) carbonate with dilute nitric acid.

$$\text{CuCO}_3 + 2\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + \text{H}_2\text{O} + \text{CO}_2$$

The dilute nitric acid the student uses contains 6.30 g of $\text{HNO}_3$.

Relative atomic masses $(A_r)$: H = 1 C = 12 N = 14 O = 16 Cu = 63.5

(a) Calculate the minimum mass of copper(II) carbonate needed to react with all of the nitric acid.

Give your answer to 3 significant figures. [4]

(b) The student adds more copper(II) carbonate than the mass calculated in (a).

Give one reason why. [1]

(a) $M_r$ of $\text{HNO}_3 = 63$ and $M_r$ of $\text{CuCO}_3 = 123.5$(1)
(a) moles of $\text{HNO}_3 = \dfrac{6.30}{63} = 0.100$ (mol)(1)
(a) moles of $\text{CuCO}_3 = \dfrac{0.100}{2} = 0.0500$ (mol) allow ECF from incorrect moles of $\text{HNO}_3$(1)
(a) mass $= 0.0500 \times 123.5 = 6.18$ g (must be 3 s.f.) allow ECF; allow 6.175 for 3 marks only (not 3 s.f.); alternative method: 126 g of $\text{HNO}_3$ reacts with 123.5 g $\text{CuCO}_3$ (1), so 6.30 g reacts with $\frac{6.30}{126} \times 123.5$ (1) = 6.18 g (1); correct answer with no working scores 4(1)
(b) to make sure all of the (nitric) acid reacts / is used up allow so the copper nitrate is not contaminated with acid / so the excess can be filtered off leaving only the salt solution; ignore 'to speed up the reaction'; ignore 'to make more product' unless linked to all acid reacting(1)
(c) $M_r$ of $\text{Cu(NO}_3)_2 = 187.5$(1)
(c) mass $= 0.0500 \times 187.5 = 9.38$ g allow 9.375 g / 9.4 g; allow ECF from moles in (a); correct answer with no working scores 2(1)
Answer: (a) 6.18 g (b) so all the acid reacts (c) 9.38 g
Examiner tip: Most students lose a mark by forgetting the 1 : 2 ratio — divide the moles of acid by 2 before multiplying by the M_r of the carbonate.

8Grade 6–76 marks

A student investigated how much magnesium oxide reacts with 50 cm³ of warm dilute sulfuric acid.

$$\text{MgO} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\text{O}$$

This is the method used.

Warm 50 cm³ of dilute sulfuric acid in a beaker.
Add a known mass of magnesium oxide and stir for 5 minutes.
Filter the mixture. Dry the solid left in the filter paper and weigh it.
Repeat with different masses of magnesium oxide and fresh acid each time.

The graph shows the results and the student's lines of best fit.

(a) Use the graph to determine the mass of magnesium oxide that reacts exactly with 50 cm³ of the acid. [1]

(b) Explain the shape of the graph. [2]

Give the mass of magnesium oxide added for the anomalous result. Suggest one reason for the anomalous result. [2]

(d) Predict the mass of magnesium oxide that would react exactly with 100 cm³ of the same acid. [1]

(a) 2.2 g allow 2.1–2.3 g(1)
(b) (up to 2.2 g) all the magnesium oxide reacts / there is no solid left because the acid is in excess / not all the acid has reacted(1)
(b) (after 2.2 g) all the acid has been used up, so any extra magnesium oxide does not react / stays as solid allow the mass left increases by the same amount as the extra mass added(1)
(c) 3.0 g (added) allow the point at 1.05 g unreacted(1)
(c) the solid was not completely dry / still contained water (so its mass was too high) allow the student misread / incorrectly recorded the balance; allow less acid was used / the acid was measured inaccurately (less than 50 cm³); ignore 'human error' or 'did not stir' unqualified(1)
(d) 4.4 g allow 2 × answer to (a)(1)
Answer: (a) 2.2 g (b) acid in excess so all MgO reacts until 2.2 g; then acid used up so extra MgO is left (c) 3.0 g; solid not fully dried (d) 4.4 g
Examiner tip: In (b) you must explain both sections of the graph in terms of which reactant is in excess — just describing 'it stays flat then goes up' scores nothing.

9Grade 6–76 marks

A student wanted to make pure, dry crystals of zinc sulfate from zinc carbonate and dilute sulfuric acid.

Zinc carbonate is an insoluble white solid.

This is the student's method.

Pour 40 cm³ of dilute sulfuric acid into a beaker.
Add one spatula of zinc carbonate and stir.
Pour all of the mixture into an evaporating basin.
Heat the evaporating basin strongly with a Bunsen burner until no liquid is left.

The student did not get pure, dry crystals of zinc sulfate.

Explain how the student's method should be improved to make pure, dry crystals of zinc sulfate. [6]

Step 2: one spatula may not be enough / some acid will be left unreacted, which would contaminate the salt(1)
Step 2 improvement: add zinc carbonate a little at a time until it is in excess / until fizzing stops / until solid remains (so all the acid has reacted)(1)
warm the acid (gently) first / stir, so the reaction is faster (credit as part of a good method)(1)
Step 3: the mixture was not filtered, so excess / unreacted zinc carbonate would be left in the product(1)
Step 3 improvement: filter the mixture and keep the filtrate (zinc sulfate solution)(1)
Step 4: heating strongly to dryness gives a powder (not crystals) / the solid may spit / decompose(1)
Step 4 improvement: heat gently / use a water bath or electric heater to evaporate only some of the water / until crystals start to form (e.g. on a glass rod)(1)
then leave the solution to cool so crystals form slowly(1)
remove the crystals (e.g. by filtering) and dry them between filter paper / in a warm oven / desiccator(1)
safety: wear eye protection, as hot acid / solution may spit(1)

Level	Marks	What it looks like
Level 3	5–6	Identifies the faults in at least two of steps 2, 3 and 4, explains why each stops the product being pure or dry, and gives a clear, logically ordered improved method that would produce pure, dry crystals (excess, filter, partial evaporation, crystallise, dry).
Level 2	3–4	Gives improvements to at least two steps, with some explanation of why they are needed, but the explanation or sequence is incomplete, so the improved method may not fully produce pure, dry crystals.
Level 1	1–2	Identifies one or more faults or suggests one or more simple improvements (e.g. 'filter it' or 'do not boil it dry') with little or no explanation.
No relevant content	0	Nothing worthy of credit.

Answer: Add ZnCO₃ until in excess (fizzing stops, solid left); filter off excess; warm filtrate gently (water bath) only until crystals start to form; cool to crystallise; filter off and dry crystals between filter paper.
Examiner tip: Saying 'add more zinc carbonate' is not enough — say until it is in excess / fizzing stops and explain that this uses up all the acid.

10Grade 4–55 marks

Soluble salts can be made by adding an insoluble substance to an acid until it is in excess, then filtering.

(a) A student tries to make copper(II) chloride by adding copper metal to dilute hydrochloric acid. No reaction happens.

(i) Explain why copper does not react with dilute hydrochloric acid. [1]

(ii) Name a copper compound the student could add to dilute hydrochloric acid to make copper(II) chloride. [1]

(b) Another student wants to make sodium chloride. She suggests adding sodium metal to dilute hydrochloric acid.

Give one reason why this is not a suitable method. [1]

Explain why the method of adding the sodium hydroxide in excess and then filtering would not produce pure sodium chloride. [2]

(i) copper is less reactive than hydrogen / copper is below hydrogen in the reactivity series allow copper is too unreactive / not reactive enough; ignore 'copper is a transition metal' / 'copper is unreactive' without comparison(1)
(ii) copper(II) oxide / copper(II) hydroxide / copper(II) carbonate allow copper oxide / CuO etc.; do not accept copper sulfate / copper chloride(1)
(b) the reaction is too violent / dangerous / explosive allow sodium is too reactive (with acid) ignore 'sodium reacts with water' unqualified(1)
(c) the excess sodium hydroxide dissolves / is in solution, so it cannot be removed by filtering(1)
(c) so the sodium chloride (solution / crystals) would be contaminated with sodium hydroxide / would contain sodium hydroxide as an impurity allow the solution would be alkaline (and impure)(1)
Answer: (a)(i) Cu is less reactive than hydrogen (ii) copper(II) oxide / hydroxide / carbonate (b) reaction too violent (c) excess NaOH is dissolved, can't be filtered off, so salt contaminated with NaOH
Examiner tip: In (c) the key word is soluble: filtering only removes solids, so a dissolved excess stays in the salt solution.

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