pH, neutralisation, titrations (RP2) & strong/weak acids – GCSE Chemistry exam questions with mark scheme

1Grade 1–34 marks

(a) Solution X has a pH of 2. Solution Y has a pH of 7. Solution Z has a pH of 12.

Which solution is: (i) neutral (ii) alkaline? [2]

(b) Name the ion that all acids produce in aqueous solution. [1]

$\text{H}^+ + \ldots\ldots \rightarrow \text{H}_2\text{O}$

(a)(i) Y(1)
(a)(ii) Z(1)
(b) hydrogen ion / $\text{H}^+$ allow $\text{H}^+(\text{aq})$; do not accept hydrogen / H / $\text{H}_2$ (no ion or charge)(1)
(c) $\text{OH}^-$ allow hydroxide / hydroxide ion; ignore state symbols; do not accept OH or OH⁺ (missing / wrong charge)(1)
Answer: (a) Y neutral; Z alkaline (b) H⁺ (c) OH⁻
Examiner tip: Acids produce H⁺, alkalis produce OH⁻ — write the charge on the ion.

2Grade 8–9Higher5 marks

Hydrochloric acid is a strong acid. Ethanoic acid is a weak acid.

(a) Explain the difference between a strong acid and a weak acid. [2]

(b) Solution F has pH 2 and solution G has pH 4. How many times greater is the hydrogen ion concentration in F than in G? [1]

Identify the species that is oxidised. Explain your answer. [2]

(a) a strong acid is completely / fully ionised in aqueous solution allow dissociated for ionised(1)
(a) a weak acid is only partially ionised allow dissociated; do not accept answers in terms of dilute / concentrated(1)
(b) 100 (times) allow $10^2$(1)
(c) magnesium / Mg is oxidised allow magnesium atoms; do not accept $\text{Mg}^{2+}$ / magnesium ions(1)
(c) because it loses (two) electrons allow $\text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^-$; do not accept gains oxygen(1)
Answer: (a) strong fully ionised, weak partially ionised (b) 100 times (c) Mg — loses electrons
Examiner tip: Each pH unit is a factor of 10, so 2 units is ×100, not ×2 or ×20.

3Grade 1–3Foundation4 marks

A student added a few drops of universal indicator to samples of three household solutions.

Solution	Colour with universal indicator
vinegar	orange
washing-up liquid	blue
oven cleaner	purple

The diagram shows the colour chart for universal indicator.

(a) Use the colour chart to give the pH range of vinegar. [1]

(b) Which solution is the most alkaline? Give a reason for your answer. [1]

What colour was the universal indicator in the neutral mixture? [1]

(d) Give one advantage of using a pH probe instead of universal indicator to measure pH. [1]

(a) 3–4 allow 3 to 4 / between 3 and 4; do not accept a single value(1)
(b) oven cleaner and it has the highest pH / purple is the highest pH on the chart (11–14) allow pH 11–14; ignore 'strongest'; do not accept oven cleaner with no reason(1)
(c) green(1)
(d) any one from: gives a more accurate / precise (numerical) value; does not rely on judging a colour; can measure pH within a colour band / gives pH to decimal places ignore 'easier' / 'quicker' unqualified; do not accept 'more reliable' alone(1)
Answer: (a) pH 3–4 (b) oven cleaner — purple, pH 11–14, highest pH (c) green (d) e.g. gives a more precise numerical pH value
Examiner tip: In (b) the reason must be about the highest pH (or its colour on the chart) — naming oven cleaner alone scores nothing.

4Grade 4–55 marks

A student put 25.0 cm³ of dilute hydrochloric acid into a conical flask.

The student added sodium hydroxide solution to the acid 1 cm³ at a time and measured the pH with a pH probe after each addition.

The graph shows the results.

(a) Use the graph to give the pH of the hydrochloric acid before any sodium hydroxide was added. [1]

(b) Use the graph to determine the volume of sodium hydroxide solution needed to exactly neutralise the acid. [1]

(d) Name the ion, other than $\text{Na}^+$ and $\text{Cl}^-$, that is present in excess in the flask after 35 cm³ of sodium hydroxide solution has been added. [1]

(a) 1(1)
(b) 25 (cm³) allow 24.5–25.5(1)
(c) pH increases slowly at first / only rises to about 3 by 24 cm³ allow pH stays low / below 3 up to about 24 cm³(1)
(c) then rises sharply / rapidly at about 25 cm³ (and then levels off at about pH 12) levelling off is not required for this mark; do not accept 'the pH increases' with no description of the shape for either mark(1)
(d) hydroxide (ion) / $\text{OH}^-$ allow $\text{OH}^-(\text{aq})$; do not accept OH or sodium hydroxide(1)
Answer: (a) 1 (b) 25 cm³ (c) rises slowly, then very steeply at about 25 cm³, then levels off at about pH 12 (d) OH⁻
Examiner tip: 'The pH goes up' is not a description — you must say how fast it changes at the start, near 25 cm³ and at the end.

5Grade 4–5Triple only6 marks

A student carried out a titration to find the volume of dilute nitric acid needed to neutralise 25.0 cm³ of potassium hydroxide solution.

(a) Name the piece of apparatus used to measure exactly 25.0 cm³ of potassium hydroxide solution into the conical flask. [1]

(b) The student placed the conical flask on a white tile. Give the reason why. [1]

(c) The student wanted to use universal indicator. Explain why universal indicator is not suitable for a titration. [1]

(d) The burette reading at the start was 0.20 cm³. The diagram shows part of the burette at the end point.

(i) Give the burette reading at the end point. [1]

(ii) Calculate the volume of nitric acid added. [1]

(e) Near the end point the student added the acid one drop at a time. Explain why. [1]

(a) (volumetric) pipette allow pipette and filler; do not accept measuring cylinder / burette / dropper(1)
(b) so the colour change (of the indicator) is easier to see allow to see the end point clearly(1)
(c) the colour change is gradual / it has several colours so there is no sharp / single end point allow the end point is difficult to see; ignore 'not accurate' unqualified(1)
(d)(i) 23.35 (cm³) allow 23.30–23.40(1)
(d)(ii) 23.15 (cm³) allow ecf: their (d)(i) − 0.20(1)
(e) so that the end point is not overshot / so the exact volume at which the indicator changes colour is found allow so that too much acid is not added; ignore 'to make it accurate' unqualified(1)
Answer: (a) pipette (b) to see the colour change clearly (c) gradual colour change, no sharp end point (d)(i) 23.35 cm³ (ii) 23.15 cm³ (e) so the end point is not overshot
Examiner tip: Burettes read downwards and to 0.05 cm³ — read the bottom of the curved liquid surface, not the top.

6Grade 6–7HigherTriple only4 marks

A student titrated 25.0 cm³ of sodium hydroxide solution with sulfuric acid of concentration 0.0750 mol/dm³.

The mean volume of sulfuric acid needed to neutralise the sodium hydroxide was 18.40 cm³.

$$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

Calculate the concentration of the sodium hydroxide solution in mol/dm³.

Give your answer to 3 significant figures. [4]

moles of $\text{H}_2\text{SO}_4 = 0.0750 \times \dfrac{18.40}{1000} = 0.00138$ (mol) allow $1.38 \times 10^{-3}$(1)
moles of NaOH $= 2 \times 0.00138 = 0.00276$ (mol) allow ecf from their moles of acid(1)
concentration $= \dfrac{0.00276}{0.0250} = 0.1104$ allow ecf from their moles of NaOH(1)
0.110 (mol/dm³) to 3 significant figures allow ecf for a correctly rounded wrong value; correct answer with no working scores 4; if mole ratio not used (answer 0.0552) max 3(1)
Answer: 0.110 mol/dm³
Examiner tip: Use the 1 : 2 ratio from the equation — forgetting it gives 0.0552 mol/dm³, which loses a mark.

7Grade 8–9HigherTriple only6 marks

Limewater is a solution of calcium hydroxide, $\text{Ca(OH)}_2$.

A student titrated 25.0 cm³ samples of limewater with hydrochloric acid of concentration 0.0500 mol/dm³.

$$\text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}$$

The table shows the results.

Titration	Rough	1	2	3
Volume of hydrochloric acid in cm³	21.00	20.15	20.45	20.25

Concordant results are within 0.10 cm³ of each other.

Calculate the concentration of calcium hydroxide in the limewater in g/dm³.

Use only the concordant results.

Relative atomic masses ($A_r$): H = 1 O = 16 Ca = 40 [6]

mean titre $= \dfrac{20.15 + 20.25}{2} = 20.20$ (cm³) do not accept a mean that includes the rough or 20.45(1)
moles of HCl $= 0.0500 \times \dfrac{20.20}{1000} = 0.00101$ (mol) allow ecf from their mean(1)
moles of $\text{Ca(OH)}_2 = \dfrac{0.00101}{2} = 0.000505$ (mol) allow ecf(1)
concentration $= \dfrac{0.000505}{0.0250} = 0.0202$ (mol/dm³) allow ecf(1)
$M_r$ of $\text{Ca(OH)}_2 = 40 + 2(16 + 1) = 74$(1)
concentration $= 0.0202 \times 74 = 1.49$ (g/dm³) allow 1.4948 / 1.495 / 1.5; allow ecf; correct answer with no working scores 6(1)
Answer: 1.49 g/dm³
Examiner tip: Pick the concordant pair (20.15 and 20.25) — averaging all three titres or including the rough gives the wrong mean.

8Grade 6–7Higher4 marks

Stomach acid contains hydrochloric acid. A sample of stomach acid has a pH of 1.

An antacid tablet containing magnesium hydroxide was added to the sample. The pH changed to 4.

(a) By what factor did the concentration of hydrogen ions change? State whether it increased or decreased. [1]

(b) Solution P has a hydrogen ion concentration 10 000 times greater than solution Q.

Solution Q has a pH of 6. What is the pH of solution P? [1]

(a) decreased by a factor of 1000 allow $10^3$ / divided by 1000; do not accept 3 / factor of 3; both factor and direction needed(1)
(b) 2(1)
(c) hydroxide ions / $\text{OH}^-$ (from the antacid) react with / neutralise hydrogen ions / $\text{H}^+$ (to form water) allow $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$(1)
(c) so the concentration of hydrogen ions / $\text{H}^+$ decreases allow fewer $\text{H}^+$ ions (in the solution); do not accept 'the acid becomes weaker'(1)
Answer: (a) decreased by a factor of 1000 (b) pH 2 (c) OH⁻ ions react with H⁺ ions forming water, so [H⁺] falls
Examiner tip: Each pH unit is a factor of 10, so a change of 3 pH units is a factor of 10 × 10 × 10 = 1000, not 3 or 30.

9Grade 6–7Triple only6 marks

A student wants to find the volume of dilute sulfuric acid that reacts exactly with 25.0 cm³ of potassium hydroxide solution.

Describe a titration method the student could use to find this volume accurately.

In your answer you should name the apparatus and indicator used. [6]

use a (volumetric) pipette (and pipette filler) to measure 25.0 cm³ of potassium hydroxide solution(1)
put the alkali into a conical flask(1)
add a few drops of a suitable indicator, e.g. phenolphthalein or methyl orange (not universal indicator)(1)
fill a burette with sulfuric acid (using a funnel; remove the funnel; read with the eye level with the bottom of the meniscus)(1)
record the starting burette reading(1)
place the flask on a white tile(1)
add the acid from the burette to the alkali while swirling the flask(1)
add the acid drop by drop near the end point(1)
stop when the indicator changes colour (permanently), e.g. phenolphthalein pink → colourless, or methyl orange yellow → orange allow yellow → red(1)
record the final burette reading and subtract to find the volume of acid added(1)
do a rough titration first, then repeat until concordant results (within 0.10 cm³) are obtained(1)
calculate the mean of the concordant titres(1)

Level	Marks	What it looks like
Level 3	5–6	A logically sequenced, coherent method that would allow the reacting volume to be found accurately. Names the correct measuring apparatus for each solution and a suitable indicator, and includes steps that improve accuracy (e.g. dropwise near end point, swirling, repeating to get concordant results and finding a mean).
Level 2	3–4	A method that would give a reacting volume, with most key steps in a sensible order. Apparatus is mostly correct, but an accuracy step or the indicator/colour change may be missing or vague.
Level 1	1–2	Some relevant steps or apparatus are given, but the method is incomplete or poorly ordered and would not produce a valid result.
No relevant content	0	Nothing worthy of credit.

Answer: Pipette 25.0 cm³ KOH into a conical flask with a few drops of phenolphthalein on a white tile; add H₂SO₄ from a burette with swirling, dropwise near the end point, until pink → colourless; record the volume; repeat to get concordant titres and take the mean.
Examiner tip: Name an indicator with a sharp colour change and give its colour change — 'add an indicator' or using universal indicator caps the answer.

10Grade 6–7Higher5 marks

The table shows four acid solutions.

Solution	Acid	Concentration in mol/dm³
A	hydrochloric acid	2.0
B	hydrochloric acid	0.010
C	ethanoic acid	2.0
D	ethanoic acid	0.010

(a) Which solution, A, B, C or D, is a concentrated solution of a weak acid? [1]

(b) Solutions B and D have the same concentration. Explain why solution B has a lower pH than solution D. [2]

Explain why the student is wrong. [2]

(a) C(1)
(b) hydrochloric acid is a strong acid / is completely ionised (in aqueous solution) and ethanoic acid is a weak acid / is only partially ionised allow dissociated for ionised(1)
(b) so solution B has a higher concentration of hydrogen ions / $\text{H}^+$ allow more $\text{H}^+$ ions (per dm³) in B; do not accept 'more hydrogen' / 'more H'(1)
(c) concentration is the amount of acid in a given volume (of solution) allow how much acid is dissolved in the water(1)
(c) strength depends on how much the acid ionises in water, and ethanoic acid only partially ionises (whatever its concentration) allow dissociates; do not accept answers that confuse strong with concentrated(1)
Answer: (a) C (b) HCl fully ionised, ethanoic acid partially ionised, so B has a higher H⁺ concentration (c) concentrated means a lot of acid per volume; strength is about the degree of ionisation — ethanoic acid is weak at any concentration
Examiner tip: Never use 'strong' to mean 'concentrated': strong/weak is about ionisation, concentrated/dilute is about amount per volume.

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