Acids, alkalis, indicators and titrations – IGCSE Chemistry exam questions with mark scheme

1Grade 1–34 marks

Indicators change colour depending on whether a solution is acidic or alkaline.

Complete the table by giving the colour of each indicator in each solution. One has been done for you. [4]

indicator	colour in dilute hydrochloric acid	colour in sodium hydroxide solution
litmus	red	(i) ..................
phenolphthalein	(ii) ..................	(iii) ..................
methyl orange	(iv) ..................	yellow

(i) blue (litmus in alkali) (reject purple)(1)
(ii) colourless (phenolphthalein in acid) (reject clear; reject white)(1)
(iii) pink (phenolphthalein in alkali) (accept magenta; ignore qualifiers such as pale/bright)(1)
(iv) red (methyl orange in acid) (accept pink; reject orange)(1)
Answer: (i) blue; (ii) colourless; (iii) pink; (iv) red
Examiner tip: Writing 'clear' for phenolphthalein in acid loses the mark: clear means transparent, and the answer needed is colourless.

2Grade 1–34 marks

The diagram shows a pH scale. The letters A to E show the pH values of five different solutions.

(a) Give the letter of the solution that is strongly alkaline. [1]

(b) Give the letter of the solution that is weakly acidic. [1]

(d) Give the letter of the solution that contains the highest concentration of hydrogen ions, H⁺. [1]

(a) E(1)
(b) B(1)
(c) neutral (ignore 'water'; reject 'weak')(1)
(d) A(1)
Answer: (a) E (pH 13); (b) B (pH 5); (c) neutral; (d) A (pH 1)
Examiner tip: In (d) students pick E because 13 is the biggest number, but the lower the pH, the more H⁺ ions there are.

3Grade 4–54 marks

A gardener wants to find the pH of the soil in a vegetable patch. A student shakes a sample of the soil with distilled water and filters the mixture to obtain a colourless solution.

(a) Describe how the student could use universal indicator to estimate the pH of the solution. [2]

(b) Give one reason why universal indicator is more useful than litmus for this test. [1]

(c) A second soil solution turns universal indicator purple. Describe this solution using a term from the pH scale. [1]

(a) add (a few drops of) universal indicator to (a sample of) the solution (accept dip universal indicator paper into the solution / put a drop of solution on UI paper)(1)
(a) compare the colour with a (pH) colour chart (and read off the pH) (ignore 'see what colour it turns' without comparison)(1)
(b) it gives an (approximate) pH value / shows how acidic or alkaline the solution is (accept shows the strength of acid or alkali; accept it has a range of colours) (reject 'it shows if it is acid or alkali' alone, as litmus does that too)(1)
(c) strongly alkaline (both words needed; reject 'alkaline' alone)(1)
Answer: (a) Add a few drops of universal indicator and match the colour against a pH colour chart. (b) It gives a pH value / degree of acidity, not just acid or alkali. (c) Strongly alkaline.
Examiner tip: The second mark in (a) needs the colour chart: just saying 'look at the colour' does not explain how the pH is found.

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4Grade 4–55 marks

Dilute nitric acid, HNO₃, is neutralised by potassium hydroxide solution, KOH.

(a) Give the formula of the ion that is present in all acids in aqueous solution. [1]

(b) Give the formula of the ion that makes potassium hydroxide solution alkaline. [1]

(d) Write the ionic equation for the neutralisation reaction. Include state symbols. [2]

(a) H⁺ (accept H⁺(aq); accept H₃O⁺; reject H, H₂, 'hydrogen' without charge)(1)
(b) OH⁻ (accept HO⁻; reject OH without charge)(1)
(c) potassium nitrate (reject potassium nitrite; ignore formula KNO₃ if name also correct)(1)
(d) H⁺ + OH⁻ → H₂O (correct species, balanced; ignore spectator ions K⁺/NO₃⁻ if shown on both sides)(1)
(d) state symbols (aq), (aq) → (l) (dependent on correct equation)(1)
Answer: (a) H⁺; (b) OH⁻; (c) potassium nitrate; (d) H⁺(aq) + OH⁻(aq) → H₂O(l)
Examiner tip: Writing the full equation KOH + HNO₃ → KNO₃ + H₂O in (d) scores nothing: the ionic equation shows only the ions that react.

5Grade 4–56 marks

A student places 25.0 cm³ of dilute hydrochloric acid in a beaker. The student adds sodium hydroxide solution a little at a time, stirring and measuring the pH with a pH meter after each addition.

The graph shows the results.

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

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

(c) The experiment is repeated with a few drops of phenolphthalein added to the acid.
Give the colour of the mixture after 10 cm³ of sodium hydroxide solution has been added and after 30 cm³ has been added. [2]

(d) Explain, in terms of ions, why the pH increases as sodium hydroxide solution is added. [2]

(a) 1.1 (allow any value from 1.0 to 1.2)(1)
(b) 20.0 cm³ (allow 19.8 to 20.2; read at the vertical section / pH 7)(1)
(c) after 10 cm³: colourless (reject clear)(1)
(c) after 30 cm³: pink (accept magenta)(1)
(d) hydroxide ions / OH⁻ react with hydrogen ions / H⁺ (to form water) (accept H⁺ + OH⁻ → H₂O)(1)
(d) so the concentration of H⁺ ions decreases (accept fewer H⁺ ions; then excess OH⁻ ions after neutralisation) (ignore 'acid is neutralised' without ions)(1)
Answer: (a) about 1.1; (b) 20.0 cm³; (c) colourless, then pink; (d) OH⁻ ions react with H⁺ ions to form water, so [H⁺] falls (and OH⁻ is in excess after 20 cm³).
Examiner tip: In (d) 'the alkali neutralises the acid' is not enough: you must name the H⁺ and OH⁻ ions and say H⁺ is used up.

6Grade 4–5Triple only4 marks

A student carries out a titration to find the volume of dilute nitric acid needed to neutralise 25.0 cm³ of lithium hydroxide solution. The acid is added from a burette to the alkali in a conical flask.

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

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

(d) Explain why universal indicator is not suitable for this titration. [1]

(a) (volumetric) pipette (reject measuring cylinder; allow burette; ignore 'with pipette filler')(1)
(b) to see the colour change (of the indicator) clearly (accept to make the end point easier to see; ignore 'to see better' without colour)(1)
(c) to mix the acid and alkali (so that they react completely) (accept to make sure the reaction happens throughout; ignore 'to speed up the reaction' alone)(1)
(d) it changes colour gradually / has many colours so there is no sharp / sudden colour change at the end point (accept end point hard to see; reject 'it does not change colour')(1)
Answer: (a) pipette; (b) to see the colour change clearly; (c) to mix the solutions so they react fully; (d) its colour changes gradually, so there is no sharp end point.
Examiner tip: Measuring cylinder is the most common wrong answer in (a): it is not accurate enough to measure exactly 25.0 cm³.

7Grade 6–7Triple only5 marks

A student titrates 25.0 cm³ portions of potassium hydroxide solution with dilute sulfuric acid. The diagram shows part of the burette at the end of titration 3.

	rough	titration 1	titration 2	titration 3
final burette reading in cm³	24.20	24.80	25.55	(a) ..........
initial burette reading in cm³	0.50	1.20	2.10	1.00
volume of acid added in cm³	23.70	23.60	23.45	(b) ..........

(a) Use the diagram to give the final burette reading for titration 3. [1]

(b) Calculate the volume of acid added in titration 3. [1]

(c) Concordant results are within 0.10 cm³ of each other. Identify the concordant results and use them to calculate the mean volume of acid added. [2]

(d) Give a reason why the student does not use the rough titration when calculating the mean. [1]

(a) 24.35 (cm³) (must be to 2 d.p.; reject 24.3, 24.4, 24.45)(1)
(b) 23.35 (cm³) (allow ecf from (a): (a) − 1.00, given to 2 d.p.)(1)
(c) titrations 2 and 3 chosen / 23.45 and 23.35 used (allow ecf from (b) if a different pair is then concordant)(1)
(c) mean = 23.40 (cm³) (accept 23.4; reject an answer that includes titration 1 or the rough value)(1)
(d) the acid is added quickly / it is only used to find the approximate end point, so the end point is overshot / it is less accurate (accept it is a trial run)(1)
Answer: (a) 24.35 cm³; (b) 23.35 cm³; (c) titrations 2 and 3; mean = 23.40 cm³; (d) the rough titration is done quickly to find the approximate end point, so it overshoots.
Examiner tip: Burette readings must be read from the bottom of the meniscus and recorded to 2 decimal places, ending in 0 or 5.

8Grade 6–7Triple only5 marks

Limewater is a solution of calcium hydroxide, Ca(OH)₂. A student titrates 25.0 cm³ of limewater with dilute hydrochloric acid of concentration 0.0500 mol/dm³, using methyl orange indicator. The mean titre is 18.40 cm³.

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

(a) Give the colour change at the end point. [1]

(b) Calculate the concentration of calcium hydroxide in the limewater in g/dm³.
[$M_r$ of Ca(OH)₂ = 74] [4]

(a) yellow to orange (accept yellow to red; colours must be in the correct order; reject orange to yellow)(1)
(b) moles HCl = 0.0500 × 18.40 / 1000 = 9.20 × 10⁻⁴ (mol)(1)
(b) moles Ca(OH)₂ = 9.20 × 10⁻⁴ ÷ 2 = 4.60 × 10⁻⁴ (mol) (allow ecf: their moles HCl ÷ 2)(1)
(b) concentration = 4.60 × 10⁻⁴ ÷ 0.0250 = 0.0184 (mol/dm³) (allow ecf)(1)
(b) 0.0184 × 74 = 1.36 (g/dm³) (accept 1.3616, 1.4; correct final answer with no working scores 4)(1)
Answer: (a) yellow to orange (or red); (b) 1.36 g/dm³
Examiner tip: The most common slip is forgetting to halve the moles of HCl: the equation shows 2 HCl react with each Ca(OH)₂.

9Grade 6–7Triple only6 marks

A student wants to find the exact volume of dilute hydrochloric acid needed to neutralise 25.0 cm³ of sodium hydroxide solution. This is the student's method.

Measure 25.0 cm³ of sodium hydroxide solution using a measuring cylinder and pour it into a beaker.
Add a few drops of universal indicator.
Fill a burette with the acid, reading the top of the meniscus.
Pour the acid into the beaker quickly until the colour changes.
Record the volume of acid added. Do the experiment once.

Evaluate the student's method and describe how it could be improved to give an accurate and reliable result. [6]

measuring cylinder is not accurate enough: use a (volumetric) pipette (with a pipette filler) to measure the 25.0 cm³ (allow burette)(1)
use a conical flask (instead of a beaker) so it can be swirled to mix without spilling (allow place the flask on a white tile to see the colour change)(1)
universal indicator changes colour gradually / has no sharp end point: use methyl orange / phenolphthalein (reject universal indicator / litmus)(1)
read the burette at the bottom of the meniscus (at eye level) (allow rinse the burette with the acid before filling)(1)
add the acid dropwise near the end point (while swirling) so the end point is not overshot / stop at the first permanent colour change(1)
repeat the titration until concordant results (within 0.10 cm³) are obtained and calculate the mean of the concordant titres (allow do a rough titration first; ignore 'repeat' alone without concordant or mean)(1)
Answer: Use a pipette (not a measuring cylinder), a conical flask on a white tile, methyl orange or phenolphthalein (not UI), read the bottom of the meniscus, add acid dropwise near the end point while swirling, and repeat to get concordant titres and take a mean.
Examiner tip: Each mark needs the improvement AND what it fixes: 'repeat it' alone scores nothing, but 'repeat until titres are concordant and take a mean' does.

10Grade 8–9Triple only5 marks

A drain cleaner contains concentrated potassium hydroxide solution, KOH. A technician dilutes 10.0 cm³ of the drain cleaner with distilled water to make exactly 250.0 cm³ of solution.

25.0 cm³ portions of the diluted solution are titrated with sulfuric acid of concentration 0.100 mol/dm³. The mean titre is 21.60 cm³.

$$2\text{KOH(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{K}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$$

Calculate the concentration of potassium hydroxide in the original drain cleaner, in g/dm³.
[$M_r$ of KOH = 56] [5]

moles H₂SO₄ = 0.100 × 21.60 / 1000 = 2.16 × 10⁻³ (mol)(1)
moles KOH in 25.0 cm³ = 2 × 2.16 × 10⁻³ = 4.32 × 10⁻³ (mol) (allow ecf)(1)
moles KOH in 250.0 cm³ (= in 10.0 cm³ of drain cleaner) = 4.32 × 10⁻³ × 10 = 0.0432 (mol) (allow ecf)(1)
concentration of drain cleaner = 0.0432 ÷ 0.0100 = 4.32 (mol/dm³) (allow ecf; alternative: diluted conc 0.1728 mol/dm³ × 25 dilution factor = 4.32)(1)
4.32 × 56 = 242 (g/dm³) (accept 241.92, 241.9, 240 to 2 s.f.; allow ecf; correct answer with no working scores 5)(1)
Answer: 4.32 mol/dm³ = 242 g/dm³
Examiner tip: Students often stop at the diluted solution: the 10.0 cm³ was made up to 250.0 cm³, so the original is 25 times more concentrated.

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