Edexcel International A Level Chemistry: Acid-base equilibria: titration curves and buffers (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers outcomes 14.15 to 14.23 of Topic 14: Acid-base Equilibria in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, including Core Practical 11. The content is in Unit 4: Rates, Equilibria and Further Organic Chemistry, an International A2 (IA2) unit. If you have not yet worked through the study guide and the revision notes, start there. Log secure outcomes on the Edexcel Chemistry checklist; other units are on the course hub.

Assume 298 K and Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. Use the data given in each question.

Questions

1. Define a buffer solution, and name the two types of species present in an acidic buffer. [2]

2. 20.0 cm³ of 0.150 mol dm⁻³ nitric acid is titrated with 0.120 mol dm⁻³ potassium hydroxide. Calculate the volume of alkali at the equivalence point, the starting pH, and state the pH at the equivalence point. [3]

3. Use these indicator ranges: methyl orange 3.1–4.4; bromothymol blue 6.0–7.6; phenolphthalein 8.2–10.0; alizarin yellow 10.1–12.0.

(a) A titration curve has a steep section from pH 7.0 to pH 11.0. Choose the indicator and explain why the others are unsuitable. [2] (b) Another curve has a steep section from pH 2.9 to pH 6.6. Choose the indicator and give the type of titration. [2] (c) A student titrates propanoic acid with aqueous ammonia using phenolphthalein. Her three titres differ by more than 2 cm³. Explain this result and suggest a better way to find the equivalence volume. [2]

4. 50.0 cm³ of 0.240 mol dm⁻³ weak acid HM (Ka = 3.2 × 10⁻⁵ mol dm⁻³) is mixed with 30.0 cm³ of 0.160 mol dm⁻³ NaOH. Calculate the pH of the mixture. [5]

5. A mixture of aqueous ammonia and ammonium chloride is a buffer. Explain, with equations, how it resists a change in pH when small amounts of (i) acid and (ii) alkali are added. [4]

6. 25.0 cm³ of sodium carbonate solution is titrated with 0.0950 mol dm⁻³ HCl. The curve has two steep sections; the first is centred on 13.20 cm³.

(a) Write ionic equations for the reactions in each stage. [2] (b) Calculate the concentration of the sodium carbonate, and state the volume at the second steep section. [2] (c) Name a suitable indicator for each end point. [2]

7. A weak acid HR has Ka = 4.5 × 10⁻⁵ mol dm⁻³. Its sodium salt NaR has Mr = 110.0.

(a) Calculate the mass of NaR that must be dissolved in 200 cm³ of 0.125 mol dm⁻³ HR to give a buffer of pH 4.60. State one assumption. [5] (b) Instead, equal volumes of 0.150 mol dm⁻³ HR and a NaR solution are mixed. Calculate the concentration of NaR solution needed for pH 4.60. [2]

8. In Core Practical 11, a student finds Ka for a weak monoprotic acid.

(a) Describe how the student could prepare a half-neutralised solution of the acid, starting from the acid, NaOH solution of unknown concentration and phenolphthalein. [3] (b) The half-neutralised solution has pH 4.12. Calculate Ka. [2] (c) Explain why the student does not need to know the concentration of either solution, and state why the pH meter is first calibrated. [2]

9. 20.0 cm³ of 0.0750 mol dm⁻³ weak acid HG (pKa = 4.31) is titrated with 0.0600 mol dm⁻³ NaOH.

(a) Calculate the pH before any alkali is added, stating the assumption made. [3] (b) State the pH when 12.50 cm³ of NaOH has been added. [1] (c) Calculate the pH after 30.0 cm³ of NaOH has been added. [3] (d) Suggest why the solution in the flask is alkaline at the equivalence point. [1]

10. Blood and many foods contain buffers.

(a) Write the equilibrium for the main buffer in blood. [1] (b) Explain how this buffer limits the rise in pH if a small amount of base enters the blood. [2] (c) Explain why some foods contain added buffers. [2]

11. 250 cm³ of a buffer contains 0.120 mol dm⁻³ weak acid HA (Ka = 2.6 × 10⁻⁵ mol dm⁻³) and 0.0900 mol dm⁻³ NaA.

(a) Calculate the pH of the buffer. [2] (b) The buffer receives 3.00 cm³ of 2.00 mol dm⁻³ HCl. Find the new pH. [3] (c) Find the pH if that HCl went into 250 cm³ of pure water instead. [2] (d) The original buffer is diluted to 500 cm³ with water. State and explain the effect on pH. [2]

12. A diprotic weak acid H₂X is titrated with NaOH. Steep sections are centred on 11.0 cm³ and 22.0 cm³. The pH is 2.90 at 5.50 cm³ and 6.10 at 16.50 cm³. State pKa₁ and pKa₂ and explain how you identified each. [3]

Answers

1. A solution whose pH changes very little when a small amount of acid or alkali is added [1]. It contains a weak acid and its conjugate base (or a salt of the weak acid) [1]. Examiner insight: “keeps the pH constant” overstates what a buffer does; say the pH changes only slightly, and only for small additions.

2. Moles HNO₃ = 0.0200 × 0.150 = 3.00 × 10⁻³ mol; volume = 3.00 × 10⁻³/0.120 = 25.0 cm³ [1]. Starting pH = −log 0.150 = 0.82 [1]. Equivalence pH = 7 (strong acid with strong base) [1]. Examiner insight: quote pH to two decimal places unless told otherwise.

3. (a) Phenolphthalein: its range lies wholly within pH 7.0–11.0 [1]. Bromothymol blue starts below 7.0, alizarin yellow ends above 11.0 and methyl orange changes far too early [1]. (b) Methyl orange [1]; strong acid with weak base [1]. (c) With no steep section, the colour changes gradually over several cm³ [1]; use a pH meter, plot the curve and read the equivalence volume from it [1]. Examiner insight: refer to the whole range lying inside the steep section, not to the equivalence pH alone.

4. Moles HM = 0.0500 × 0.240 = 0.0120 mol [1]; moles OH⁻ = 0.0300 × 0.160 = 4.80 × 10⁻³ mol [1]. HM left = 7.20 × 10⁻³ mol, M⁻ = 4.80 × 10⁻³ mol [1]. [H⁺] = 3.2 × 10⁻⁵ × 7.20/4.80 = 4.8 × 10⁻⁵ mol dm⁻³ [1]; pH = 4.32 [1]. Examiner insight: show the subtraction; a ratio built from starting moles is wrong.

5. (i) Added H⁺ reacts with ammonia [1]: NH₃ + H⁺ → NH₄⁺ [1]. (ii) Added OH⁻ reacts with ammonium ions [1]: NH₄⁺ + OH⁻ → NH₃ + H₂O [1]. Examiner insight: an explanation with no equations is unlikely to earn full credit; write one equation for each addition.

6. (a) CO₃²⁻ + H⁺ → HCO₃⁻ [1]; HCO₃⁻ + H⁺ → H₂O + CO₂ [1]. (b) Moles HCl = 0.01320 × 0.0950 = 1.254 × 10⁻³ mol = moles CO₃²⁻; concentration = 1.254 × 10⁻³/0.0250 = 0.0502 mol dm⁻³ [1]. Second steep section at 26.40 cm³ [1]. (c) Phenolphthalein for the first [1]; methyl orange for the second [1]. Examiner insight: use the first end point (1 : 1) for the carbonate moles; using the second without halving doubles the answer.

7. (a) [H⁺] = 10^-4.60 = 2.51 × 10⁻⁵ mol dm⁻³ [1]. [R⁻] = Ka × [HR]/[H⁺] [1] = 4.5 × 10⁻⁵ × 0.125/2.51 × 10⁻⁵ = 0.224 mol dm⁻³ [1]. Moles = 0.224 × 0.200 = 0.0448 mol [1]; mass = 0.0448 × 110.0 = 4.93 g, assuming the volume does not change on dissolving [1]. (b) Ratio [R⁻]/[HR] = Ka/[H⁺] = 1.79, unchanged by mixing equal volumes [1]; [NaR] = 1.79 × 0.150 = 0.269 mol dm⁻³ [1]. Examiner insight: in (b), halving only one concentration gives a wrong ratio.

8. (a) Titrate a measured portion (e.g. 25.0 cm³) of the acid with NaOH using phenolphthalein to the first permanent pink [1]. Take an identical portion of acid [1] and add exactly half the titre of NaOH (or mix the neutralised solution with an equal portion of acid) [1]. (b) pKa = pH = 4.12 [1]; Ka = 10^-4.12 = 7.6 × 10⁻⁵ mol dm⁻³ [1]. (c) At half-neutralisation [HA] = [A⁻], so they cancel in the Ka expression whatever their values [1]. Calibration with buffers of known pH makes the readings accurate [1]. Examiner insight: name the apparatus (pipette, burette); “add some NaOH” is too vague.

9. (a) Ka = 10^-4.31 = 4.90 × 10⁻⁵ [1]; [H⁺] = √(4.90 × 10⁻⁵ × 0.0750) = 1.92 × 10⁻³, pH = 2.72 [1]; assume [HG] at equilibrium equals 0.0750 because dissociation is slight [1]. (b) 12.50 cm³ is half of the 25.0 cm³ equivalence volume, so pH = pKa = 4.31 [1]. (c) Excess OH⁻ = (30.0 − 25.0) × 10⁻³ × 0.0600 = 3.00 × 10⁻⁴ mol [1]; [OH⁻] = 3.00 × 10⁻⁴/0.0500 = 6.0 × 10⁻³ mol dm⁻³ [1]; pH = 14.00 − 2.22 = 11.78 [1]. (d) The salt’s anion G⁻ is a weak base that accepts protons from water, forming OH⁻ [1]. Examiner insight: in (c), divide by the total volume (50.0 cm³), not the volume of alkali.

10. (a) H₂CO₃ ⇌ H⁺ + HCO₃⁻ [1]. (b) H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O [1]; H₂CO₃ and HCO₃⁻ far exceed the base added, so their ratio and the pH (close to 7.4) barely change [1]. (c) Bacterial or fungal activity can change the pH of food [1]; a buffer resists the change, slowing deterioration [1]. Examiner insight: name the actual species, HCO₃⁻ and H₂CO₃; “the buffer soaks up the base” is too vague.

11. (a) [H⁺] = 2.6 × 10⁻⁵ × 0.120/0.0900 = 3.47 × 10⁻⁵ [1]; pH = 4.46 [1]. (b) Moles: HA 0.0300, A⁻ 0.0225; H⁺ added 6.00 × 10⁻³ [1]. HA 0.0360, A⁻ 0.0165 [1]. [H⁺] = 2.6 × 10⁻⁵ × 0.0360/0.0165 = 5.67 × 10⁻⁵; pH = 4.25 [1]. (c) [H⁺] = 6.00 × 10⁻³/0.253 = 0.0237 mol dm⁻³ [1]; pH = 1.62 [1]. (d) pH stays almost the same [1]: doubling the volume halves both concentrations, which leaves [HA]/[A⁻] at 0.120/0.0900 [1]. Examiner insight: in (b), work in moles so the change in volume cancels.

12. pKa₁ = 2.90: 5.50 cm³ is half the first equivalence volume, so [H₂X] = [HX⁻] [1]. pKa₂ = 6.10 [1]: 16.50 cm³ is midway between the equivalence points, so [HX⁻] = [X²⁻] [1]. Examiner insight: explain each value with the pair of species at equal concentration.

Where marks are usually lost

• Assuming every equivalence point is at pH 7.
• Picking an indicator that begins or ends its colour change outside the steep section.
• Using starting moles, not moles left after reaction, in a buffer ratio.
• Inverting the ratio in [H⁺] = Ka × [HA]/[A⁻].
• Taking pKa at the equivalence volume instead of half of it.
• Buffer explanations with no equation.
• Forgetting the two-step stoichiometry for diprotic species.

Next steps

• Recap with the titration curves and buffers revision notes.
• Reread the worked examples in the study guide.
• See every topic on the course hub.
• Mark your progress on the Edexcel Chemistry checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 14: Acid-base Equilibria, outcomes 14.15 to 14.23, including Core Practical 11.