Acids and Bases MCQs — 40 Practice Questions with Answers & Explanations

Test yourself with 40 free, original acids and bases MCQs, each with a worked answer and a short explanation. The questions start with GCSE and IGCSE core ideas — the pH scale, indicators, neutralisation and salts — and build up to the A-Level and AP stretch topics of pH calculations, Ka, pKa and buffer solutions. Work through them in order, or jump straight to the section you are revising.

Quick formula recap: need the key equations before you start? Open the formula sheet for your exam board:

Section 1: Acids, Alkalis & the pH Scale

Q1. What does the pH of a solution actually measure?

  • (A) The mass of acid dissolved in the solution
  • (B) The concentration of hydrogen ions in the solution
  • (C) The volume of alkali needed to neutralise the solution
  • (D) The temperature change during neutralisation
Show Answer

Answer: B — pH is a measure of the hydrogen ion concentration, [H+]. The lower the pH, the higher the [H+] and the more acidic the solution.

Q2. A few drops of universal indicator are added to a solution of pH 13. What colour is seen?

  • (A) Red
  • (B) Green
  • (C) Purple
  • (D) Orange
Show Answer

Answer: C — pH 13 is strongly alkaline, so universal indicator turns purple (violet). It is green at pH 7 and red or orange in acids.

Q3. What colour is phenolphthalein in an acidic solution?

  • (A) Pink
  • (B) Colourless
  • (C) Blue
  • (D) Yellow
Show Answer

Answer: B — Phenolphthalein is colourless in acid and only turns pink in alkali, changing over roughly pH 8.2–10.

Q4. What colour is methyl orange in an alkaline solution?

  • (A) Red
  • (B) Pink
  • (C) Colourless
  • (D) Yellow
Show Answer

Answer: D — Methyl orange is red in acid, orange near the end point, and yellow in alkali.

Q5. Which ionic equation correctly represents any acid–alkali neutralisation?

  • (A) H+(aq) + OH−(aq) → H2O(l)
  • (B) Na+(aq) + Cl−(aq) → NaCl(s)
  • (C) H2O(l) → H+(aq) + OH−(aq)
  • (D) 2H+(aq) + CO32−(aq) → H2O(l) + CO2(g)
Show Answer

Answer: A — In every neutralisation the hydrogen ions from the acid react with the hydroxide ions from the alkali to form water. The other ions are spectators and do not appear in the ionic equation.

Q6. Which salt is formed when potassium hydroxide solution is neutralised by sulfuric acid?

  • (A) Potassium chloride
  • (B) Potassium nitrate
  • (C) Potassium sulfate
  • (D) Potassium carbonate
Show Answer

Answer: C — The salt takes its second name from the acid: sulfuric acid makes sulfates, hydrochloric acid makes chlorides and nitric acid makes nitrates.

Q7. Dilute hydrochloric acid is added to calcium carbonate. What is produced?

  • (A) Calcium chloride and water only
  • (B) Calcium chloride and hydrogen
  • (C) Calcium chloride, water and carbon dioxide
  • (D) Calcium hydroxide and carbon dioxide
Show Answer

Answer: C — Acid + carbonate → salt + water + carbon dioxide. The carbon dioxide causes the fizzing and turns limewater milky.

Q8. Zinc is added to dilute sulfuric acid. Which gas is given off?

  • (A) Oxygen
  • (B) Carbon dioxide
  • (C) Chlorine
  • (D) Hydrogen
Show Answer

Answer: D — Acid + metal → salt + hydrogen. Hydrogen is tested with a lighted splint, which gives a squeaky pop.

Q9. Sodium hydroxide solution is neutralised by nitric acid. What are the products?

  • (A) Sodium nitrate and water
  • (B) Sodium chloride and water
  • (C) Sodium nitrate and hydrogen
  • (D) Sodium sulfate and water
Show Answer

Answer: A — Alkali + acid → salt + water, and nitric acid forms nitrate salts. No hydrogen is produced — hydrogen only forms when an acid reacts with a metal.

Q10. Excess black copper(II) oxide powder is warmed with dilute sulfuric acid and the mixture is filtered. What is observed?

  • (A) The black solid dissolves and a blue solution of copper(II) sulfate forms
  • (B) A green solution of copper(II) chloride forms
  • (C) Bubbles of hydrogen gas are given off
  • (D) Nothing happens and the mixture stays colourless
Show Answer

Answer: A — Acid + metal oxide → salt + water. The black CuO dissolves and copper(II) sulfate solution is blue. No gas is produced in this reaction.

Section 2: Strong vs Weak Acids & Bases

Q11. What is meant by a strong acid?

  • (A) An acid that ionises completely in aqueous solution
  • (B) An acid that has a very high concentration
  • (C) An acid that reacts dangerously with every metal
  • (D) An acid with a pH above 7
Show Answer

Answer: A — Strength describes the degree of ionisation: a strong acid ionises (dissociates) fully, a weak acid only partially. Strength is not the same as concentration.

Q12. Which of the following is a weak acid?

  • (A) Hydrochloric acid
  • (B) Nitric acid
  • (C) Sulfuric acid
  • (D) Ethanoic acid
Show Answer

Answer: D — Ethanoic acid is a weak acid: only about 1% of its molecules ionise in a 0.1 mol/dm³ solution. Hydrochloric, nitric and sulfuric acids are all strong.

Q13. Four solutions all have a concentration of 0.10 mol/dm³. Which one has the lowest pH?

  • (A) Ethanoic acid
  • (B) Citric acid
  • (C) Ammonia solution
  • (D) Hydrochloric acid
Show Answer

Answer: D — Hydrochloric acid is fully ionised, so [H+] = 0.10 mol/dm³ and pH = 1. The weak acids give a much lower [H+], and ammonia is alkaline.

Q14. A solution is described as a concentrated weak acid. This means it has…

  • (A) few moles per dm³ and ionises completely
  • (B) many moles per dm³ and ionises completely
  • (C) many moles per dm³ but ionises only partially
  • (D) few moles per dm³ and ionises only partially
Show Answer

Answer: C — “Concentrated” is about how many moles are dissolved per dm³, while “weak” is about partial ionisation. The two ideas are independent of each other.

Q15. A 1.0 mol/dm³ solution of hydrochloric acid conducts electricity much better than a 1.0 mol/dm³ solution of ethanoic acid because…

  • (A) ethanoic acid contains no ions at all
  • (B) hydrochloric acid molecules are smaller and move faster
  • (C) hydrochloric acid produces a much higher concentration of mobile ions
  • (D) the ethanoic acid is more concentrated
Show Answer

Answer: C — Solutions conduct using mobile ions. Fully ionised HCl gives far more ions per dm³ than partially ionised ethanoic acid at the same concentration, so it conducts better.

Q16. Identical pieces of magnesium ribbon are added to equal volumes of 0.10 mol/dm³ hydrochloric acid and 0.10 mol/dm³ ethanoic acid. What is observed at the start?

  • (A) Much faster fizzing in the hydrochloric acid
  • (B) Much faster fizzing in the ethanoic acid
  • (C) Exactly the same rate of fizzing in both
  • (D) No reaction in either acid
Show Answer

Answer: A — The reaction rate depends on the [H+] the magnesium actually meets. At the same concentration the strong acid has a far higher [H+], so it fizzes much faster.

Q17. Which oxide is amphoteric, reacting with both acids and alkalis?

  • (A) Sodium oxide
  • (B) Sulfur dioxide
  • (C) Aluminium oxide
  • (D) Calcium oxide
Show Answer

Answer: C — Aluminium oxide is amphoteric. Sodium oxide and calcium oxide are basic oxides, and sulfur dioxide is an acidic oxide.

Q18. Which oxide is acidic?

  • (A) Magnesium oxide
  • (B) Sodium oxide
  • (C) Copper(II) oxide
  • (D) Sulfur dioxide
Show Answer

Answer: D — Non-metal oxides such as sulfur dioxide are acidic (SO2 dissolves in water to form an acidic solution). Metal oxides are basic.

Q19. Which statement about a 0.10 mol/dm³ ammonia solution is correct?

  • (A) Its pH is lower than that of 0.10 mol/dm³ sodium hydroxide, because ammonia is only partially ionised
  • (B) It is fully ionised, like sodium hydroxide
  • (C) Its pH is exactly 13, the same as 0.10 mol/dm³ sodium hydroxide
  • (D) It contains no hydroxide ions at all
Show Answer

Answer: A — Ammonia is a weak base: NH3 + H2O ⇌ NH4+ + OH− lies well to the left, so [OH−] is much lower than in NaOH of the same concentration (pH about 11 rather than 13).

Q20. Samples of 25 cm³ of 0.10 mol/dm³ hydrochloric acid and 25 cm³ of 0.10 mol/dm³ ethanoic acid are each titrated with sodium hydroxide solution. Which statement is true?

  • (A) The ethanoic acid needs a larger volume of sodium hydroxide
  • (B) Both samples need the same volume of sodium hydroxide to reach the end point
  • (C) The hydrochloric acid needs twice the volume of sodium hydroxide
  • (D) The ethanoic acid cannot be neutralised
Show Answer

Answer: B — Both samples contain the same amount of acid (0.0025 mol). As OH− removes H+, more ethanoic acid molecules ionise to replace it, until eventually every molecule has reacted — so the same titre is needed, even though the starting pH values differ.


Section 3: pH Calculations

Q21. What is the pH of a 0.010 mol/dm³ solution of hydrochloric acid?

  • (A) 1.0
  • (B) 2.0
  • (C) 12.0
  • (D) 0.01
Show Answer

Answer: B — Hydrochloric acid is strong and fully ionised, so [H+] = 0.010 = 1.0 × 10−2 mol/dm³. pH = −log(1.0 × 10−2) = 2.0.

Q22. A solution has a pH of 4.50. What is its hydrogen ion concentration?

  • (A) 1.0 × 10−4 mol/dm³
  • (B) 3.2 × 10−4 mol/dm³
  • (C) 3.2 × 10−5 mol/dm³
  • (D) 4.5 mol/dm³
Show Answer

Answer: C — [H+] = 10−pH = 10−4.50 = 3.2 × 10−5 mol/dm³. A pH halfway between 4 and 5 gives a concentration halfway between 10−4 and 10−5 on the log scale, not exactly 10−4.

Q23. A hydrochloric acid solution has pH 1.0. It is diluted with water until its concentration is 100 times smaller. What is the new pH?

  • (A) 1.0
  • (B) 2.0
  • (C) 11.0
  • (D) 3.0
Show Answer

Answer: D — Every tenfold dilution of a strong acid raises the pH by exactly 1 unit. A 100-fold dilution is two tenfold steps, so the pH rises from 1.0 to 3.0.

Q24. What is the pH of a 0.050 mol/dm³ solution of nitric acid, a strong acid?

  • (A) 1.30
  • (B) 2.30
  • (C) 0.30
  • (D) 12.70
Show Answer

Answer: A — [H+] = 0.050 mol/dm³, so pH = −log(0.050) = 1.30.

Q25. 25 cm³ of 0.20 mol/dm³ hydrochloric acid is mixed with 25 cm³ of 0.10 mol/dm³ sodium hydroxide. What is the pH of the resulting mixture?

  • (A) 1.00
  • (B) 1.30
  • (C) 2.30
  • (D) 7.00
Show Answer

Answer: B — n(H+) = 0.025 × 0.20 = 5.0 × 10−3 mol; n(OH−) = 0.025 × 0.10 = 2.5 × 10−3 mol. Excess H+ = 2.5 × 10−3 mol in a total volume of 50 cm³ (0.050 dm³), so [H+] = 0.050 mol/dm³ and pH = −log(0.050) = 1.30. The dilution on mixing must not be forgotten.

Q26. At 25 °C, Kw = 1.0 × 10−14 mol2/dm6. A solution has [H+] = 1.0 × 10−3 mol/dm³. What is its hydroxide ion concentration?

  • (A) 1.0 × 10−3 mol/dm³
  • (B) 11.0 mol/dm³
  • (C) 1.0 × 10−17 mol/dm³
  • (D) 1.0 × 10−11 mol/dm³
Show Answer

Answer: D — [OH−] = Kw / [H+] = (1.0 × 10−14) / (1.0 × 10−3) = 1.0 × 10−11 mol/dm³. (11 is the pH of this solution, not a concentration.)

Q27. What is the pH of a 0.010 mol/dm³ solution of sodium hydroxide at 25 °C?

  • (A) 2.0
  • (B) 10.0
  • (C) 12.0
  • (D) 14.0
Show Answer

Answer: C — [OH−] = 1.0 × 10−2 mol/dm³, so [H+] = Kw / [OH−] = 1.0 × 10−12 mol/dm³ and pH = 12.0. (2.0 is the pOH, a common trap.)

Q28. At 25 °C, a solution has pH 9.0. What is its pOH?

  • (A) 23.0
  • (B) 5.0
  • (C) 9.0
  • (D) −5.0
Show Answer

Answer: B — At 25 °C, pH + pOH = 14.0, so pOH = 14.0 − 9.0 = 5.0.

Q29. What is the pH of a 0.20 mol/dm³ solution of hydrochloric acid?

  • (A) 0.70
  • (B) 1.70
  • (C) 0.20
  • (D) 13.30
Show Answer

Answer: A — pH = −log(0.20) = 0.70. Concentrations above 0.1 mol/dm³ give pH values below 1, and very concentrated strong acids can even have negative pH values.

Q30. 50 cm³ of 0.10 mol/dm³ hydrochloric acid is mixed with 25 cm³ of 0.10 mol/dm³ sodium hydroxide. What is the pH of the resulting solution?

  • (A) 1.00
  • (B) 2.48
  • (C) 1.30
  • (D) 1.48
Show Answer

Answer: D — n(H+) = 0.050 × 0.10 = 5.0 × 10−3 mol; n(OH−) = 0.025 × 0.10 = 2.5 × 10−3 mol. Excess H+ = 2.5 × 10−3 mol, now in 75 cm³ (0.075 dm³), so [H+] = 0.033 mol/dm³ and pH = −log(0.033) = 1.48.

Section 4: Ka, pKa & Buffer Solutions (A-Level / AP Stretch)

Q31. Ethanoic acid ionises as shown: CH3COOH(aq) ⇌ CH3COO−(aq) + H+(aq). Which expression gives the acid dissociation constant, Ka?

  • (A) Ka = [CH3COOH] / ([H+][CH3COO−])
  • (B) Ka = [H+][CH3COO−] / [CH3COOH]
  • (C) Ka = [H+]2 / [CH3COO−]
  • (D) Ka = [CH3COO−] / [CH3COOH]
Show Answer

Answer: B — Ka is the equilibrium constant for the ionisation, so it is products over reactants, each raised to its stoichiometric power. Option A is the expression upside down, which would give 1/Ka.

Q32. A 0.10 mol/dm³ solution of a weak acid has Ka = 1.8 × 10−5 mol/dm³. What is the pH of the solution?

  • (A) 2.87
  • (B) 4.74
  • (C) 1.87
  • (D) 5.74
Show Answer

Answer: A — For a weak acid, [H+] ≈ √(Ka × c) = √(1.8 × 10−5 × 0.10) = √(1.8 × 10−6) = 1.34 × 10−3 mol/dm³. pH = −log(1.34 × 10−3) = 2.87. (4.74 is the pKa of the acid, not the pH.)

Q33. A weak acid has Ka = 1.8 × 10−5 mol/dm³. What is its pKa?

  • (A) −4.74
  • (B) 5.26
  • (C) 4.74
  • (D) 9.26
Show Answer

Answer: C — pKa = −log Ka = −log(1.8 × 10−5) = 4.74. The minus sign in the definition makes pKa a positive number for a typical weak acid.

Q34. A weak acid has pKa = 3.40. What is its Ka?

  • (A) 3.4 × 10−4 mol/dm³
  • (B) 4.0 × 10−4 mol/dm³
  • (C) 2.5 × 10−11 mol/dm³
  • (D) 4.0 × 10−3 mol/dm³
Show Answer

Answer: B — Ka = 10−pKa = 10−3.40 = 4.0 × 10−4 mol/dm³. Simply copying the 3.40 into the value, as in option A, is a common error.

Q35. A buffer solution contains 0.20 mol/dm³ ethanoic acid (pKa = 4.74) and 0.30 mol/dm³ sodium ethanoate. What is the pH of the buffer?

  • (A) 4.56
  • (B) 4.74
  • (C) 4.92
  • (D) 5.24
Show Answer

Answer: C — Using the Henderson–Hasselbalch equation: pH = pKa + log([salt] / [acid]) = 4.74 + log(0.30 / 0.20) = 4.74 + 0.18 = 4.92. Using the ratio upside down gives 4.56, the trap in option A.

Q36. A few drops of dilute hydrochloric acid are added to an ethanoic acid / sodium ethanoate buffer. Why does the pH stay almost unchanged?

  • (A) The added acid is diluted by the large volume of water present
  • (B) The ethanoic acid molecules neutralise the added acid
  • (C) The buffer contains no ions that are able to react with H+
  • (D) The added H+ ions combine with ethanoate ions to form weakly ionised ethanoic acid
Show Answer

Answer: D — The buffer holds a large reserve of ethanoate ions, A−. Added H+ reacts with them, A− + H+ → HA, shifting the ionisation equilibrium to the left, so the free [H+] and the pH barely change.

Q37. A weak acid is titrated with sodium hydroxide solution. At the half-neutralisation point the pH of the mixture is 4.90. What is the Ka of the acid?

  • (A) 4.9 × 10−5 mol/dm³
  • (B) 1.3 × 10−5 mol/dm³
  • (C) 1.3 × 10−9 mol/dm³
  • (D) 7.9 × 10−10 mol/dm³
Show Answer

Answer: B — At half-neutralisation exactly half the acid has been converted to its salt, so [HA] = [A−] and pH = pKa = 4.90. Therefore Ka = 10−4.90 = 1.3 × 10−5 mol/dm³.

Q38. A small amount of sodium hydroxide is added to an ethanoic acid buffer. Which reaction mainly removes the added hydroxide ions?

  • (A) CH3COOH + OH− → CH3COO− + H2O
  • (B) CH3COO− + OH− → CH3COOH + O2−
  • (C) Na+ + OH− → NaOH
  • (D) CH3COO− + H2O → CH3COOH + OH−
Show Answer

Answer: A — The added OH− reacts with the large reserve of undissociated ethanoic acid molecules, forming ethanoate ions and water. Because the acid reservoir absorbs the alkali, the pH rises only very slightly.

Q39. Which mixture forms an acidic buffer solution?

  • (A) 50 cm³ of 0.10 mol/dm³ HCl mixed with 50 cm³ of 0.10 mol/dm³ NaCl
  • (B) 50 cm³ of 0.10 mol/dm³ ethanoic acid mixed with 25 cm³ of 0.10 mol/dm³ NaOH
  • (C) 50 cm³ of 0.10 mol/dm³ HCl mixed with 50 cm³ of 0.10 mol/dm³ NaOH
  • (D) 50 cm³ of 0.10 mol/dm³ ethanoic acid mixed with 50 cm³ of 0.10 mol/dm³ NaOH
Show Answer

Answer: B — The smaller amount of NaOH neutralises only half of the ethanoic acid, leaving a mixture of unreacted weak acid (HA) and its conjugate base (A−) — the two components a buffer needs. In option D the acid is fully neutralised, leaving no reserve of HA.

Q40. A 0.050 mol/dm³ solution of a weak acid has Ka = 6.3 × 10−6 mol/dm³. What is the pH of the solution?

  • (A) 2.25
  • (B) 5.20
  • (C) 4.25
  • (D) 3.25
Show Answer

Answer: D — [H+] ≈ √(Ka × c) = √(6.3 × 10−6 × 0.050) = √(3.15 × 10−7) = 5.6 × 10−4 mol/dm³. pH = −log(5.6 × 10−4) = 3.25. (5.20 is the pKa, the distractor for students who stop one step early.)

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