This set covers indicators, hydrogen ions, neutralisation, dilution, choosing a salt method and titration results. The questions run from easier to harder, and every answer is worked in full. All numbers and observations are invented for practice.
Attempt each question before opening the answer. Keep a calculator and periodic table beside you, and use the mole and equation-ratio tutor or the equation balance reasoning trainer to check ratios and atom counts afterwards. This set belongs to acids, bases and salts.
Questions
Q1. Solution A turns litmus blue. Solution B turns universal indicator red. Classify each.
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Blue litmus means A is alkaline. Red universal indicator is the strongly acidic end, so B is strongly acidic.
Q2. Four solutions have pH 2, 7, 9 and 13. Put them in order from most acidic to most alkaline and classify each.
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Order: pH 2 (acidic), pH 7 (neutral), pH 9 (weakly alkaline), pH 13 (strongly alkaline). The lower the pH below 7, the more acidic it is, and the higher above 7, the more alkaline.
Q3. A solution is colourless with phenolphthalein. A student says it must be an acid. Comment.
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Phenolphthalein is colourless in acid and in neutral solutions, so the solution is acidic or neutral, and certainly not alkaline. Another indicator, such as litmus or universal indicator, could separate the two.
Q4. State the particle that makes a solution acidic and the particle that makes it alkaline. Give the ionic equation for neutralisation.
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Acidic solutions contain excess hydrogen ions, H⁺(aq). Alkaline solutions contain excess hydroxide ions, OH⁻(aq). Neutralisation: H⁺(aq) + OH⁻(aq) → H₂O(l).
Q5. Balance: H₂SO₄ + NaOH → Na₂SO₄ + H₂O
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Sodium needs 2 on the left, so use 2NaOH. That gives 2 H and 2 O from NaOH plus H₂SO₄. Hydrogen on the left is 2 + 2 = 4, so use 2H₂O.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Check: H 4 and 4, O 4 + 2 = 6 and 4 + 2 = 6, Na 2 and 2, S 1 and 1.
Q6. Equal volumes (25.0 cm³) of 0.100 mol/dm³ hydrochloric acid and 0.100 mol/dm³ ethanoic acid react with excess magnesium. Compare the initial rate and the final volume of hydrogen, and explain.
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Hydrochloric acid is strong, so it has a higher hydrogen ion concentration and reacts faster at the start. Both contain 0.0250 × 0.100 = 0.00250 mol of acid and each reacts in a 2:1 ratio with hydrogen, so the final volume of hydrogen is the same (0.00125 mol).
Q7. 10.0 cm³ of 1.00 mol/dm³ hydrochloric acid is made up to 100 cm³ with water. Find the new concentration and the new pH in the simple model for a strong acid.
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Amount = 0.0100 × 1.00 = 0.0100 mol. New concentration = 0.0100 ÷ 0.100 = 0.100 mol/dm³. The H⁺ concentration fell by a factor of 10, so the pH rises from 0 to 1. The solution is still acidic.
Q8. A student says, “If I keep diluting an acid with water, it will become neutral and then alkaline.” Explain the error.
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Water does not remove hydrogen ions. It spreads them through a larger volume, so the pH moves toward 7 but never passes it. Only a base that reacts with hydrogen ions can neutralise an acid.
Q9. Find the volume of 0.100 mol/dm³ hydrochloric acid needed to neutralise 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide.
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NaOH amount = 0.0250 × 0.200 = 0.00500 mol. HCl + NaOH → NaCl + H₂O is 1:1, so HCl = 0.00500 mol. Volume = 0.00500 ÷ 0.100 = 0.0500 dm³ = 50.0 cm³.
Q10. Choose a preparation principle for each salt and justify in one line: (a) magnesium sulfate, from magnesium oxide and sulfuric acid, (b) sodium sulfate, from sodium hydroxide and sulfuric acid, (c) barium sulfate.
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(a) Magnesium sulfate is soluble and magnesium oxide is an insoluble base, so use excess solid, filter, crystallise. MgO + H₂SO₄ → MgSO₄ + H₂O.
(b) Sodium hydroxide is a soluble alkali, so use titration, then crystallise. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
(c) Barium sulfate is insoluble, so use precipitation. BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq).
Q11. 25.0 cm³ of sodium hydroxide of unknown concentration needs a mean titre of 15.00 cm³ of 0.100 mol/dm³ sulfuric acid (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O). A second student had titres of 23.60 (rough), 22.40, 22.35 and 22.30 cm³ with a different acid. (a) Find the NaOH concentration. (b) Give the mean titre for the second student.
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(a) H₂SO₄ amount = 0.01500 × 0.100 = 0.00150 mol. The ratio is 1:2, so NaOH = 0.00300 mol. Concentration = 0.00300 ÷ 0.0250 = 0.120 mol/dm³.
(b) Ignore the rough titre. The other three lie between 22.30 and 22.40, so they are concordant. Mean = (22.40 + 22.35 + 22.30) ÷ 3 = 67.05 ÷ 3 = 22.35 cm³.
If you got these wrong
| Error | Go back to |
|---|---|
| Wrong class from an indicator colour (Q1 to Q3) | Use an indicator result to classify a solution |
| Missing or vague particle explanations, or strength mixed with concentration (Q4, Q6) | Relate acidity to the relevant particles |
| Dilution treated as neutralisation, or wrong moles (Q7 to Q9) | Distinguish neutralisation from dilution |
| Wrong salt method, or unbalanced equations (Q5, Q10) | Choose a suitable salt-preparation principle |
| Titre choice, mean or concentration steps (Q11) | Explain an endpoint from supplied observations |
Record each repeated error in the mistake log and retest queue, and return to the module overview if the order of topics is unclear.
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