Acids & Bases
The complete Grade 11 reference: Arrhenius and Lowry-Brønsted theories, conjugate acid-base pairs, strong vs weak and concentrated vs dilute, neutralisation reactions, indicators and titrations, pH calculations, and stoichiometric titration calculations. Everything tested in NSC Paper 1, in one place.
Section 1
Defining Acids and Bases
You already know acids and bases by taste and use — lemon juice, vinegar, soap, antacids. These theories give a precise, testable definition for any acid or base.
Arrhenius Theory
Arrhenius Acid
A compound that increases the concentration of H3O+ ions in solution.
Arrhenius Base
A compound that increases the concentration of OH− ions in solution.
Water dissociates very slightly:
2H2O(l) → H3O+(aq) + OH−(aq)
Worked Example 1
Show that HCl is an Arrhenius acid and NaOH is an Arrhenius base.
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) — [H3O+] increases → HCl is an acid.
NaOH(s) → Na+(aq) + OH−(aq) — [OH−] increases → NaOH is a base.
HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) — [H3O+] increases → HCl is an acid.
NaOH(s) → Na+(aq) + OH−(aq) — [OH−] increases → NaOH is a base.
Lowry-Brønsted Theory
Lowry-Brønsted Acid
A proton (H+) donor. The Arrhenius theory only applies to water — in 1923, Lowry and Brønsted proposed this broader definition.
Lowry-Brønsted Base
A proton (H+) acceptor.
Worked Example 2
Identify the acid and base: HCl(aq) + NH3(aq) → NH4+(aq) + Cl−(aq)
HCl → Cl− (lost a proton); NH3 → NH4+ (gained a proton).
∴ HCl is the acid (donor); NH3 is the base (acceptor).
HCl → Cl− (lost a proton); NH3 → NH4+ (gained a proton).
∴ HCl is the acid (donor); NH3 is the base (acceptor).
Ampholytes / Amphiprotic Substances
Definition
A substance that can act as either an acid or a base, depending on what it reacts with. Water, HCO3− and HSO4− are common examples.
Worked Example 3 — water is amphiprotic
Reacting with an acid: HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) — water accepts a proton → acts as a base.
Reacting with a base: NH3(aq) + H2O(l) → NH4+(aq) + OH−(aq) — water donates a proton → acts as an acid.
Reacting with a base: NH3(aq) + H2O(l) → NH4+(aq) + OH−(aq) — water donates a proton → acts as an acid.
Practice: strength and classification
You already met conjugate pairs before this course — question 1 leans on that prior knowledge; questions 2 and 3 test the strong/weak and Lowry-Brønsted ideas directly.
- 1.In each reaction, label the conjugate acid-base pairs.
(a) H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4−(aq)
(b) NH4+(aq) + F−(aq) → HF(aq) + NH3(aq) - 2.Classify each as a strong or weak acid, and explain your classification: sulfuric acid, hydrochloric acid, carbonic acid.
- 3.State whether each of the following can act as a Lowry-Brønsted acid or base, giving a reason.
(a) Ca(OH)2 (b) HBr - 4.Hydrogen sulfide gas, H2S, dissolves in water and donates a proton to water. Write the balanced equation and label the conjugate acid-base pairs.
- 5.Classify ammonia (NH3) as a strong or weak base, giving a reason.
- 6.Identify the conjugate acid-base pairs in: HSO4−(aq) + OH−(aq) ⇌ SO42−(aq) + H2O(l)
Section 2
Strong/Weak & Concentrated/Dilute
Two independent ideas: how completely an acid or base forms ions (strength), and how much of it is dissolved (concentration).
Strong, Weak, Concentrated, Dilute
Strong
Ionises/dissociates completely in water → high [H3O+] or [OH−].
Weak
Ionises/dissociates incompletely in water → low [H3O+] or [OH−].
| Strong acids | Strong bases | Weak acids | Weak bases |
|---|---|---|---|
| HCl (monoprotic) | NaOH | CH3COOH (ethanoic/acetic) | NH3 |
| HNO3 (monoprotic) | KOH | (COOH)2 (oxalic) | Zn(OH)2 |
| H2SO4 (diprotic) | LiOH | H2CO3 (carbonic) | Ca(OH)2 |
| H3PO4 (triprotic) | Ba(OH)2 | — | Mg(OH)2 |
A monoprotic acid donates one proton (HCl); a diprotic acid donates two (H2SO4); a triprotic acid donates three (H3PO4).
Misconception check
Your intuition says: "concentrated" and "strong" must mean the same thing.
Actually: they're independent. Concentrated/dilute = moles of acid/base per volume of water. Strong/weak = how completely it ionises. A weak acid can be concentrated; a strong acid can be dilute. A strong acid/base of the same concentration conducts electricity better and reacts faster, because it produces more ions.
Actually: they're independent. Concentrated/dilute = moles of acid/base per volume of water. Strong/weak = how completely it ionises. A weak acid can be concentrated; a strong acid can be dilute. A strong acid/base of the same concentration conducts electricity better and reacts faster, because it produces more ions.
Practice: definitions and conjugate pairs
Now that conjugate pairs are fully formalised, this multiple-choice set tests the whole of Sections 1–2 together.
- 1.Which ONE is the conjugate acid of HCO3−? A. CO32− B. OH− C. H2CO3 D. H3CO3
- 2.An aqueous solution that contains more hydronium ions than hydroxide ions is: A. Basic B. Acidic C. Neutral D. Amphiprotic
- 3.According to the Lowry-Brønsted theory, a base: A. dissociates in aqueous solution B. raises the OH− concentration above 10−7 mol·dm−3 C. tastes bitter and feels slippery D. accepts a proton during a collision with an acid
- 4.Which ONE of the following species cannot act as both a Lowry-Brønsted acid and base? A. H2PO4− B. H2O C. HSO4− D. CH3COOH
- 5.Consider: CH3COOH(aq) + OH−(aq) ⇌ CH3COO−(aq) + H2O(l). The correct conjugate acid-base pair is: A. CH3COOH and CH3COO− B. CH3COOH and OH− C. CH3COO− and H2O D. CH3COOH and H2O
- 6.Ammonia only partially reacts with water to produce a low concentration of OH− ions. This shows that ammonia is a: A. strong acid B. weak acid C. strong base D. weak base
- 7.Which indicator is most suitable for the titration of ethanoic acid (CH3COOH) against sodium hydroxide (NaOH)? A. Methyl orange B. Bromothymol blue C. Phenolphthalein D. Any of the above works equally well
- 8.Which ONE of the following species can act as both a Lowry-Brønsted acid and base? A. Cl− B. NH4+ C. HSO4− D. Na+
- 9.Which statement about Kw is correct? A. Kw increases as a solution becomes more acidic. B. Kw is the same at every temperature. C. Kw = [H3O+][OH−] = 1×10−14 at 25 °C for any aqueous solution. D. Kw only applies to solutions of strong acids.
Section 3
Conjugate Acid-Base Pairs
Acids and bases don't just react — they transform into each other by gaining or losing a single proton.
Conjugate Acid-Base Pairs
Definition
When an acid HA loses a proton, it forms its conjugate base A−. When a base A− gains a proton, it forms its conjugate acid HA. HA and A− differ by one proton.
Worked Example 4
Identify the two conjugate acid-base pairs in: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
NH3 (base 1) gains a proton → NH4+ (acid 1) → pair: NH3/NH4+
H2O (acid 2) loses a proton → OH− (base 2) → pair: H2O/OH−
NH3 (base 1) gains a proton → NH4+ (acid 1) → pair: NH3/NH4+
H2O (acid 2) loses a proton → OH− (base 2) → pair: H2O/OH−
Quick rules: to find a conjugate acid, add a proton (conjugate acid of HCO3− is H2CO3). To find a conjugate base, remove a proton (conjugate base of HCO3− is CO32−).
Section 4
Neutralisation Reactions
Put an acid and a base in the same beaker: three reaction types you must recognise and balance.
Neutralisation — Definition
Definition
Neutralisation is a reaction between an acid and a base that produces a salt (and usually water). A salt is made of the cation from the base and the anion from the acid.
Acid + Metal Hydroxide → Salt + Water
General equation
nH+(aq) + M(OH)n(aq) → nH2O(l) + Mn+(aq)
Worked Example 5:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
2HBr(aq) + Mg(OH)2(aq) → 2H2O(l) + MgBr2(aq)
Acid + Metal Oxide → Salt + Water
General equation
2yH+(aq) + MxOy(aq) → yH2O(l) + xMn+(aq)
Worked Example 6:
2HCl(aq) + Na2O(aq) → H2O(l) + 2NaCl(aq)
6HCl(aq) + Al2O3(aq) → 3H2O(l) + 2AlCl3(aq)
Acid + Metal Carbonate → Salt + Water + CO₂
Worked Example 7
2HNO3(aq) + Na2CO3(aq) → 2NaNO3(aq) + CO2(g) + H2O(l)
H2SO4(aq) + CaCO3(aq) → CaSO4(s) + CO2(g) + H2O(l)
Everyday Uses of Neutralisation
- Agriculture: lime (CaO / CaCO3) is added to acidic soil.
- Biology: antacids (e.g. Mg(OH)2, Al(OH)3, NaHCO3) neutralise excess stomach acid.
- Industry: limewater absorbs acidic SO2 gas released by power stations.
Practice: writing neutralisation equations
- 1.Write a balanced equation for the reaction between HNO3 and KOH.
- 2.Write a balanced equation for the reaction between HBr and K2O.
- 3.Write a balanced equation for the reaction between HCl and K2CO3.
- 4.Write a balanced equation for the reaction between H2SO4 and calcium oxide, CaO.
- 5.Write a balanced equation for the reaction between H3PO4 and magnesium hydroxide, Mg(OH)2.
- 6.Write a balanced equation for the reaction between HNO3 and calcium carbonate, CaCO3.
Section 5
Indicators and Titrations
Indicators can't see the reaction "finish" for you — you have to choose one whose colour change lands at the right pH.
Titration Key Terms & Indicator Choice
Titration
A technique to determine the concentration of an unknown solution by reacting it with a standard solution (accurately known concentration).
Equivalence vs End Point
Equivalence point — the acid has exactly reacted with the base. End point — the indicator changes colour (should be as close to the equivalence point as possible).
| Acid | Base | Best indicator | Colour-change range |
|---|---|---|---|
| Strong | Strong | Bromothymol blue | pH 6.0 – 7.6 |
| Strong | Weak | Methyl orange | pH 3.2 – 4.4 |
| Weak | Strong | Phenolphthalein | pH 8.2 – 10.0 |
Section 6
The pH Scale and pH Calculations
Indicators only tell you acidic/neutral/basic in broad strokes. pH gives an exact number.
Kw and the pH Formula
Ionisation constant of water
Water auto-ionises: 2H2O(l) ⇌ H3O+(aq) + OH−(aq)
\(K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1\times10^{-14}\) at 25 °C
Neutral: [H3O+] = [OH−] = 1×10−7 mol·dm−3. Acidic: [H3O+] > 1×10−7. Basic: [H3O+] < 1×10−7.
\(\text{pH} = -\log[\text{H}_3\text{O}^+]\) — a scale from 0 to 14. At Grade 11 level you only calculate the pH of strong acids and bases (complete ionisation/dissociation, so the mole ratio gives [H3O+] or [OH−] directly).
\(K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1\times10^{-14}\) at 25 °C
Neutral: [H3O+] = [OH−] = 1×10−7 mol·dm−3. Acidic: [H3O+] > 1×10−7. Basic: [H3O+] < 1×10−7.
\(\text{pH} = -\log[\text{H}_3\text{O}^+]\) — a scale from 0 to 14. At Grade 11 level you only calculate the pH of strong acids and bases (complete ionisation/dissociation, so the mole ratio gives [H3O+] or [OH−] directly).
Method for an acid
Write the ionisation equation → mole ratio gives [H3O+] → substitute into pH = −log[H3O+].
Method for a base
Write the dissociation equation → mole ratio gives [OH−] → use Kw for [H3O+] → substitute into the pH formula.
Worked Examples 8–11
WE8 — monoprotic acid
pH of 0.2 mol·dm−3 HNO3.
1:1 ratio → [H3O+] = 0.2
pH = −log(0.2) = 0.70
1:1 ratio → [H3O+] = 0.2
pH = −log(0.2) = 0.70
WE9 — diprotic acid
pH of 0.25 mol·dm−3 H2SO4.
1:2 ratio → [H3O+] = 0.5
pH = −log(0.5) = 0.30
1:2 ratio → [H3O+] = 0.5
pH = −log(0.5) = 0.30
WE10 — base
pH of 0.4 mol·dm−3 NaOH.
[OH−] = 0.4 → [H3O+] = 10−14/0.4 = 2.5×10−14
pH = 13.60
[OH−] = 0.4 → [H3O+] = 10−14/0.4 = 2.5×10−14
pH = 13.60
WE11 — concentration from pH
[HCl] with pH = 4.5.
[H3O+] = 10−4.5 = 3.16×10−5
[HCl] = 3.16×10−5 mol·dm−3 (1:1)
[H3O+] = 10−4.5 = 3.16×10−5
[HCl] = 3.16×10−5 mol·dm−3 (1:1)
Misconception check
Your intuition says: a more concentrated acid should have a higher pH number.
Actually: pH is a negative logarithmic scale — more H3O+ gives a smaller pH. Compare WE8 and WE9: the more concentrated H2SO4 (0.25 mol·dm−3, diprotic) has a lower pH (0.30) than the less concentrated HNO3 (0.20 mol·dm−3, pH 0.70). Same trap in reverse for bases: more concentrated → higher pH.
Actually: pH is a negative logarithmic scale — more H3O+ gives a smaller pH. Compare WE8 and WE9: the more concentrated H2SO4 (0.25 mol·dm−3, diprotic) has a lower pH (0.30) than the less concentrated HNO3 (0.20 mol·dm−3, pH 0.70). Same trap in reverse for bases: more concentrated → higher pH.
Guided practice — your turn
pH of a strong monoprotic acid. Calculate the pH of a 0.1 mol·dm−3 HBr solution (strong, monoprotic).
HBr(aq) + H2O(l) → H3O+(aq) + Br−(aq) — ratio 1:1 → [H3O+] = ____ mol·dm−3
pH = −log(____) = ____
Check your answer: pH = 1.00
HBr(aq) + H2O(l) → H3O+(aq) + Br−(aq) — ratio 1:1 → [H3O+] = ____ mol·dm−3
pH = −log(____) = ____
Check your answer: pH = 1.00
Practice: pH calculations
These questions remove the scaffolding of the guided box above — attempt them independently. Use Kw = 1×10−14 at 25 °C.
- 1.Calculate the pH of a 0.05 mol·dm⁻³ HCl solution.
- 2.Calculate the pH of a 0.1 mol·dm⁻³ H₂SO₄ solution.
- 3.Calculate the pH of a 0.3 mol·dm⁻³ KOH solution.
Need a hint?
Hint 1: KOH is a base, so you're solving for [OH−] first, not [H3O+] directly.
Hint 2: Once you have [OH−], use Kw = [H3O+][OH−] to find [H3O+] before applying the pH formula.
- 4.A solution of HNO₃ has a pH of 2.5. Calculate its concentration.
- 5.A student must prepare 300 cm³ of a 0.2 mol·dm⁻³ KOH solution.
(a) Calculate the mass of KOH required.
(b) Calculate the pH of this solution.Need a hint?
Hint 1 (part a): Use n = cV to find the moles of KOH needed, then m = nM to convert to mass.
Hint 2 (part b): This is the same solution and concentration as WE10 — but with KOH instead of NaOH. Compare your method.
- 6.Calculate the pH of a 0.1 mol·dm⁻³ H3PO4 solution, assuming complete ionisation of all three protons.
- 7.25 cm³ of a 2.0 mol·dm⁻³ HCl solution is diluted to a final volume of 500 cm³. Calculate the pH of the diluted solution.
Need a hint?
Hint: Two-step problem. First use c1V1 = c2V2 to find the new concentration after dilution, then apply the usual pH method.
- 8.A 0.2 mol·dm⁻³ solution of HCl and a 0.2 mol·dm⁻³ solution of H2SO4 are prepared separately. Calculate the pH of each, and state which is more acidic — explain why, given equal concentrations.
Section 7
Stoichiometric (Titration) Calculations
You calculate pH of a known concentration. Titrations flip this: a measured volume and a known solution reveal an unknown concentration, mass, or purity.
Titration Toolkit
Formulae
\(n = cV \qquad n = \dfrac{m}{M} \qquad c = \dfrac{n}{V} \qquad \dfrac{c_AV_A}{a} = \dfrac{c_BV_B}{b}\)
where a and b are stoichiometric coefficients of A and B, and V is always in dm³ (1 dm³ = 1000 cm³). If a reagent is in excess: ninitial = nreacted + nexcess.
where a and b are stoichiometric coefficients of A and B, and V is always in dm³ (1 dm³ = 1000 cm³). If a reagent is in excess: ninitial = nreacted + nexcess.
Worked Examples 12–13
Worked Example 12 — basic titration
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l). 25 cm³ of 0.2 mol·dm−3 HCl is titrated with NaOH; 15 cm³ of NaOH neutralises it. Find c(NaOH).
(0.2)(0.025)/1 = cNaOH(0.015)/1 → 0.005 = 0.015 cNaOH
cNaOH = 0.33 mol·dm−3
(0.2)(0.025)/1 = cNaOH(0.015)/1 → 0.005 = 0.015 cNaOH
cNaOH = 0.33 mol·dm−3
Guided practice — diprotic acid titration
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l). 20 cm³ of 0.15 mol·dm−3 H2SO4 is titrated with NaOH; 24 cm³ neutralises it.
n(H2SO4) = cV = (0.15)(0.020) = ____ mol
Mole ratio H2SO4:NaOH = 1:2 → n(NaOH) = ____ mol
c(NaOH) = n/V = ____ / 0.024 = ____ mol·dm−3
Check your answer: c(NaOH) = 0.25 mol·dm−3
n(H2SO4) = cV = (0.15)(0.020) = ____ mol
Mole ratio H2SO4:NaOH = 1:2 → n(NaOH) = ____ mol
c(NaOH) = n/V = ____ / 0.024 = ____ mol·dm−3
Check your answer: c(NaOH) = 0.25 mol·dm−3
Worked Example 13 — excess / back-titration
A sample of CaCO3 is added to 50.0 cm³ of 1.0 mol·dm−3 HCl (acid in excess). CaCO3 + 2HCl → CaCl2 + CO2 + H2O. The excess HCl is neutralised by 28.0 cm³ of 0.5 mol·dm−3 NaOH.
1: n(HCl)initial = (1.0)(0.050) = 0.05 mol
2: n(NaOH) = (0.5)(0.028) = 0.014 mol = n(HCl)excess
3: n(HCl)reacted with CaCO₃ = 0.05 − 0.014 = 0.036 mol
4: ratio CaCO3:HCl = 1:2 → n(CaCO3) = 0.018 mol
5: m(CaCO3) = nM = (0.018)(100) = 1.8 g
1: n(HCl)initial = (1.0)(0.050) = 0.05 mol
2: n(NaOH) = (0.5)(0.028) = 0.014 mol = n(HCl)excess
3: n(HCl)reacted with CaCO₃ = 0.05 − 0.014 = 0.036 mol
4: ratio CaCO3:HCl = 1:2 → n(CaCO3) = 0.018 mol
5: m(CaCO3) = nM = (0.018)(100) = 1.8 g
Exam technique
Draw the limiting reagent (usually the one common to both reactions, e.g. the acid) in the middle of your working, with the two reactions branching off it — makes it easy to see which reagent has "complete" data and which one you're solving for.
Practice: titration and stoichiometric calculations
Same idea as the pH practice — the guided box gave support, these are independent. Use hints only if you get stuck.
- 1.KOH(aq) + HNO3(aq) → KNO3(aq) + H2O(l). 20 cm³ of 1.3 mol·dm⁻³ KOH is titrated with HNO₃; 17 cm³ neutralises it. Calculate c(HNO₃).
- 2.3Ca(OH)2(aq) + 2H3PO4(aq) → Ca3(PO4)2(aq) + 6H2O(l). 10 cm³ of 0.4 mol·dm⁻³ Ca(OH)₂ is titrated with H₃PO₄; 11 cm³ neutralises it. Calculate c(H₃PO₄).
Need a hint?
Hint 1: The mole ratio is not 1:1 — read it off the balanced equation: Ca(OH)₂:H₃PO₄ = 3:2.
Hint 2: n(H₃PO₄) = n(Ca(OH)₂) × 2/3
- 3.A 3.7 g sample of antacid (pure CaCO₃) is dissolved and made up to 500 cm³. A 25 cm³ sample is titrated with HCl; 20 cm³ of HCl is needed. CaCO₃(aq) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). Calculate c(HCl).
Need a hint?
Hint 1: First find the concentration of the CaCO₃ stock solution (the full 500 cm³) — a smaller 25 cm³ sample does not change the concentration.
Hint 2: Use that concentration with the 25 cm³ sample volume to get moles of CaCO₃ actually titrated, then apply the 1:2 ratio for moles of HCl.
- 4.25 cm³ of 0.3 mol·dm⁻³ H₂SO₄ is mixed with 25 cm³ of 0.3 mol·dm⁻³ NaOH. H₂SO₄(aq) + 2NaOH(aq) ⇌ Na₂SO₄(aq) + 2H₂O(l). Calculate the pH of the resulting mixture.
Need a hint?
Hint 1: Check whether the two reagents are actually in a 1:2 ratio as supplied — equal volumes and concentrations does not mean the exact mole ratio the equation requires.
Hint 2: Whichever reagent runs out first is limiting — the other is left over ("in excess"). The leftover acid or base determines the final pH.
- 5.Warm-up. 15 cm³ of 0.5 mol·dm⁻³ NaOH is exactly neutralised by 25 cm³ of HCl of unknown concentration. Calculate c(HCl).
- 6.Percentage purity. A 2.5 g impure sample of NaOH is dissolved and made up to 250 cm³. A 25 cm³ portion requires 22.5 cm³ of 0.20 mol·dm⁻³ HCl for complete neutralisation. HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l). Calculate the percentage purity of the NaOH sample.
Need a hint?
Hint 1: Find the concentration of NaOH in the 25 cm³ sample first (same method as question 3 above), then scale up to the full 250 cm³ to get total moles of pure NaOH.
Hint 2: % purity = (mass of pure substance / mass of impure sample) × 100
- 7.Excess/back-titration. A sample of Na₂CO₃ reacts with 40.0 cm³ of 1.2 mol·dm⁻³ HCl (acid in excess). Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g). The excess HCl is neutralised by 18.0 cm³ of 0.80 mol·dm⁻³ NaOH. Calculate the mass of Na₂CO₃ in the original sample.
Challenge questions — back-titration
Challenge 1
HCl + KOH → KCl + H2O. A solution of KOH is made by dissolving an unknown mass of KOH in a 250 cm³ beaker. 12 cm³ of this base is neutralised by 15 cm³ of acid of concentration 0.3 mol·dm−3. The pH of the final solution is 12.76. Determine the mass of KOH used.
Need a hint?
Hint 1: pH 12.76 is strongly basic — decide first whether the acid or base was in excess. This tells you which species' concentration the given pH describes.
Hint 2: Convert the pH to [OH−]excess (via Kw) in the total combined volume (12 + 15 cm³) to get moles of excess base. Add this to the moles that did react (equal to moles of acid used) for the total moles of KOH in the 12 cm³ sample — then scale up to the full 250 cm³ stock.
Challenge 2 — percentage-purity back-titration
A 2.00 g impure sample of calcium carbonate is reacted with 100 cm³ of 0.600 mol·dm−3 HCl (acid in excess). CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g). The excess acid requires 52.0 cm³ of 0.500 mol·dm−3 NaOH for complete neutralisation. Calculate the percentage purity of the calcium carbonate sample.
Need a hint?
Hint: Same pattern as Worked Example 13 and question 7 of the previous section — find moles of HCl initially present, subtract the moles neutralised by the NaOH (the excess), and the remainder reacted with the CaCO₃. Convert that to a mass of pure CaCO₃, then divide by the impure sample mass.
Summary
Mind-Map
The whole schema, from "what is an acid" through to multi-step titration calculations.
- Definitions: Arrhenius (H₃O⁺/OH⁻ producer) vs Lowry-Brønsted (proton donor/acceptor)
- Strength: strong = ionises/dissociates completely; weak = incompletely
- Concentration: amount of acid/base per volume of water (independent of strength)
- Conjugate pairs: differ by one proton; ampholytes act as both acid and base
- Neutralisation: acid + metal hydroxide / metal oxide / carbonate → salt (+ water) (+ CO₂)
- Titrations: equivalence point vs end point; choose indicator to match acid/base strength
- pH: Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴; pH = −log[H₃O⁺] (strong acids/bases only)
- Stoichiometry: n = cV, mole ratio, cAVA/a = cBVB/b, excess/back-titration
| Section | Topic |
|---|---|
| A | Definitions & conjugate pairs (MCQ) |
| B | Strength & classification |
| C | Neutralisation equations |
| D | pH calculations |
| E | Titration & stoichiometry |
| F | Challenge: back-titration |
Real NSC exam questions
From ExamBank's own database
Genuine matric-level questions, not adapted or simplified, pulled straight from ExamBank's database of past papers.
Grade 11 Physical Sciences Paper 1 — Question 4
Draws on definitions (Section 1), indicators (Section 5) and stoichiometric calculations (Section 7) in one multi-part question — exactly how acids and bases shows up in a real exam.
Question 4.1 (Lowry-Brønsted definitions, strong/weak, ampholytes) and Question 4.2 (titration of sulfuric acid against a standard potassium hydroxide solution).
Open this question page on ExamBank → | ▶ Watch the video solution
Open this question page on ExamBank → | ▶ Watch the video solution
Question 4.3 (only the top portion of this page is relevant; a Question 5 on gases follows below it, unrelated).
Open this question page on ExamBank → | ▶ Watch the video solution
Open this question page on ExamBank → | ▶ Watch the video solution
Need a hint on 4.2.2?
This is exactly Worked Example 12's method: find n(KOH) from the average titre volume, use the 2:1 mole ratio from the balanced equation, then c = n/V for the acid.
Need a hint on 4.3?
This is an excess/back-titration problem, just like Worked Example 13 — find which reagent is in excess, then use the excess concentration to get [H3O+] and pH.
Copyright reserved. Grade 11 Physical Sciences Paper 1, September 2025, KwaZulu-Natal — sourced from the ExamBank question database (exambank.org).
Grade 12 preview
These three items are sourced from a Grade 12 paper, but every one tests knowledge you already have from this page — no Ka/Kb, no weak-acid pH, no equilibrium constants. They preview what this content looks like one grade up, and a couple of ideas slightly beyond this page's core scope (titre averaging, an indicator's own equilibrium).
G12-1 — indicator equilibrium (Grade 12 P2, June 2025, Gauteng, Q1.7)
HIn(aq) + H2O(l) ⇌ H3O+(aq) + In−(aq), ΔH < 0. At equilibrium the solution is red (HIn) on the left and yellow (In−) on the right. Which will change the colour from red to yellow? A. Increasing [H3O+] B. Increasing temperature C. Adding a base D. Adding an acid
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G12-2 — equivalence point vs end point (Grade 12 P2, June 2025, Gauteng, Q1.9)
Which statement best describes the difference between the end point and the equivalence point? A. The end point occurs when the acid or base has completely reacted, while the equivalence point is when the indicator changes colour. B. The equivalence point occurs when the acid or base has completely reacted, while the end point is when the indicator changes colour. C. They occur at different times with no connection. D. They are always exactly the same.
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G12-3 — titration calculation with concordant titres (Grade 12 P2, June 2025, Gauteng, Q1.10)
A titration used a 0.1 mol·dm⁻³ NaOH standard solution against HCl of unknown concentration. In each titration, 20 cm³ of NaOH was used. Burette readings for HCl: Titration 1 = 26.66 cm³, Titration 2 = 26.50 cm³, Titration 3 = 26.60 cm³. What is c(HCl), in mol·dm⁻³? A. 0.0752 B. 0.0750 C. 0.754 D. 0.0755
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▶ Watch the video solution | Open on ExamBank →
Copyright reserved. Grade 12 Physical Sciences Paper 2, June 2025, Gauteng — sourced from the ExamBank question database (exambank.org). Full marking guidelines →