6L

Acid/base reactions review

A summary of Module 6 Acid/Base Reactions, how the ideas connect, the key formulas and words, and a mixed HSC-style test covering acid reactions, indicators, enthalpy of neutralisation, models of acids, pH and Ka calculations, titrations and buffers.

Use this page to check the whole module. Read the summary, make sure you can use every formula in the list without looking it up, then do the test in exam conditions.

Module summary

TopicWhat you need to know
6A Properties and reactionsNames and formulae of common acids and bases. Acid + base → salt + water; + carbonate → salt + water + CO₂; + metal → salt + H₂. Full and net ionic equations (keep weak acids, solids and gases whole). Acidic, basic and amphoteric oxides. Uses of neutralisation: liming soil, antacids, spills, mine drainage.
6B IndicatorsHIn ⇌ H⁺ + In⁻ with two colours; Le Chatelier explains the change; range ≈ p$K_a$ ± 1. Universal indicator. Natural indicators (anthocyanins) and the preparation practical. pH meters need calibration.
6C Enthalpy of neutralisationΔ$H$ per mole of water; ≈ −57 kJ/mol for every strong–strong pair; less for weak acids and bases (ionisation absorbs energy). $q = mc\Delta T$ with total mass; Δ$H = -q/n$. Polystyrene, lid, extrapolation.
6D Changing modelsLavoisier (oxygen), Davy (replaceable H), Arrhenius (H⁺ and OH⁻ in water), Brønsted–Lowry (proton donor and acceptor), Lewis (electron pairs). Limitations of each with evidence: HCl, NH₃, NH₃(g) + HCl(g), acidic and basic salts, solvent effects, BF₃.
6E Conjugate pairs and amphiprotic speciesPairs differ by one H⁺. Strong acid ↔ negligible conjugate base. Proton transfer favours the weaker acid. Polyprotic acids ionise stepwise. Amphiprotic: H₂O, HCO₃⁻, H₂PO₄⁻ (two equations each). Acidic, basic and neutral salts.
6F pH, pOH and $K_w$$K_w$ = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C; increases with temperature (neutral pH < 7 when hot). pH, pOH, pH + pOH = 14. Dilution ($c_1V_1 = c_2V_2$) and mixing (excess moles ÷ total volume). Measuring pH practical.
6G Strong and weak, $K_a$Strength ≠ concentration. $K_a$ = [H₃O⁺][A⁻]/[HA]; p$K_a$ = −log₁₀$K_a$. ICE table, $x = \sqrt{K_ac}$, 5% check. % ionisation rises on dilution. $K_a$ from measured pH. $K_b$ and $K_aK_b = K_w$.
6H Standard solutions and titrationPrimary standards (pure, stable, non-hygroscopic, high M): Na₂CO₃, KHP. Standardising NaOH. Rinsing rules. Meniscus to 0.05 mL; concordant titres within 0.10 mL. Calculations with dilution factors. Systematic and random errors.
6I Curves and conductivitySA/SB pH 7; WA/SB ≈ 8.7 (phenolphthalein); SA/WB ≈ 5.3 (methyl orange); WA/WB no jump. Half-equivalence pH = p$K_a$. Na₂CO₃ has two end points. Conductometric graphs: H₃O⁺ and OH⁻ most mobile; equivalence where lines meet.
6J Back titrations and analysisExcess reagent, titrate the leftover: reacted = total − excess. For insoluble, slow or volatile samples. Household analysis. Titration in wine, food, pharmaceuticals, environment; strengths and limitations.
6K BuffersWeak acid + conjugate base; equations and Le Chatelier for added acid and base. Capacity (concentration, 1 : 1 ratio), range p$K_a$ ± 1. pH = p$K_a$ + log([A⁻]/[HA]). Blood (H₂CO₃/HCO₃⁻, 7.35–7.45), cells (phosphate), ocean (carbonate; acidification).

Almost everything in this module comes from one idea: an acid–base reaction is a proton moving from one particle to another. Arrhenius saw only the special case of H⁺ and OH⁻ in water; Brønsted and Lowry saw that any particle with a proton to give can be an acid and any particle with a lone pair to take it can be a base. That picture explains why NH₄Cl is acidic and Na₂CO₃ is basic, why water is amphiprotic, and why every strong acid–strong base pair releases the same 57 kJ/mol.

The second idea is equilibrium. A weak acid is an equilibrium between molecules and ions, described by $K_a$; water is an equilibrium described by $K_w$; an indicator is an equilibrium whose two sides are different colours; and a buffer is an equilibrium with large reserves on both sides. Le Chatelier's principle explains all four. When you're stuck on a "why" question in this module, write the equilibrium equation and ask which way it shifts.

The third idea is stoichiometry. Strength decides the pH, but the mole ratio alone decides how much base an acid neutralises. That is why titration measures concentration regardless of strength, and why the titration curve's shape (set by $K_a$) only matters for choosing the indicator. Back titrations, standardisation and buffer calculations are all the same bookkeeping: count the moles of H⁺ that move, using $n = cV$.

Key formulas

QuantityFormula
pH and pOH$\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}]$, $\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}]$, pH + pOH = 14.00 (25 °C)
ionisation of water$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ (25 °C)
acid dissociation constant$K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$, $\mathrm{p}K_a = -\log_{10}K_a$
weak acid (approximation)$[\mathrm{H_3O^+}] = \sqrt{K_a c}$ if under 5% ionised
conjugate pair$K_a \times K_b = K_w$
moles in solution; dilution$n = cV$; $c_1V_1 = c_2V_2$
heat and enthalpy$q = mc\Delta T$; $\Delta H = -q/n$
buffer (extension)$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}$

Key words

TermMeaning
acida substance that donates a proton (Brønsted–Lowry)
basea substance that accepts a proton (Brønsted–Lowry)
alkalia base that dissolves in water
conjugate basethe species formed when an acid loses a proton
amphiproticable to act as a proton donor or a proton acceptor
amphotericable to react as an acid or a base, e.g. ZnO, Al₂O₃
strong acidionises completely in water
weak acidionises only partially, setting up an equilibrium
acid dissociation constantthe equilibrium constant for the ionisation of an acid in water
standard solutiona solution whose concentration is accurately known
primary standarda pure, stable substance used to make a standard solution by mass
titrethe volume of titrant delivered from the burette
aliquotan accurately measured volume delivered by a pipette
equivalence pointwhen reactants have been mixed in the exact mole ratio of the equation
end pointwhen the indicator changes colour
back titrationadding a known excess of reagent and titrating what is left over
buffera solution that resists changes in pH when small amounts of acid or base are added
buffer capacitythe amount of acid or base a buffer can absorb before its pH changes significantly

Test yourself

  1. Which is the net ionic equation for the reaction of solid magnesium carbonate with dilute nitric acid?

  2. Which species is the conjugate acid of HPO₄²⁻?

  3. What is the pH of the solution formed when 10.0 mL of 0.200 mol/L Ba(OH)₂ is added to 30.0 mL of 0.100 mol/L HCl?

  4. A 0.050 mol/L solution of a weak acid has pH 3.00. What is the percentage ionisation?

  5. Which indicator is suitable for titrating methanoic acid with potassium hydroxide?

  6. Which statement about a buffer made from equal amounts of NH₃ and NH₄Cl is correct?

  7. Which observation could the Arrhenius model explain?

Label the titration curve.

Strong acid + strong base071402550volume added (mL)pHequivalence pH 7.0equivalence pointWeak acid + strong base071402550volume added (mL)pHequivalence pH 8.7half-equivalence: pH = pKa = 4.74buffer regionStrong acid + weak base071402550volume added (mL)pHequivalence pH 5.3steep jump is shortWeak acid + weak base071402550volume added (mL)pHequivalence pH 7.0no sharp jump: no indicator works

A student titrates an unknown acid (pipetted into the flask) with standard NaOH from the burette. Sort each mistake by its effect.

In the Brønsted–Lowry model an acid is a proton . A species that can donate or accept a proton is . At 25 °C, pH + pOH = . At the half-equivalence point of a weak acid titration, pH = . A pure, stable substance used to make a standard solution is a . A mixture of a weak acid and its conjugate base is a .

7 marks

A student investigated the acidity of a brand of apple juice, which contains mainly malic acid (C₄H₆O₅, a diprotic weak acid, M = 134.09 g/mol). 25.00 mL of juice was diluted to 100.0 mL. 20.00 mL aliquots of the diluted juice were titrated with 0.1040 mol/L NaOH using phenolphthalein. Titres: 15.20 (rough), 14.65, 14.70, 14.60 and 14.95 mL. (a) Calculate the concentration of malic acid in the juice in g/L. (b) Justify the choice of indicator. (c) The student's teacher suggested using a pH probe instead because apple juice is coloured. Evaluate this suggestion.