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3.1 Proton transfer reactions

Syllabus
First assessment 2025
Topic
3.1
Level
SL

Learning objectives

3.1.1Brønsted-Lowry theory• Acid = proton donor• Base = proton acceptor• Deduce acid and base in a reaction; distinguish base and alkali3.1.2Conjugate acid-base pairs• Differ by one proton• Deduce conjugate acid/base formulas3.1.3Amphiprotic species• Can act as both acid and base• Write equations showing acid and base behaviour3.1.4pH scale• pH = −log₁₀[H⁺]• [H⁺] = 10⁻ᵖᴴ• Calculate pH and [H+]; include universal indicator and pH probe use3.1.5Ion product of water (Kw)• Kw = [H⁺][OH⁻]• Acidic: [H⁺] > [OH⁻]• Neutral: [H⁺] = [OH⁻]• Basic: [H⁺] < [OH⁻]• Recognize acidic, neutral, and basic solutions from [H+] and [OH-]3.1.6Strong vs. weak acids/bases• Differ in extent of ionization• Equilibrium favors weaker conjugate• Distinguish strong/weak from concentrated/dilute3.1.7Neutralization reactions• Acid + metal oxide/hydroxide/carbonate/hydrogencarbonate• Formulate equations and identify parent acids/bases of salts3.1.8pH curves• Strong acid-strong base titrations• Characteristic shapes• Sketch and interpret intercept and equivalence point for monoprotic titrations

Brønsted–Lowry Acids and Bases

A Brønsted–Lowry acid donates H+ and a Brønsted–Lowry base accepts H+. An alkali is a base that is soluble in water.

HCl(g) is the proton donor and Brønsted–Lowry acid; NH3(g) is the proton acceptor and Brønsted–Lowry base; the proton-transfer cue runs from HCl to NH3; the overall product is unambiguously NH4Cl(s), composed of NH4+ and Cl−.

Follow the proton: the species losing it is the acid and the species gaining it is the base.

Pair species that differ by exactly one H⁺ to identify conjugate acid–base pairs. Charge alone does not decide the role: in NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, NH₄⁺ is the proton donor and water is the acceptor.

Assigning Acid and Base Roles

2 marks

Describe whether ammonia acts as a Brønsted-Lowry acid or base in its reaction with water. Include an equation in your answer.

Conjugate Acid–Base Pairs

A conjugate base is what remains after an acid donates one proton. A conjugate acid is formed when a base accepts one proton; the pair differs by exactly one H+.

Remove H+ to find the conjugate base or add H+ to find the conjugate acid, then check the charge changes by one unit.

NH₄⁺/NH₃ and H₂CO₃/HCO₃⁻ are conjugate pairs because each pair differs by one H⁺. Removing H⁺ lowers charge by one; adding H⁺ raises it by one. Do not pair species merely because they occur on opposite sides of an equation—trace the specific proton transfer.

Deducing Conjugate Formulae

2 marks

A solution of nitrous acid contains two conjugate acid-base pairs.

State the formulas of the conjugate acid and conjugate base in each pair.

Conjugate acid:
Conjugate base:
Conjugate acid:
Conjugate base:

Amphiprotic Species

An amphiprotic species can donate H+ in one reaction and accept H+ in another.

water donates a proton in H2O(l) ⇌ H+(aq) + OH−(aq); water accepts a proton in H2O(l) + H+(aq) ⇌ H3O+(aq); acid means proton donor and base means proton acceptor; amphiprotic is stated as the ability to donate or accept H+.

Write one equation in which the species becomes its conjugate base and another in which it becomes its conjugate acid.

For HCO₃⁻, donation gives CO₃²⁻ whereas acceptance gives H₂CO₃. Showing both reactions is the evidence for amphiprotic behaviour; one acid–base equation alone is insufficient.

Showing Amphiprotic Behaviour

2 marks

Formulate two equations to show the amphiprotic nature of H2PO4−\mathrm{H}_{2} \mathrm{PO}_{4}^{-}.

pH and Hydrogen-Ion Concentration

pH=−log10[H+];[H+]=10(−pH)pH = −log10[H+]; [H+] = 10^(−pH)

pH is logarithmic: a one-unit change represents a tenfold concentration change. Universal indicator gives a colour range; a pH probe gives an instrumental pH measurement.

For [H⁺] = 2.0 × 10⁻³ mol dm⁻³, pH = 2.70; the leading 2 makes the answer non-integer. A colour indicator estimates a range, whereas a calibrated probe supports a numerical measurement.

Calculating pH and [H+]

1 mark

A solution has a pH of 3.0 . What is the hydrogen ion concentration in the solution in moldm−3\mathrm{mol} \mathrm{dm}^{-3} ?

The Ion Product of Water

Kw=[H+][OH−]Kw = [H+][OH−]

Solution Ion comparison
acidic [H+] > [OH−]
neutral [H+] = [OH−]
basic [H+] < [OH−]

At 25 °C, Kw = 1.0 × 10⁻¹⁴, so a neutral solution has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³. Neutrality always means equal ion concentrations; neutral pH is not necessarily 7 when temperature changes.

At a fixed temperature, Kw is constant, so [OH-] = Kw/[H+]: a higher [H+] means a lower [OH-]. For example, at pH 9.3 and 25 C, [OH-] = 2.0 x 10^-5 mol dm^-3. Classify a solution from the ion comparison; do not assume neutral pH is 7 at every temperature.

Classifying Solutions with Kw

1 mark

Calculate the concentration of hydroxide ions in an ammonia solution with pH=9.3. Use sections 1 and 2 of the data booklet.

Strong and Weak Acids and Bases

A strong acid or base ionizes completely in aqueous solution; a weak acid or base ionizes only partially. The equilibrium favours the weaker conjugate species.

same-concentration strong acid shows six H3O+/A− pairs and no undissociated acid; same-concentration weak acid shows five HB plus one H3O+ and one B−, totalling six acid formula units; same-strength concentration comparison uses six versus two ion pairs in equal volumes; strength and concentration are explicitly independent.

Strength is the extent of ionization, whereas concentration is the amount of solute per volume. A concentrated weak acid can be more acidic than a dilute strong acid.

Represent a strong acid with essentially complete ionization and a weak acid with an equilibrium containing substantial undissociated acid. Strength is an equilibrium property, while concentration is an initial amount per volume; pH depends on both, so strength alone cannot rank arbitrary solutions.

Distinguishing Strength from Concentration

2 marks

Explain the difference in pH .

Neutralization Reactions

Acids neutralize metal oxides and hydroxides to form salt and water. Carbonates and hydrogencarbonates also produce carbon dioxide when the reaction requires it; balance all formulae and coefficients.

Identify the parent acid and parent base of a salt by tracing its anion and cation back to the neutralization reactants.

Balance proton capacity as well as atoms: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, while an acid–carbonate reaction also releases CO₂. To identify parents of Na₂SO₄, trace SO₄²⁻ to the acid and Na⁺ to the base rather than treating the salt name as a reaction equation.

Writing Neutralization Equations

1 mark

Write two equations showing how these antacids neutralize excess hydrochloric acid.

Magnesium carbonate:

Aluminium hydroxide:

Strong-Acid–Strong-Base Titration Curves

The equivalence point is where stoichiometric amounts of analyte and titrant have reacted. A monoprotic strong-acid–strong-base curve has a steep neutral region centred at the equivalence point.

HCl is the strong-acid analyte and NaOH is the titrant; the curve rises from low pH and the equivalence point is on pH 7; HCl plus NaCl appears before equivalence and NaCl plus NaOH after equivalence; the steep region is identified as a pH jump without inventing a volume.
NaOH is the strong-base analyte and HCl is the titrant; the curve falls from high pH and the equivalence point is on pH 7; NaOH plus NaCl appears before equivalence and NaCl plus HCl after equivalence; the steep region is identified as a pH drop without inventing a volume.

Read the initial pH, steep intercept region and final plateau; curve direction depends on whether acid or base is added.

For a strong acid titrated with strong base at 25 °C, calculate the initial pH from excess acid, locate equivalence from stoichiometric moles, and place the steep section around pH 7. Equivalence is a mole condition; it is not the same as equal solution volumes unless concentrations and stoichiometry make it so.

Interpreting a Strong Titration Curve

1 mark

Which graph would be obtained by adding 0.10moldm−3HCl(aq)0.10 \mathrm{moldm}^{-3} \mathrm{HCl}(\mathrm{aq}) to 25 cm325 \mathrm{~cm}^{3} of 0.10moldm−3NaOH(aq)0.10 \mathrm{moldm}^{-3} \mathrm{NaOH}(\mathrm{aq}) ?

Proton Transfer Reactions Summary

Retrieve the route: track proton transfer and conjugates, calculate pH and Kw, distinguish strength, balance neutralization, read titration curves, use Ka/Kb and hydrolysis, select indicators, and explain and calculate buffer behaviour.

Check donor versus acceptor, one-proton differences, logarithm direction, ion comparison, strength versus concentration, equivalence versus endpoint, pKa landmarks, conjugate equations and dilution ratios.