3.1 Proton transfer reactions
- Syllabus
- First assessment 2025
- Topic
- 3.1
- Level
- SL
A Brønsted–Lowry acid donates H+ and a Brønsted–Lowry base accepts H+. An alkali is a base that is soluble in water.
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.
Representative question
Describe whether ammonia acts as a Brønsted-Lowry acid or base in its reaction with water. Include an equation in your answer.
base AND accepts H+/hydrogen ion/proton NH3( g)+H2O(I)⇌NH4OH(aq)ORNH3( g)+H2O(l)⇌NH4+(aq)+OH−(aq)↓
Accept either type of arrow
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.
Representative question
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:
conjugate acid H3O+«(aq)» AND conjugate base H2O<(l)»
conjugate acid HNO2 «(aq)» AND conjugate base NO2−«(aq)»
An amphiprotic species can donate H+ in one reaction and accept H+ in another.
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.
Representative question
Formulate two equations to show the amphiprotic nature of H2PO4−.
H2PO4−(aq)+H+(aq)→H3PO4(aq)H2PO4−(aq)+OH−(aq)→HPO42−(aq)+H2O(l)
Accept reactions of H2PO4− with any acidic, basic or amphiprotic species, such as H3O+, NH3 or H2O.
Accept:
H2PO4−(aq)→HPO42−(aq)+H+(aq)
for M2.
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.
Representative question
A solution has a pH of 3.0 . What is the hydrogen ion concentration in the solution in moldm−3 ?
3.0×10−3
1.0×10−3
1.0×103
3.0×103
B
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.
Representative question
Calculate the concentration of hydroxide ions in an ammonia solution with pH=9.3. Use sections 1 and 2 of the data booklet.
[OH−]⟨⟨=[H+]Kw=10−9.310−14=10−4.7⟩⟩=2.0×10−5⟨⟨ moldm−3⟩⟩
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.
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.
Representative question
Explain the difference in pH .
nitrous acid/ HNO2 is not fully dissociated/is a weak acid.
OR
HCI is fully dissociated/is a strong acid.
«HNOX2» lower concentration of H+ions
OR
«HCl» higher concentration of H+ions.
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.
Representative question
Write two equations showing how these antacids neutralize excess hydrochloric acid.
Magnesium carbonate:
Aluminium hydroxide:
MgCO3(s)+2HCl(aq)→MgCl2(aq)+H2O(l)+CO2(g)Al(OH)3(s)+3HCl(aq)→AlCl3(aq)+3H2O(l)
Accept appropriate ionic equations.
Do not accept H2CO3 as a product of the first reaction.
Ignore equilibrium arrows.
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.
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.
Representative question
Which graph would be obtained by adding 0.10moldm−3HCl(aq) to 25 cm3 of 0.10moldm−3NaOH(aq) ?
B
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.