5 higher-level sections hidden.
Educerie · IB Diploma · Chemistry
Reactivity 3 What are the mechanisms of chemical change? · R3.1 Proton transfer reactions
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Deduce the Brønsted–Lowry acid and base in a reaction, and tell a base from an alkali | SL, HL | "Identify the Brønsted–Lowry base in the forward reaction" (1 mark) |
| Deduce the conjugate acid or base of any species | SL, HL | "State the formula of the conjugate base of H₂PO₄⁻" (1 mark) |
| Write equations showing a species acting as an acid and as a base | SL, HL | "Formulate two equations to show that HCO₃⁻ is amphiprotic" (2 marks) |
| Calculate pH from [H⁺] and [H⁺] from pH | SL, HL | Paper 2 calculation, 1–2 marks, or a Paper 1A comparison of two solutions |
| Use Kw to classify solutions and find [H⁺] or [OH⁻] | SL, HL | "Calculate [H⁺] in 0.020 mol dm⁻³ NaOH(aq)" (2 marks) |
| Distinguish strong from weak and concentrated from dilute, and say which way an acid–base equilibrium lies | SL, HL | "Outline one experiment that distinguishes …" (2–3 marks), often Paper 1B |
| Write neutralization equations for oxides, hydroxides, carbonates, hydrogencarbonates, ammonia and amines, and name the parent acid and base of a salt | SL, HL | "Formulate the equation for …" (1–2 marks) |
| Sketch and interpret the pH curve for a strong acid and a strong base | SL, HL | "Sketch the pH curve, labelling the equivalence point" (2–3 marks) |
| Interconvert [H⁺], [OH⁻], pH and pOH | HL only | Paper 2, 1–2 marks |
| Interpret and calculate with Ka, Kb, pKa and pKb, including Ka × Kb = Kw | HL only | "Calculate the pH of 0.10 mol dm⁻³ …" (3 marks) |
| Write hydrolysis equations for ions in a salt and predict the pH | HL only | "Explain, with an equation, why sodium ethanoate solution is basic" (2 marks) |
| Interpret the four pH curves: start, equivalence point, buffer region, pH = pKa | HL only | Paper 1B or 2: a curve is given, 3–5 marks |
| Explain how an indicator works and choose one; distinguish end point from equivalence point | HL only | "Identify a suitable indicator, giving a reason" (2 marks) |
| Describe and explain buffers, and calculate their pH and composition | HL only | "Explain, using equations, how this buffer resists …" (3 marks) plus a calculation |
Before you start
You need concentration in mol dm⁻³ and titration calculations from Structure 1.4, because a pH curve is a titration drawn as a graph. You need the equilibrium law and Le Châtelier's principle from Reactivity 2.3: weak acids, Kw, indicators and buffers are all equilibria. You need the formulas of the polyatomic ions from Structure 2.1 (sulfate, nitrate, carbonate, hydrogencarbonate, phosphate, ammonium). The only new maths is the logarithm to base 10, and your calculator does it.
1The idea in one paragraph
An acid is a species that gives away a proton, H⁺, and a base is a species that takes one. The proton cannot float about on its own, so every acid needs a base to receive it: acid–base chemistry is the transfer of one proton from one species to another, and that transfer can often run backwards, so there are two acids and two bases in every equilibrium. How many protons are free in a solution is measured on the pH scale, which counts powers of ten. A strong acid gives up all its protons in water; a weak acid gives up only a few, which is a question of strength, not of how much acid is in the bottle. When an acid meets a base they neutralize each other and make a salt. At HL you put numbers on the strength of weak acids and bases (Ka and Kb), follow the pH through a titration of any combination, and use the fact that a weak acid with its partner base can soak up added acid or alkali: a buffer.
2Brønsted–Lowry acids and bases
A Brønsted–Lowry acid is a proton donor. A Brønsted–Lowry base is a proton acceptor.
A hydrogen atom without its electron is just a proton, so H⁺ and "proton" mean the same thing. When hydrogen chloride dissolves in water, each HCl molecule hands its proton to a water molecule:
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
HCl is the acid (it donated the proton) and H₂O is the base (it accepted it). The ion made, H₃O⁺, is the oxonium ion (also called the hydronium ion). A bare proton is too small and too highly charged to exist alone in water; it is always bonded to a water molecule through a lone pair on the oxygen. So H⁺(aq) and H₃O⁺(aq) are two ways of writing the same thing, and either is accepted.
To deduce which species is the acid, follow the proton: the species that ends up with one H fewer was the acid; the one that ends up with one H more was the base. In NH₃ + HCl → NH₄⁺ + Cl⁻, HCl is the acid and NH₃ the base, although no water appears at all.
Base or alkali? A base is any proton acceptor: copper(II) oxide, the carbonate ion, ammonia, the hydroxide ion. An alkali is a base that dissolves in water to give hydroxide ions, OH⁻(aq). Sodium hydroxide and aqueous ammonia are alkalis; copper(II) oxide is a base but not an alkali, because it does not dissolve. Every alkali is a base; not every base is an alkali.
Why the definition keeps changing (nature of science). Lavoisier thought oxygen made things acidic (its name means "acid-former"); hydrochloric acid, with no oxygen, broke that model. Arrhenius defined acids as substances releasing H⁺ in water, which tied acidity to water. In 1923 Brønsted and Lowry, independently, defined acids by proton transfer, so a reaction between two gases could count, and in the same year Lewis widened it to electron pairs (Reactivity 3.4). Each model was replaced because it failed to explain something new, not because it was useless.
3Conjugate acid–base pairs
When an acid gives away its proton, what is left can take a proton back, so it is a base. When a base takes a proton, the product can give it away again, so it is an acid. A conjugate acid–base pair is two species that differ by a single proton. Figure 1 marks the two pairs in the reactions of ethanoic acid and ammonia with water.
To find a conjugate base, take one H⁺ off the formula: remove one H and lower the charge by one. To find a conjugate acid, add one H⁺: add one H and raise the charge by one.
| Species | Conjugate base (remove H⁺) | Conjugate acid (add H⁺) |
|---|---|---|
| H₂O | OH⁻ | H₃O⁺ |
| NH₃ | NH₂⁻ | NH₄⁺ |
| H₂SO₄ | HSO₄⁻ | — |
| HSO₄⁻ | SO₄²⁻ | H₂SO₄ |
| H₂PO₄⁻ | HPO₄²⁻ | H₃PO₄ |
| CO₃²⁻ | — | HCO₃⁻ |
| CH₃NH₂ | — | CH₃NH₃⁺ |
The polyatomic anions from Structure 2.1 are all conjugate bases: sulfate of HSO₄⁻, nitrate of HNO₃, carbonate of HCO₃⁻, phosphate of HPO₄²⁻. The commonest slip is changing the charge the wrong way: taking away a positive charge makes the ion more negative.
4Species that can be both
Water was the base when HCl dissolved and the acid when NH₃ dissolved. A species that can donate or accept a proton is amphiprotic. To be amphiprotic it must have a hydrogen it can give up and a lone pair that can accept a proton. The hydrogencarbonate ion is the standard example, and "formulate two equations" is the standard question:
- as a base, with an acid: HCO₃⁻(aq) + H₃O⁺(aq) → H₂CO₃(aq) + H₂O(l)
- as an acid, with a base: HCO₃⁻(aq) + OH⁻(aq) → CO₃²⁻(aq) + H₂O(l)
H₂PO₄⁻, HPO₄²⁻ and HSO₄⁻ (as a base only weakly) behave the same way. A related word is amphoteric: able to react as an acid and as a base. Every amphiprotic species is amphoteric, but aluminium oxide is amphoteric without being amphiprotic, since it has no hydrogen to donate. It reacts with acids and with hydroxide solutions.
This links to Structure 3.1: across period 3 the oxides change from basic (Na₂O + H₂O → 2NaOH) through amphoteric (Al₂O₃) to acidic (SO₃ + H₂O → H₂SO₄). Non-metal oxides are acidic, which is why sulfur dioxide from fuels and nitrogen oxides from engines cause acid rain: SO₂ dissolves to give H₂SO₃ (or is oxidized on to H₂SO₄), and NO₂ reacts with water to give HNO₃ and HNO₂.
5pH, and the ion product of water
Concentrations of H⁺ in real solutions run from about 1 mol dm⁻³ down to 10⁻¹⁴ mol dm⁻³, a range too wide for everyday numbers. The pH scale compresses it with a logarithm:
pH = −log₁₀[H⁺] and [H⁺] = 10⁻ᵖᴴ
Square brackets mean "concentration in mol dm⁻³". Both equations are in the data booklet. Three things follow from the log.
- A lower pH means more H⁺. The minus sign turns small concentrations into positive numbers.
- One pH unit is a factor of ten. A solution at pH 3 has ten times the [H⁺] of one at pH 4 and a hundred times that of one at pH 5.
- The relationship is a curve, not a line. Figure 2 is the sketch graph the guide asks for: halving [H⁺] does not halve or double the pH; it raises it by only 0.30.
Worked example 1. Find the pH of 0.0250 mol dm⁻³ HCl(aq), and [H⁺] in a solution of pH 3.40.
Give a pH to two decimal places when the data allow; a pH has no unit. Diluting that HCl a hundredfold makes [H⁺] = 2.50 × 10⁻⁴ mol dm⁻³ and pH 3.60: two units up for a factor of a hundred.
Measuring pH. Universal indicator, a mixture of indicators compared against a colour chart, gives an estimate to about one pH unit, and a coloured sample can mislead it. A pH meter or probe, calibrated with buffers of known pH, reads to 0.01. A digital probe with a data logger records continuously, which is what you want during a titration, with a coloured solution, or when you need precision; indicator paper is quicker and cheaper for a rough check.
Water itself ionizes, very slightly. One water molecule passes a proton to another:
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), or simply H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
The equilibrium constant for this, with the nearly constant [H₂O] folded in, is the ion product constant of water:
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K
This is the typical figure; use the value printed in your data booklet. Because the product is fixed, [H⁺] and [OH⁻] move in opposite directions: raise one tenfold and the other falls tenfold. Figure 3 sets both against the pH scale.
The classification follows directly, and it is a statement about the two concentrations, not about the number 7:
- acidic: [H⁺] > [OH⁻]
- neutral: [H⁺] = [OH⁻]
- basic (alkaline): [OH⁻] > [H⁺]
In pure water at 298 K, [H⁺] = [OH⁻] = √(1.00 × 10⁻¹⁴) = 1.00 × 10⁻⁷ mol dm⁻³, so pH = 7.00.
Worked example 2. Find [H⁺] and the pH of 0.0200 mol dm⁻³ NaOH(aq) at 298 K.
Kw depends on temperature. The ionization of water breaks an O–H bond, so it is endothermic. By Le Châtelier's principle, heating pushes the equilibrium to the right and Kw rises: at 323 K it is about 5.5 × 10⁻¹⁴. Pure water at that temperature has [H⁺] = √(5.5 × 10⁻¹⁴) = 2.3 × 10⁻⁷ mol dm⁻³ and a pH of about 6.6. It is still exactly neutral, because [H⁺] still equals [OH⁻]. "pH 7 means neutral" is true only at 298 K.
6Strong and weak; concentrated and dilute
A strong acid ionizes completely in water; a weak acid ionizes only partially, so most of its molecules stay whole and an equilibrium is set up. The same goes for bases.
- HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq) (single arrow: complete)
- CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) (equilibrium: in 0.10 mol dm⁻³ ethanoic acid only about 1% of molecules are ionized)
- NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) (a weak base)
The strong acids you must know are HCl, HBr, HI, HNO₃ and H₂SO₄. The strong bases are the group 1 hydroxides (LiOH, NaOH, KOH …). Treat everything else in this course as weak unless you are told otherwise: carboxylic acids, carbonic acid, ammonia and the amines are all weak.
Why equilibrium lies one way. Every acid–base equilibrium contains two acids competing to hold the proton, and it lies towards the side with the weaker acid and the weaker base. HCl is a much stronger acid than H₃O⁺, so the proton ends up on water and the reaction goes to completion. Ethanoic acid is a weaker acid than H₃O⁺, so the proton mostly stays on the ethanoic acid and the equilibrium lies to the left. It follows that the stronger an acid, the weaker its conjugate base: Cl⁻ has almost no tendency to take a proton back, while CH₃COO⁻ takes one readily.
Strength also follows structure. Down group 17 the hydrogen halides get more acidic: HF is weak, and HCl, HBr and HI are strong, increasingly so. The halogen atom gets larger, the H–X bond gets longer and weaker (lower bond enthalpy, Reactivity 1.2), and the proton is released more easily.
Strong is not concentrated. Concentration is how many moles of acid are dissolved per dm³; strength is what fraction of them ionize. They are independent, as Figure 4 shows. A dilute solution of a strong acid can have a higher pH than a concentrated solution of a weak one.
Telling them apart in the laboratory. Compare a strong and a weak acid at the same concentration, say 0.10 mol dm⁻³ HCl and CH₃COOH. The strong acid has a higher [H⁺], so:
| Test | Strong acid | Weak acid | Why |
|---|---|---|---|
| pH (probe or indicator) | lower (about 1.0) | higher (about 2.9) | more H⁺ ions |
| Electrical conductivity | higher | lower | more ions carry the current |
| Rate with magnesium ribbon or CaCO₃ | faster bubbling | slower | more frequent collisions with H⁺ |
| Volume of NaOH to neutralize | the same | the same | same amount of acid; the weak acid keeps ionizing as H⁺ is removed |
| Total volume of gas with excess Mg | the same | the same | the same reason |
The last two rows catch students every year: a weak acid holds its protons back, but they are all still there to be neutralized.
7Neutralization and salts
A neutralization reaction is one between an acid and a base that forms a salt and, usually, water. Strip out the spectator ions from an acid–alkali reaction and what is left is the same every time:
H⁺(aq) + OH⁻(aq) → H₂O(l)
That is why neutralization is always exothermic (Reactivity 1.1): the reaction forms a new O–H bond and breaks none. The four families of reactions you must write:
| Acid + | Products | Example |
|---|---|---|
| metal oxide | salt + water | 2HNO₃(aq) + CuO(s) → Cu(NO₃)₂(aq) + H₂O(l) |
| metal hydroxide | salt + water | H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l) |
| carbonate | salt + water + carbon dioxide | 2HCl(aq) + Na₂CO₃(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g) |
| hydrogencarbonate | salt + water + carbon dioxide | CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O(l) + CO₂(g) |
Ammonia and amines are bases because the lone pair on nitrogen accepts a proton. With them the salt forms and no water is made:
- NH₃(aq) + HCl(aq) → NH₄Cl(aq)
- CH₃CH₂NH₂(aq) + HNO₃(aq) → CH₃CH₂NH₃⁺NO₃⁻(aq), ethylammonium nitrate
Organic acids behave like any other acid, only weakly: 2CH₃COOH + Mg(OH)₂ → (CH₃COO)₂Mg + 2H₂O, giving magnesium ethanoate.
Parent acid and base. Every salt is a cation from a base and an anion from an acid. To find the parents, give the anion back its proton and the cation back its hydroxide (or its lone pair).
| Salt | Parent acid | Parent base |
|---|---|---|
| KNO₃ | HNO₃ (strong) | KOH (strong) |
| NH₄Cl | HCl (strong) | NH₃ (weak) |
| CH₃COONa | CH₃COOH (weak) | NaOH (strong) |
| Na₂CO₃ | H₂CO₃, carbonic acid (weak) | NaOH (strong) |
At HL the strengths of the parents give the pH of the salt solution (section 10). An acid with a reactive metal, Mg + 2HCl → MgCl₂ + H₂, also gives a salt, but it is not a neutralization: magnesium is oxidized and H⁺ reduced, a redox reaction (Reactivity 3.2).
8The pH curve for a strong acid and a strong base
A pH curve is a titration watched with a pH probe: the pH of the flask is plotted against the volume of titrant added. Figure 5 is the curve for 0.100 mol dm⁻³ NaOH added to 25.0 cm³ of 0.100 mol dm⁻³ HCl, calculated point by point.
Read four features from it.
- The intercept with the pH axis is the pH of the acid alone: [H⁺] = 0.100 mol dm⁻³, so pH 1.00. The intercept tells you the concentration of a strong acid.
- The slow rise. At first each cm³ of alkali removes only a small fraction of the acid, and [H⁺] falls only slowly on a log scale. After 24.9 cm³, 99.6% of the acid has gone and the pH is still only 3.7.
- The equivalence point, where the amounts of acid and base are exactly stoichiometric, which is why it is also called the stoichiometric point. Here n(NaOH) = n(HCl) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol, reached at 25.0 cm³. The solution contains only NaCl and water, so the pH is 7. The curve is almost vertical here: from 24.9 cm³ to 25.1 cm³, about four drops, the pH jumps from 3.7 to 10.3.
- The flattening after equivalence, as excess OH⁻ builds up towards the pH of the alkali itself. After 30.0 cm³ there is 5.0 × 10⁻⁴ mol of excess OH⁻ in 55.0 cm³, pH 11.96; the curve approaches, but never reaches, 13.
The vertical section is why a titration works: a drop either side of equivalence changes the pH hugely, so an indicator changing anywhere between about pH 4 and 10 marks it sharply. Only monoprotic reactions are assessed.
9HLpOH, Ka and Kb
SL students can skip to section 14.
pOH does for hydroxide ions what pH does for hydrogen ions:
pOH = −log₁₀[OH⁻] [OH⁻] = 10⁻ᵖᴼᴴ pH + pOH = 14.00 at 298 K
The last equation is Kw written as logs: −log Kw = pKw = 14.00. With it, any one of [H⁺], [OH⁻], pH and pOH gives the other three. For 0.0350 mol dm⁻³ KOH: pOH = −log 0.0350 = 1.46, so pH = 14.00 − 1.46 = 12.54.
How weak is weak? For a weak acid HA, the equilibrium HA(aq) ⇌ H⁺(aq) + A⁻(aq) has an acid dissociation constant:
Ka = [H⁺][A⁻] ÷ [HA] pKa = −log₁₀ Ka
For a weak base B, with B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq), the base dissociation constant is Kb = [BH⁺][OH⁻] ÷ [B], and pKb = −log₁₀ Kb. A larger Ka means more ionization and a stronger acid; because of the minus sign, a smaller pKa means a stronger acid. The same for bases.
| Species (typical values: use the figures in your data booklet) | pKa or pKb at 298 K | Strength |
|---|---|---|
| methanoic acid, HCOOH | pKa 3.75 | strongest of these acids |
| benzoic acid, C₆H₅COOH | pKa 4.20 | |
| ethanoic acid, CH₃COOH | pKa 4.76 | weakest of these acids |
| methylamine, CH₃NH₂ | pKb 3.34 | stronger base |
| ammonia, NH₃ | pKb 4.75 | weaker base |
Each step of 1 in pKa is a factor of 10 in Ka.
Calculating the pH of a weak acid. Two approximations make it a one-line calculation, and the guide does not expect quadratics. Because each HA that ionizes makes one H⁺ and one A⁻, [H⁺] = [A⁻]. Because so little ionizes, [HA] at equilibrium is taken as the concentration you started with. Then Ka = [H⁺]² ÷ [HA], so [H⁺] = √(Ka × [HA]).
Worked example 3. Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, pKa = 4.76.
Check: 1.32 × 10⁻³ is only 1.3% of 0.100, so the approximation was fair. The reverse is just as common: a measured pH gives Ka. If 0.0500 mol dm⁻³ propanoic acid has pH 3.09, then [H⁺] = 10⁻³·⁰⁹ = 8.13 × 10⁻⁴, Ka = (8.13 × 10⁻⁴)² ÷ 0.0500 = 1.32 × 10⁻⁵ and pKa = 4.88.
Worked example 4. Calculate the pH of 0.200 mol dm⁻³ NH₃(aq), pKb = 4.75.
Ka × Kb = Kw. Take ammonia and its conjugate acid, the ammonium ion. Write both constants and multiply:
This is true for every conjugate pair, and it puts a number on section 6: the stronger the acid, the weaker its conjugate base. For NH₄⁺, pKa = 14.00 − 4.75 = 9.25, and Ka = 5.62 × 10⁻¹⁰. For the ethanoate ion, pKb = 14.00 − 4.76 = 9.24.
10HLThe pH of a salt solution
A salt is made of the conjugates of its parents. The conjugate of a strong acid or base (Cl⁻, NO₃⁻, SO₄²⁻, Na⁺, K⁺) is too weak to react with water and leaves the pH alone. The conjugate of a weak parent is a real acid or base in its own right, and it reacts with water: hydrolysis.
- Ammonium ion, acidic. NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq). So NH₄Cl(aq) is acidic.
- Carboxylate ion, basic. RCOO⁻(aq) + H₂O(l) ⇌ RCOOH(aq) + OH⁻(aq). So CH₃COONa(aq) is basic.
- Carbonate ion, basic. CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq). Sodium carbonate solution is distinctly alkaline.
- Hydrogencarbonate ion, weakly basic. HCO₃⁻ is amphiprotic, so two reactions compete: HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻ and HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺. Its Kb is larger than its Ka, so the first wins and NaHCO₃(aq) is slightly alkaline.
| Parent acid | Parent base | Salt solution | Example |
|---|---|---|---|
| strong | strong | neutral, pH 7 | NaCl, KNO₃ |
| strong | weak | acidic | NH₄Cl, NH₄NO₃ |
| weak | strong | basic | CH₃COONa, Na₂CO₃ |
| weak | weak | depends on Ka against Kb | CH₃COONH₄ ≈ 7, since 9.25 ≈ 9.24 |
The numbers come from section 9. For 0.100 mol dm⁻³ sodium ethanoate, Kb(CH₃COO⁻) = Kw ÷ Ka = 5.75 × 10⁻¹⁰, [OH⁻] = √(5.75 × 10⁻¹⁰ × 0.100) = 7.58 × 10⁻⁶ mol dm⁻³, pOH = 5.12 and pH = 8.88. You do not need the acidity of hydrated metal ions such as Al³⁺(aq).
11HLpH curves for weak acids and bases
Figure 6 puts all four combinations side by side, each with 0.100 mol dm⁻³ base added to 25.0 cm³ of 0.100 mol dm⁻³ acid. Compare them feature by feature.
| (a) strong acid + strong base | (b) weak acid + strong base | (c) strong acid + weak base | (d) weak acid + weak base | |
|---|---|---|---|---|
| Starting pH | 1.0 | 2.9 (weak acid, few H⁺) | 1.0 | 2.9 |
| Pattern before equivalence | slow rise | quick rise, then a gentle buffer region | slow rise | buffer region |
| Vertical section | about 3.7 to 10.3 | about 7 to 10 | about 3.7 to 7 | none: a gentle bend |
| pH at equivalence | 7 | about 8.7: the salt's anion is basic | about 5.3: the salt's cation is acidic | about 7 here, because pKa ≈ pKb |
| Final pH | about 12.5 | about 12.5 | levels at about 9.3, a weak base | about 9.3 |
The equivalence point is always at 25.0 cm³, whatever the strengths. Strength moves the pH at which it happens, never the volume, because the amount of base needed is set by the amount of acid, weak or strong.
Figure 7 takes curve (b) and labels what an examiner asks for.
The buffer region and the half-equivalence point. After the first few cm³, the flask contains ethanoic acid and ethanoate ions together, which is a buffer (section 13), so the pH climbs only slowly. Halfway to equivalence, at 12.5 cm³, exactly half of the acid has been converted to its conjugate base, so [HA] = [A⁻]. Put that into Ka = [H⁺][A⁻] ÷ [HA] and the concentrations cancel: Ka = [H⁺], so pH = pKa. That is the standard way to find pKa from a titration: read the pH at half the equivalence volume. For a weak base, the matching point is where the base is half converted into its conjugate acid, and there pOH = pKb. In curve (c), where NH₃ is being added to acid, that happens after equivalence, at 50.0 cm³, when the flask holds equal amounts of NH₃ and NH₄⁺ and the pH is 14.00 − 4.75 = 9.25.
Collecting the data (Tool 1). Add titrant in 1 cm³ portions while the pH changes slowly, then about 0.1 cm³ at a time near equivalence, or you will skip straight over the steep section that fixes the equivalence volume.
12HLIndicators
An acid–base indicator is a weak acid whose undissociated form and conjugate base are different colours. Write it as HInd:
HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq) colour A colour B
Add acid and, by Le Châtelier's principle, the equilibrium shifts left: most of the indicator is HInd and you see colour A. Add alkali and OH⁻ removes H⁺, the equilibrium shifts right, and you see colour B. The switch-over is governed by Ka(HInd):
Ka(HInd) = [H⁺][Ind⁻] ÷ [HInd], so [Ind⁻] ÷ [HInd] = Ka ÷ [H⁺]
When [Ind⁻] = [HInd] the eye sees the halfway colour, and at that point [H⁺] = Ka, so the end point of an indicator is at pH ≈ pKa(HInd). Because the eye needs roughly a tenfold excess of one form to see its colour clearly, the visible change runs over about pKa ± 1. Figure 8 shows this for an indicator with pKa = 7.0.
Typical ranges from the data booklet (check your own copy): methyl orange 3.1–4.4, methyl red 4.4–6.2, bromothymol blue 6.0–7.6, phenolphthalein 8.2–10.0. Universal indicator is a mixture of many indicators, so it changes colour steadily over a wide range. That makes it good for estimating pH and useless for a titration, which needs one sharp change.
End point is not equivalence point. The equivalence point is where the acid and base have reacted in exactly stoichiometric amounts; it is a property of the reaction. The end point is where the indicator changes colour; it is a property of the indicator. A good choice makes them coincide to within a drop.
Choose an indicator whose colour-change range lies within the vertical section of the pH curve, around the pH at the equivalence point.
| Titration | Salt at equivalence | pH at equivalence | Suitable indicator |
|---|---|---|---|
| strong acid + strong base | neutral | 7 | almost any; phenolphthalein or methyl orange |
| weak acid + strong base | basic | above 7 (≈ 8.7) | phenolphthalein |
| strong acid + weak base | acidic | below 7 (≈ 5.3) | methyl red (or methyl orange) |
| weak acid + weak base | ≈ neutral | no vertical section | none gives a sharp end point |
Figure 7 shows why. Phenolphthalein changes between 8.2 and 10.0, inside the steep section, so it turns pink within a drop of 25.0 cm³. Methyl orange would finish changing at pH 4.4, deep in the buffer region, after only about 8 cm³.
13HLBuffer solutions
A buffer solution resists a change in pH when small amounts of acid or alkali are added. It needs two ingredients in similar amounts: something to remove added H⁺ and something to remove added OH⁻.
- An acidic buffer (pH below 7) is a weak acid and its conjugate base, for example ethanoic acid with sodium ethanoate.
- A basic buffer (pH above 7) is a weak base and its conjugate acid, for example ammonia with ammonium chloride.
You can also make one by partly neutralizing a weak acid with a strong base: the buffer region of Figure 7 is exactly that.
How it works. In the ethanoic acid buffer, the large reservoir of ethanoate ions mops up added acid, and the large reservoir of ethanoic acid molecules neutralizes added alkali:
- add H⁺: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq)
- add OH⁻: CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)
For the ammonia buffer: NH₃ + H⁺ → NH₄⁺ and NH₄⁺ + OH⁻ → NH₃ + H₂O. In each case the added ion is replaced by a weak acid or weak base, which barely ionizes, so [H⁺] hardly changes. A strong acid and its salt cannot do this: Cl⁻ is such a weak base that it will not take up added H⁺, and there are no HCl molecules in solution to react with added OH⁻.
The pH of a buffer depends on two things, both visible when you rearrange Ka:
The pKa sets roughly where the buffer works (within about one unit of it) and the ratio fine-tunes it. Both species share one volume, so concentrations or amounts in moles give the same ratio.
Worked example 5. 100 cm³ of buffer contains 0.100 mol dm⁻³ CH₃COOH and 0.150 mol dm⁻³ CH₃COONa (pKa = 4.76). Calculate its pH, then the pH after adding 0.0010 mol of HCl.
The pH falls by less than 0.1. The same 0.0010 mol of HCl in 100 cm³ of pure water would give [H⁺] = 0.0100 mol dm⁻³ and pH 2.00, a fall of 5 units. Figure 9 shows the two changes side by side, and the matching addition of alkali.
Worked example 6: a basic buffer. Find the pH of a solution 0.200 mol dm⁻³ in NH₃ and 0.100 mol dm⁻³ in NH₄Cl (pKb = 4.75). The same algebra with Kb gives pOH = pKb + log([BH⁺] ÷ [B]) = 4.75 + log(0.100 ÷ 0.200) = 4.45, so pH = 14.00 − 4.45 = 9.55.
Designing a buffer. To make an ethanoate buffer of pH 5.00, you need log([A⁻] ÷ [HA]) = 5.00 − 4.76 = 0.24, so the ratio [A⁻] : [HA] must be 10⁰·²⁴ = 1.74 : 1.
Dilution. Adding water to a buffer lowers [HA] and [A⁻] by the same factor, so their ratio, and therefore the pH, stays almost exactly the same. What dilution does change is the buffer's capacity: there are fewer moles of each reservoir per cm³, so less added acid or alkali is needed to use one of them up.
14Where marks are lost
Getting a conjugate pair's charge wrong. Removing H⁺ makes a species one unit more negative. The conjugate base of HSO₄⁻ is SO₄²⁻, not SO₄⁻.
Calling a base an alkali, or the other way round. An alkali is a soluble base that gives OH⁻(aq). CuO is a base and not an alkali.
Treating strong as concentrated. Strength is the extent of ionization; concentration is moles per dm³. "A strong acid has a high concentration of acid" scores nothing.
Saying a weak acid needs less alkali to neutralize it. At equal concentration and volume it needs the same amount, because it keeps ionizing as its H⁺ is removed.
Thinking pH 7 always means neutral. Neutral means [H⁺] = [OH⁻]. At temperatures above 298 K, Kw is larger and neutral water has a pH below 7.
Reading a pH difference as a simple ratio. pH 2 is not twice as acidic as pH 4; it has 100 times the [H⁺].
(HL) Using the weak-acid shortcut for a strong acid or a buffer. [H⁺] = √(Ka × [HA]) applies only to a weak acid on its own. A buffer needs the ratio [A⁻] ÷ [HA]; a strong acid needs none of it.
(HL) Confusing end point with equivalence point, or saying the equivalence point is always at pH 7. It is at 7 only for strong acid with strong base. The pH at equivalence is the pH of the salt solution.
15Draw it right
The pH curve is the diagram of this subtopic. For any combination:
- Label the axes: pH (no unit) on the vertical, volume of titrant added / cm³ on the horizontal, with pH running 0 to 14.
- Start at the right place: pH 1 for 0.1 mol dm⁻³ strong acid, about 3 for a weak acid (about 13 or 11 if a strong or weak base is in the flask).
- Put the equivalence point at the volume given by the stoichiometry, and mark it. Its pH is 7, above 7 or below 7 according to the salt.
- Draw the vertical section at that volume, long for strong + strong, shorter on the weak side for one weak partner, absent for weak + weak.
- Finish flattening towards the pH of the titrant, never beyond it.
- (HL) Mark the buffer region and the half-equivalence point, and label it pH = pKa (or pOH = pKb for a weak base being converted to its conjugate acid).
- (HL) If asked about an indicator, show its range as a band and check that it lies inside the vertical section.
- pH against [H⁺] is a curve falling steeply at low [H⁺], never a straight line.
16Try it
Marks in brackets. Answers and marker's notes are at the end. Use Kw = 1.00 × 10⁻¹⁴ at 298 K.
Q1. Consider the reaction H₂PO₄⁻(aq) + CO₃²⁻(aq) ⇌ HPO₄²⁻(aq) + HCO₃⁻(aq). 3 marks
(a) Identify the Brønsted–Lowry acid and base in the forward reaction. 1 mark
(b) State the two conjugate acid–base pairs. 2 marks
Q2. A solution of nitric acid has a concentration of 0.00400 mol dm⁻³. 5 marks
(a) Calculate its pH. 1 mark
(b) A different solution has pH 9.50 at 298 K. Calculate [H⁺] and [OH⁻] in it, and state whether it is acidic, neutral or basic. 3 marks
(c) The nitric acid is diluted until its pH is 3.40. Determine by what factor it was diluted. 1 mark
Q3. Formulate a balanced equation for each reaction. 4 marks
(a) dilute nitric acid with calcium carbonate 1 mark
(b) ethanoic acid with sodium hydrogencarbonate solution 1 mark
(c) ammonia solution with sulfuric acid, and identify the parent acid and parent base of the salt 2 marks
Q4. (Data-based, invented data.) A student tests two acids, P and Q, both at 0.10 mol dm⁻³ and 25 °C. 5 marks
| pH (probe) | Conductivity / arbitrary units | Time for 2 cm of Mg ribbon to disappear / s | |
|---|---|---|---|
| P | 1.0 | 38 | 45 |
| Q | 2.9 | 2 | 610 |
(a) Deduce which acid is strong, using two pieces of evidence. 2 marks
(b) The student predicts that 25.0 cm³ of Q will need less 0.10 mol dm⁻³ NaOH to neutralize it than 25.0 cm³ of P. Evaluate this prediction. 2 marks
(c) Suggest why the conductivity of Q is not zero. 1 mark
Q5 (HL). A 0.0500 mol dm⁻³ solution of a weak monoprotic acid, HX, has a pH of 3.09 at 298 K. 5 marks
(a) Calculate Ka and pKa for HX. 3 marks
(b) Calculate pKb for X⁻. 1 mark
(c) State one assumption made in (a). 1 mark
Q6 (HL). 20.0 cm³ of 0.100 mol dm⁻³ propanoic acid is titrated with 0.100 mol dm⁻³ NaOH(aq). The pH after 10.0 cm³ of alkali is 4.87, and the pH at the equivalence point is 8.8. 8 marks
(a) State pKa for propanoic acid, explaining your answer. 2 marks
(b) Explain, with an equation, why the pH at the equivalence point is greater than 7. 2 marks
(c) Identify a suitable indicator from methyl orange (3.1–4.4) and phenolphthalein (8.2–10.0), giving a reason. 1 mark
(d) Explain, using an equation, how the mixture after 10.0 cm³ of alkali resists a change in pH when a small amount of HCl(aq) is added. 2 marks
(e) Calculate Kb for the propanoate ion. 1 mark
17In one breath
An acid donates a proton and a base accepts one; an alkali is a base that dissolves to give OH⁻. Every proton transfer has two conjugate pairs, each differing by one H⁺, and amphiprotic species such as H₂O and HCO₃⁻ can act either way. pH = −log[H⁺], so one unit is a factor of ten; Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K, and neutral means [H⁺] = [OH⁻], which is pH 7 only at 298 K. Strong acids (HCl, HBr, HI, HNO₃, H₂SO₄) and group 1 hydroxides ionize fully, weak ones partly, and equilibrium lies towards the weaker conjugate; strong is not concentrated, and a weak acid needs just as much alkali. Oxides and hydroxides give salt and water, carbonates add CO₂, ammonia and amines give the salt alone. A strong–strong pH curve starts at the acid's pH and leaps vertically at pH 7 at equivalence. HL: pH + pOH = 14; smaller pKa means stronger acid, and Ka × Kb = Kw; a weak acid's [H⁺] = √(Ka[HA]); NH₄⁺ makes salts acidic, RCOO⁻, CO₃²⁻ and HCO₃⁻ basic; a weak partner shortens the jump and shifts the equivalence pH, and pH = pKa at half-equivalence; an indicator HInd changes colour near its pKa and must change inside the vertical section; a buffer is a weak acid or base with its conjugate, pH = pKa + log([A⁻] ÷ [HA]), and dilution keeps the pH but lowers the capacity.
Answers
Q1. (a) Acid: H₂PO₄⁻ (it loses a proton); base: CO₃²⁻ (it gains one). (b) H₂PO₄⁻ / HPO₄²⁻, and HCO₃⁻ / CO₃²⁻. 1 for both acid and base correctly identified; 1 for each pair. A pair written the wrong way round (base first) is accepted; a pair of an acid with the other side's species, such as H₂PO₄⁻ / HCO₃⁻, scores 0.
Q2. (a) Strong acid, so [H⁺] = 0.00400 mol dm⁻³ and pH = −log 0.00400 = 2.40. (b) [H⁺] = 10⁻⁹·⁵⁰ = 3.16 × 10⁻¹⁰ mol dm⁻³; [OH⁻] = Kw ÷ [H⁺] = 1.00 × 10⁻¹⁴ ÷ 3.16 × 10⁻¹⁰ = 3.16 × 10⁻⁵ mol dm⁻³; basic, because [OH⁻] > [H⁺]. (c) The pH rose by 1.00, so [H⁺] fell by a factor of 10: it was diluted ten times. A1 for 2.40; A1 for [H⁺; A1 for [OH⁻] via Kw; A1 for basic with its reason ("pH above 7" is accepted at 298 K); A1 for a factor of 10. "Diluted by 1" scores 0.]
Q3. (a) 2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g). (b) CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O(l) + CO₂(g). (c) 2NH₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq); parent acid sulfuric acid, parent base ammonia. 1 for each balanced equation; 1 for both parents. State symbols are not required unless asked. An equation for (c) that produces water scores 0.
Q4. (a) P is strong. At the same concentration it has the lower pH (1.0 against 2.9), so a higher [H⁺]; its conductivity is far higher, so it contains many more ions; and it reacts with magnesium much faster, so collisions with H⁺ are more frequent. Any two. A pH of 1.0 for a 0.10 mol dm⁻³ acid also shows complete ionization. (b) The prediction is wrong. Both solutions contain the same amount of acid, 2.5 × 10⁻³ mol; as OH⁻ removes H⁺ from the weak acid, its equilibrium shifts right and more HA ionizes, until all of it has reacted. Both need the same volume, 25.0 cm³. (c) Q is weak, not un-ionized: a small fraction of its molecules ionize, so a few ions are present to carry a current. 1 for P with one piece of evidence, 1 for a second; 1 for "same volume", 1 for the equilibrium-shift reason; 1 for partial ionization. "Q is weaker so it needs less" scores 0 in (b).
Q5 (HL). (a) [H⁺] = 10⁻³·⁰⁹ = 8.13 × 10⁻⁴ mol dm⁻³ = [X⁻]. Ka = (8.13 × 10⁻⁴)² ÷ 0.0500 = 1.32 × 10⁻⁵. pKa = 4.88. (b) pKb = 14.00 − 4.88 = 9.12. (c) The equilibrium concentration of HX is taken as 0.0500 mol dm⁻³, because the ionization is so small (or: all the H⁺ comes from HX, none from water). M1 for [H⁺ from the pH, M1 for Ka = [H⁺]² ÷ [HX], A1 for 1.32 × 10⁻⁵ and 4.88; A1 for 9.12 (ECF); 1 for a valid assumption.]
Q6 (HL). (a) pKa = 4.87. 10.0 cm³ is half the equivalence volume, so half of the acid has been converted to propanoate: [HA] = [A⁻], so Ka = [H⁺] and pH = pKa. (b) At equivalence the flask contains sodium propanoate; the propanoate ion is the conjugate base of a weak acid and hydrolyses: CH₃CH₂COO⁻(aq) + H₂O(l) ⇌ CH₃CH₂COOH(aq) + OH⁻(aq), so [OH⁻] > [H⁺]. (c) Phenolphthalein, because its range lies within the vertical section around pH 8.8; methyl orange would change in the buffer region, long before equivalence. (d) The mixture contains similar amounts of propanoic acid and propanoate ions. Added H⁺ is removed by the conjugate base: CH₃CH₂COO⁻ + H⁺ → CH₃CH₂COOH, so [H⁺] hardly changes and the ratio [A⁻] ÷ [HA] changes only slightly. (e) pKb = 14.00 − 4.87 = 9.13, so Kb = 10⁻⁹·¹³ = 7.4 × 10⁻¹⁰. 1 for 4.87, 1 for the half-equivalence reason; 1 for the basic anion, 1 for the hydrolysis equation; 1 for phenolphthalein with reason; 1 for the equation, 1 for the reservoir explanation; 1 for Kb. An answer to (d) that says "the buffer neutralizes the acid" with no equation or species scores at most 1.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section Reactivity 3.1 Proton transfer reactions. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.