Eight teaching hours at both levels (outcomes 3.1.1 to 3.1.8) and nine further hours at higher level (3.1.9 to 3.1.17).
Guiding question. What happens when protons are transferred?
Reactivity 3.1 · Proton transfer reactions
1Brønsted–Lowry acids and bases 3.1.1 SL + HL
The syllabus statement: Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry base is a proton acceptor. The skill: deduce the Brønsted–Lowry acid and base in a reaction, knowing that a proton in water can be written H+(aq) or H3O+(aq), and understanding the distinction between a base and an alkali.
Lemon juice tastes sharp, soap feels slippery, and stomach acid can dissolve a steel paper clip. For centuries chemists classified acids and bases by such properties, and then by what they contained: Lavoisier thought all acids contained oxygen, and Arrhenius defined acids as substances that release H+ ions in water and bases as substances that release OH− ions. Each definition explained more than the last, but each failed somewhere. Hydrogen chloride is an acid, yet it contains no oxygen; ammonia neutralizes acids, yet it contains no hydroxide. In 1923 Johannes Brønsted and Thomas Lowry independently proposed the definition used throughout this chapter, which does not describe a substance at all. It describes an event.
The proton-transfer definition
A Brønsted–Lowry acid is a proton (H+) donor. A Brønsted–Lowry base is a proton acceptor. An alkali is a base that dissolves in water to release hydroxide ions, OH−(aq).
A hydrogen atom that has lost its only electron is a bare proton, about one hundred thousand times smaller than an atom. Its charge density is so great that it never exists free in water: it attaches at once to a lone pair on a water molecule to form the oxonium (hydronium) ion, H3O+. The guide accepts either symbol. H+(aq) is shorter; H3O+(aq) is more honest and is needed whenever water itself is written as a reactant.
In this reaction the H–Cl bond breaks so that both bonding electrons stay with chlorine; the proton moves to a lone pair on the oxygen of water. HCl is the acid (it donates the proton) and water is the base (it accepts it). Because a base must accept a proton, every Brønsted–Lowry base has a lone pair of electrons; this is the link to the Lewis theory met in Reactivity 3.4.
The same species can play either role. With ammonia, water gives up a proton:
Now ammonia is the base and water is the acid. A substance is therefore not “an acid” in isolation; it acts as an acid in a particular reaction, relative to a particular partner. To identify the roles, follow the proton: whichever species loses H+ is the acid, whichever gains it is the base.
Base or alkali?
All alkalis are bases, but not all bases are alkalis. Copper(II) oxide accepts protons from acids (the oxide ion, O2−, becomes water), so it is a base, but it is insoluble in water and produces no OH−(aq); it is not an alkali. Sodium hydroxide, potassium hydroxide and aqueous ammonia are alkalis. Ammonia is an interesting case: it contains no hydroxide, yet its solution is alkaline because ammonia molecules take protons from water, leaving OH− ions behind.
1. Find the species that has lost an H. Compare formulas across the arrow; the one that has one fewer H and one more negative charge (or one less positive charge) on the product side was the acid.
2. Find where that H+ went. The species that has gained H and one unit of positive charge was the base.
3. Check for a lone pair. The base must have one; if it does not, the reaction is not a Brønsted–Lowry acid–base reaction.
4. Say “proton” or “H+”, never “hydrogen”. A hydrogen atom carries an electron; a proton does not. Transfer of a hydrogen atom would be a redox or radical process, not an acid–base one.
A definition in science earns its place by the observations it can explain. The Arrhenius definition could not explain the basic behaviour of ammonia in water or the reaction of ammonia gas with hydrogen chloride gas, where no water or hydroxide is involved at all. The Brønsted–Lowry definition explains both, and the Lewis definition (Reactivity 3.4) is broader again. Each new model included the old one as a special case rather than overturning it.
2Conjugate acid–base pairs 3.1.2 SL + HL
The syllabus statement: A pair of species differing by a single proton is called a conjugate acid–base pair. The skill: deduce the formula of the conjugate acid or base of any Brønsted–Lowry base or acid.
When an acid HA donates a proton, what remains, A−, can in principle take a proton back: it is a base. HA and A− are called a conjugate acid–base pair. Every Brønsted–Lowry reaction contains two such pairs, because two species exchange one proton.
acid 1 base 2 base 1 acid 2
The rule for writing a conjugate is purely mechanical and applies to any species, however unfamiliar:
- Conjugate base = the acid minus one H+: remove one H and lower the charge by one. HSO4− → SO42−; H2O → OH−; NH4+ → NH3; CH3COOH → CH3COO−.
- Conjugate acid = the base plus one H+: add one H and raise the charge by one. HS− → H2S; OH− → H2O; CO32− → HCO3−; CH3NH2 → CH3NH3+.
Two checks prevent most errors. The formulas must differ by exactly one H (not two, and not an O), and the charges must differ by exactly one. H3O+ and OH− are not a conjugate pair: they differ by two protons.
| Acid | Conjugate base | Acid | Conjugate base |
|---|---|---|---|
| H2SO4 | HSO4− | H2CO3 | HCO3− |
| HSO4− | SO42− | HCO3− | CO32− |
| H3PO4 | H2PO4− | HNO3 | NO3− |
| H2PO4− | HPO42− | NH4+ | NH3 |
| HPO42− | PO43− | H3O+ | H2O |
| CH3COOH | CH3COO− | H2O | OH− |
Question. In the reaction H2PO4−(aq) + OH−(aq) ⇌ HPO42−(aq) + H2O(l), identify the two Brønsted–Lowry acids and the two conjugate pairs.
| Forward | H2PO4− loses H+ (becomes HPO42−): acid. OH− gains it (becomes H2O): base. |
| Reverse | H2O can donate a proton back to HPO42−: H2O is the acid of the reverse reaction. |
| Answer | Acids: H2PO4− and H2O. Pairs: H2PO4−/HPO42− and H2O/OH−. |
The reports record that less than 40% of the candidates were able to deduce the formula of the conjugate base of HCO3-, and that among those who knew the answer was carbonate, a significant proportion wrote the carbonate ion with an incorrect charge. In another session the vast majority of candidates gave water instead of O2- as the conjugate base of OH-. Removing one proton lowers the charge by exactly one: HCO3− → CO32−, OH− → O2−.
3Amphiprotic species 3.1.3 SL + HL
The syllabus statement: Some species can act as both Brønsted–Lowry acids and bases. The skill: interpret and formulate equations to show acid–base reactions of these species.
A species that can both donate and accept a proton is amphiprotic. To qualify it needs two features at once: a hydrogen that can leave as H+, and a lone pair that can accept one. Water is the most important example; the others are the intermediate ions of polyprotic acids, such as HCO3−, HSO4−, H2PO4− and HPO42−, together with amino acids.
HCO3−(aq) + OH−(aq) → CO32−(aq) + H2O(l) (acid)
To show amphiprotic behaviour, write two equations: one with a stronger acid (H3O+ or H+) in which the species accepts a proton, and one with a stronger base (OH−) in which it donates one.
Species that cannot do both are not amphiprotic. PO43− has no hydrogen to give; NH4+ and H3O+ have no lone pair left to accept another proton; CH4 has neither a lone pair nor an acidic hydrogen and is not a Brønsted–Lowry base at all.
Amphoteric describes a substance that reacts with both acids and bases, such as aluminium oxide or zinc hydroxide. Amphiprotic is narrower: the substance both donates and accepts protons. Every amphiprotic species is amphoteric, but not every amphoteric substance is amphiprotic: Al2O3 contains no hydrogen, so it cannot donate a proton, yet it reacts with both HCl(aq) and NaOH(aq). This links to the acid–base character of the period 3 oxides in Structure 3.1.
Exam focus · what the published papers show
One report advised that candidates should be aware of the difference between amphiprotic and amphoteric; another found that 42% of candidates could distinguish amphiprotic and amphoteric on a multiple-choice item, and a third that a few still confuse amphoteric (ability to dissolve in both acids and bases) with amphiprotic (ability to both donate and accept hydrogen ions). When asked to show amphiprotic behaviour, the mark scheme accepts an equation with any suitable acid (H3O+, H+) and any suitable base (OH−, NH3, H2O).
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Phosphoric acid, H3PO4, can undergo stepwise neutralization, forming amphiprotic species. (b) Formulate two equations to show the amphiprotic nature of H2PO4−. [2] (d) Outline the reason that sodium hydroxide is considered a Brønsted–Lowry base. [1]
Ammonium nitrate, NH4NO3, is used as a high nitrogen fertilizer. State, with a reason, whether the ammonium ion is a Brønsted–Lowry acid or base. [1]
The equation for the first dissociation of citric acid in water is C6H8O7(aq) + H2O(l) ⇌ C6H7O7−(aq) + H3O+(aq). Identify a conjugate acid–base pair in the equation. [1]
Hydrogensulfate ion, HSO4−, is amphiprotic. (a) Write equations for its reaction with OH−(aq) and with H3O+(aq). (b) State the conjugate base of HSO4− and the conjugate acid of HSO4−. [3]
Solutions and mark-scheme guidance · Set R3A
R3A.1 C
H2O can accept a proton (H3O+) and donate one (OH−). NH4+ and H3O+ cannot accept a further proton; PO43− has no proton to donate.
R3A.2 A
Add one H+ to HS−: one more H, charge raised by one, giving H2S. S2− is the conjugate base.
R3A.3 D
H2PO4− loses H+ (becomes HPO42−): acid; H2O gains it (becomes H3O+): base. Option C wrongly labels H3O+, a product, as the base. The report records that 56% of candidates could correctly identify Bronsted-Lowry acids and bases.
R3A.4 D
Forward: H2PO4− donates to OH−. Reverse: H2O donates to HPO42−. The report notes that 59% of the candidates identified the Br∅nsted-Lowry acids in the forward and reverse reactions.
R3A.5 A
Forward: HNO3 accepts a proton from H2SO4 (becoming H2NO3+), so it is the base. Reverse: HSO4− accepts a proton. Nitric acid acting as a base shows that roles depend on the partner; this is exactly what happens in the nitrating mixture of Reactivity 3.4.
R3A.6 C
HPO42− donates and accepts protons (amphiprotic) and so also reacts with both acids and bases (amphoteric).
R3A.7 [3]
(b) H2PO4−(aq) + H+(aq) → H3PO4(aq) ✔ and H2PO4−(aq) + OH−(aq) → HPO42−(aq) + H2O(l) ✔. The scheme accepts reaction with any acidic, basic or amphiprotic partner, such as H3O+, NH3 or H2O, and H2PO4− → HPO42− + H+ for the second mark. (d) The OH− ion is a proton acceptor ✔.
R3A.8 [1]
Brønsted–Lowry acid, because NH4+ can donate a proton (or: cannot accept one) ✔. It has no lone pair left to accept H+.
R3A.9 [1]
C6H8O7 and C6H7O7−, or H2O and H3O+ ✔. A pair is two species on opposite sides that differ by one proton.
R3A.10 [3]
(a) HSO4− + OH− → SO42− + H2O ✔; HSO4− + H3O+ → H2SO4 + H2O ✔. (b) Conjugate base SO42−; conjugate acid H2SO4 ✔. (In practice H2SO4 is so strong that the second reaction happens only in very acidic conditions, but the formulas are what the question tests.)
4The pH scale 3.1.4 SL + HL
The syllabus statement: The pH scale can be used to describe the [H+] of a solution: pH = –log10[H+]; [H+] = 10–pH. The skill: perform calculations involving the logarithmic relationship between pH and [H+], including estimating pH with universal indicator and measuring it precisely with a pH meter or probe.
The hydrogen ion concentration of everyday solutions spans an enormous range: about 1 mol dm−3 in laboratory hydrochloric acid, 10−7 mol dm−3 in pure water, 10−13 mol dm−3 in household oven cleaner. Numbers that span thirteen powers of ten are awkward to compare, so in 1909 Søren Sørensen, working on the brewing of beer, proposed compressing them with a logarithm.
Here [H+] is the concentration of hydrogen ions in mol dm−3, and pH has no units. The minus sign makes pH positive for most solutions, and it has a consequence students must hold on to: the higher the [H+], the lower the pH.
What a logarithmic scale means
Because pH is a logarithm, a change of one pH unit is a tenfold change in [H+]; a change of two units is a hundredfold change. A solution of pH 3 is not “a little more acidic” than one of pH 4: it has ten times the hydrogen ion concentration. Diluting a strong acid tenfold raises its pH by exactly one. The sketch graph asked for in the guide follows directly: pH against [H+] is a curve that falls steeply at low concentration and flattens at high concentration, whereas pH against log10[H+] is a straight line of gradient −1.
Question. (a) Calculate the pH of a solution in which [H+] = 2.5 × 10−4 mol dm−3. (b) Calculate [H+] in a solution of pH 3.61. (c) By what factor does [H+] change when the pH of a lake falls from 6.5 to 4.5?
| (a) | pH = −log10(2.5 × 10−4) = 3.60 |
| (b) | [H+] = 10−3.61 = 2.5 × 10−4 mol dm−3 |
| (c) | A fall of 2.0 units: [H+] rises by 102.0, a factor of 100. |
| Check | pH 3.60 lies between 3 and 4, and 2.5 × 10−4 lies between 10−4 and 10−3: consistent. The number of decimal places in a pH equals the number of significant figures in [H+]. |
Measuring pH
Universal indicator is a mixture of several indicators whose colours change at different pH values, so it passes through a spectrum from red (about pH 1) through green (about pH 7) to violet (about pH 13). It gives an estimate to about one pH unit, and the reading depends on judging a colour. A pH meter or digital pH probe measures a voltage that depends on [H+] and displays pH to 0.01 units. It must be calibrated with buffer solutions of known pH before use; an uncalibrated meter gives readings that are consistently displaced, a systematic error that repetition cannot remove.
5The ionic product of water, Kw 3.1.5 SL + HL
The syllabus statement: The ion product constant of water, Kw, shows an inverse relationship between [H+] and [OH–]. Kw = [H+] [OH–]. The skill: recognize solutions as acidic, neutral and basic from the relative values of [H+] and [OH–].
Pure water conducts electricity very slightly, which shows that it contains ions. A tiny fraction of water molecules transfer protons to one another:
Because the concentration of water itself is effectively constant, the equilibrium is described by the ionic product of water:
This relationship holds in every aqueous solution at 298 K, not only in pure water. [H+] and [OH−] are therefore inversely related: add acid and [OH−] must fall so that their product is unchanged. The classification of a solution follows from comparing them.
| Solution | Relationship | At 298 K |
|---|---|---|
| Acidic | [H+] > [OH−] | pH < 7 |
| Neutral | [H+] = [OH−] | pH = 7 |
| Basic (alkaline) | [H+] < [OH−] | pH > 7 |
Kw depends on temperature
The ionization of water breaks O–H bonds and is endothermic. By Le Châtelier’s principle (Reactivity 2.3), raising the temperature shifts the equilibrium to the right, so Kw increases: at about 373 K it is roughly 5 × 10−13. In hot pure water [H+] is larger, so the pH is below 7, yet [H+] still equals [OH−] and the water is still neutral. “pH 7 means neutral” is true only at 298 K; the definition of neutrality is equality of the two concentrations.
Question. (a) Calculate [OH−] in a solution of pH 4.00 at 298 K. (b) Calculate the pH of 1.0 × 10−3 mol dm−3 NaOH(aq) at 298 K.
| (a) | [H+] = 10−4.00; [OH−] = Kw/[H+] = 1.00 × 10−14 ÷ 1.00 × 10−4 = 1.0 × 10−10 mol dm−3 |
| (b) | NaOH is fully dissociated: [OH−] = 1.0 × 10−3. [H+] = 1.0 × 10−14 ÷ 1.0 × 10−3 = 1.0 × 10−11; pH = 11.00 |
| Check | An acidic solution has a tiny [OH−]; an alkaline one has pH > 7. The most common wrong answer to (b) is 3, which is pOH, not pH. |
Exam focus · what the published papers show
On a multiple-choice item 77% of the candidates were able to calculate the pH of the aqueous solution of NaOH, but the most commonly chosen distractor was D (pH = 13) which was probably selected because NaOH is a strong base. On another, the most commonly chosen distractor was B (pH = 3) where the students determined pOH but did not complete the calculation. Strong does not mean pH 13 or 14; the pH of an alkali depends on its concentration, through Kw.
6Strong and weak acids and bases 3.1.6 SL + HL
The syllabus statement: Strong and weak acids and bases differ in the extent of ionization. The skill: recognize that acid–base equilibria lie in the direction of the weaker conjugate, with HCl, HBr, HI, HNO3 and H2SO4 as the strong acids and the group 1 hydroxides as the strong bases, and the distinction between strong/weak and concentrated/dilute.
Two bottles on a shelf are both labelled 0.10 mol dm−3. One contains hydrochloric acid, the other ethanoic acid (the acid in vinegar). A magnesium ribbon fizzes vigorously in the first and slowly in the second; a conductivity meter reads far higher in the first; universal indicator turns red in the first and orange in the second. The acids have the same concentration, so the difference must lie in what each does in water.
Extent of ionization
A strong acid is one that is (essentially) completely ionized in aqueous solution. A weak acid is only partially ionized: an equilibrium is set up in which most molecules remain un-ionized.
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO−(aq) (partial: equilibrium arrow)
In 0.10 mol dm−3 ethanoic acid only about 1.3 % of the molecules are ionized, so [H+] is about 1.3 × 10−3 mol dm−3 instead of 0.10 mol dm−3. That explains every observation: fewer H+ ions mean a higher pH, slower reaction with magnesium or carbonates, and lower electrical conductivity. The choice of arrow in an equation is therefore a chemical claim about extent.
Strong bases are completely dissociated into ions: LiOH, NaOH, KOH and the other group 1 hydroxides. Weak bases, such as ammonia and amines, accept protons from water only partially:
Strength is not concentration
Strong and weak describe the extent of ionization, a property of the acid. Concentrated and dilute describe the amount of acid per unit volume, a property of the solution. The two are independent: 10 mol dm−3 ethanoic acid is concentrated but weak; 0.001 mol dm−3 hydrochloric acid is dilute but strong. The distinction must be made in words, and the guide requires it explicitly.
Equilibria lie towards the weaker conjugate
A strong acid gives up its proton easily, which means its conjugate base holds a proton only weakly: the stronger the acid, the weaker its conjugate base. Cl−, the conjugate base of HCl, shows essentially no tendency to take a proton back, so the ionization of HCl goes to completion. CH3COO− is a stronger base than Cl− and readily re-captures protons, which is why the ethanoic acid equilibrium lies to the left. In any acid–base equilibrium, the position lies on the side of the weaker acid and the weaker base.
| Test | Strong acid (e.g. 0.10 mol dm−3 HCl) | Weak acid (e.g. 0.10 mol dm−3 CH3COOH) |
|---|---|---|
| pH meter or universal indicator | pH 1.0 (red) | pH ≈ 2.9 (orange) |
| Electrical conductivity | high | low |
| Magnesium ribbon or calcium carbonate | rapid effervescence | slow effervescence |
| Temperature rise on neutralization with NaOH | greater | slightly smaller (energy is used to ionize the acid) |
| Volume of NaOH to neutralize | the same: both provide the same amount of H+ in the end | |
A weak acid reacts only slowly with a base at the start, but as OH− removes H+ the equilibrium shifts right and more acid ionizes, until all of it has reacted. Equal amounts of a strong and a weak monoprotic acid therefore need equal volumes of alkali for neutralization. Tests that compare strength must measure rate, pH or conductivity, not the volume of alkali required.
Exam focus · what the published papers show
The reports note that there is a difference, which candidates should note, between “not fully dissociated” and “partially dissociated” when describing a weak acid, that some candidates referred to “dissolve” rather than dissociate, and that in suggesting tests “see if it conducts” and “add pH paper” were common answers without predictions of the expected results. A method earns credit only with its expected result for each acid. Another report was surprised that a surprisingly large number of candidates thought that sulfuric acid is a weak acid.
7Neutralization reactions 3.1.7 SL + HL
The syllabus statement: Acids react with bases in neutralization reactions. The skill: formulate equations for the reactions between acids and metal oxides, metal hydroxides, hydrogencarbonates and carbonates, and identify the parent acid and base of different salts, with bases including ammonia, amines, carbonates and hydrogencarbonates and acids including organic acids.
In a neutralization an acid transfers protons to a base, forming a salt and, usually, water. The ionic equation for any strong acid with any strong base is the same, because the other ions are spectators:
The reaction is exothermic because a new O–H bond forms (Reactivity 1.1), which is why the enthalpies of neutralization of different strong acids with strong bases are almost identical. The general patterns are summarized below; the new context in an examination question only changes the formulas.
| Acid + … | Products | Example |
|---|---|---|
| metal hydroxide | salt + water | H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) |
| metal oxide | salt + water | CuO(s) + 2HCl(aq) → CuCl2(aq) + H2O(l) |
| carbonate | salt + water + carbon dioxide | CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + H2O(l) + CO2(g) |
| hydrogencarbonate | salt + water + carbon dioxide | NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + H2O(l) + CO2(g) |
| ammonia | ammonium salt only | NH3(aq) + HCl(aq) → NH4Cl(aq) |
| amine | alkylammonium salt | CH3NH2(aq) + HCl(aq) → CH3NH3Cl(aq) |
Ammonia and amines are the exceptions that produce no water: the base simply accepts the proton, and the product is a salt of the NH4+ or RNH3+ ion. With a carbonate, carbonic acid is formed first but decomposes at once, so it must never be written as a product.
Parent acid and base of a salt
Every salt can be traced back to the acid that supplied its anion and the base that supplied its cation. To identify them, add H+ to the anion and add OH− to the metal ion (or remove H+ from NH4+). Potassium ethanoate comes from ethanoic acid and potassium hydroxide; ammonium sulfate from sulfuric acid and ammonia; sodium hydrogencarbonate from carbonic acid and sodium hydroxide. At higher level (3.1.12) the strengths of the parent acid and base predict the pH of the salt’s solution.
Question. 25.00 cm3 of vinegar required 20.75 cm3 of 1.00 mol dm−3 KOH(aq) for neutralization. Calculate the concentration of ethanoic acid in the vinegar.
| Equation | CH3COOH(aq) + KOH(aq) → CH3COOK(aq) + H2O(l): ratio 1 : 1 |
| Amount of KOH | 0.02075 dm3 × 1.00 mol dm−3 = 0.02075 mol |
| Concentration | 0.02075 mol ÷ 0.02500 dm3 = 0.830 mol dm−3 |
| Check | A smaller volume of the more concentrated alkali was needed, so the acid is less concentrated than 1.00 mol dm−3. The published scheme for this question gives 0.830 mol dm−3. |
Exam focus · what the published papers show
One report found it surprising that only about a fifth of the candidates realized that sulfuric acid is diprotic, and another that only the better candidates could write an equation for the neutralisation of phosphoric(V) acid. Giving H2CO3 as a product of a carbonate and an acid was quite common. Write the correct formula of the salt first, then balance; a diprotic acid needs twice the amount of NaOH.
8pH curves for strong acids and strong bases 3.1.8 SL + HL
The syllabus statement: pH curves for neutralization reactions involving strong acids and bases have characteristic shapes and features. The skill: sketch and interpret the general shape of the pH curve, including the intercept with the pH axis and the equivalence point. Only monoprotic neutralizations are assessed.
A pH curve records the pH of the solution in the flask as titrant is added from a burette, using a pH probe connected to a data logger. For 25.0 cm3 of 0.100 mol dm−3 HCl titrated with 0.100 mol dm−3 NaOH, the curve has four features worth being able to explain.
- Intercept with the pH axis. Before any alkali is added, the flask holds 0.100 mol dm−3 of a strong acid, so [H+] = 0.100 and pH = 1.0.
- Gradual rise. Each addition removes some H+, but because pH is logarithmic, removing 90 % of the acid raises the pH by only one unit.
- Equivalence point. When the amount of OH− added equals the amount of H+ originally present (here 25.0 cm3), the flask contains only NaCl(aq) and water: pH 7 at 298 K. The very steep section around it spans roughly pH 3 to 11.
- Levelling off. Beyond equivalence the solution is increasingly dominated by excess NaOH and the curve approaches, but never reaches, the pH of the titrant (about 12.5 here, allowing for dilution).
The equivalence point is also called the stoichiometric point because it is where the reactants have been mixed in the ratio of the equation. Its volume is found from the centre of the vertical section, and it gives the concentration of the unknown through the usual titration calculation (Structure 1.4). When collecting data, smaller volumes of titrant should be added near the expected equivalence point, where the pH changes most rapidly.
Exam focus · what the published papers show
Reports describe answers that were quite poorly done with many candidates not indicating a vertical drop but rather a weak acid/weak base curve, graphs whose x-axis was incorrectly labelled as “volume of weak acid” instead of “volume of strong base”, and answers where some did not identify 25cm3 as the equivalence point and/or had a wrong pH at that volume. A mark-worthy sketch shows the correct starting pH, a near-vertical section centred on the calculated equivalence volume, pH 7 at equivalence for strong–strong, and the curve levelling towards the pH of the titrant.
Proton transfer — titration curves, computed
Not a sketch. Every point solves the charge balance for [H+], so the buffer region, the half-equivalence point and the pH at equivalence come out of the chemistry rather than being drawn in.
pH against volume of base added. The shaded band is the working range of the indicator you chose — a usable indicator is one whose band lies inside the vertical part of the curve.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Sulfur trioxide reacts with water to form a strong Brønsted–Lowry acid. State the meaning of a strong Brønsted–Lowry acid. [2]
Outline two laboratory methods of distinguishing between solutions of citric acid and hydrochloric acid of equal concentration, stating the expected observations. [2]
In a titration, 25.00 cm3 of vinegar required 20.75 cm3 of 1.00 mol dm−3 potassium hydroxide to reach the end-point. Calculate the concentration of ethanoic acid in the vinegar. [2]
Sketch the pH curve for the addition of 0.050 mol dm−3 NaOH(aq) to 20.0 cm3 of 0.050 mol dm−3 HNO3(aq), up to 40.0 cm3 of NaOH. Label the initial pH, the equivalence point and the pH it approaches. [3]
Solutions and mark-scheme guidance · Set R3B
R3B.1 B
KOH is a strong base: [OH−] = 1.0 × 10−5, [H+] = 10−14/10−5 = 10−9, pH 9. Option C would need [OH−] = 10−5, not 10−9; an acid can never give pH 9.
R3B.2 D
[H+] = 10−4; [OH−] = 10−14/10−4 = 10−10 mol dm−3.
R3B.3 A
[OH−] = 10−4 at pH 10. Diluting 10.0 cm3 to 1000.0 cm3 is a hundredfold dilution: [OH−] = 10−6, pOH 6, pH 8. Each tenfold dilution of an alkali lowers the pH by one.
R3B.4 D
[H3O+] = 10−13 gives pH 13. A and C are acidic; B has [OH−] = 10−13, so [H+] = 0.1 and it is acidic.
R3B.5 B
A strong acid donates its proton readily, so its conjugate base accepts protons poorly: good donor, weak conjugate base. The report notes this was another better answered question.
R3B.6 C
Same concentration: strong acid (HCl) lowest pH, then weak acid, then weak base (NH3), then strong base (NaOH).
R3B.7 C
Both acids supply the same total amount of H+ once the weak acid’s equilibrium has been driven over by the base, so the same volume of NaOH is needed. Initial pH, conductivity and heat evolved differ.
R3B.8 B
A weak acid is a proton donor that is only partly ionized, so it has relatively few ions and its solution conducts poorly.
R3B.9 [2]
Fully ionizes/dissociates ✔; proton/H+ donor ✔. Both points are needed: the Brønsted–Lowry part (proton donor) and the “strong” part (complete ionization).
R3B.10 [2]
Any two of: electrical conductivity, greater for HCl ✔; pH, higher for citric acid ✔; add a reactive metal, carbonate or hydrogencarbonate, faster or more vigorous effervescence with HCl ✔; greater temperature rise with HCl ✔. The scheme does not accept smell or taste, and a method without its expected result scores nothing.
R3B.11 [2]
n(KOH) = 0.02075 dm3 × 1.00 = 0.0208 mol ✔; 1 : 1 ratio, so [CH3COOH] = 0.0208 ÷ 0.02500 = 0.830 mol dm−3 ✔.
R3B.12 [3]
Start at pH = −log(0.050) = 1.3 ✔; near-vertical rise centred on 20.0 cm3 with equivalence at pH 7 ✔; curve levelling towards the pH of the titrant, 12.7 (at 40.0 cm3 the excess OH− is 0.050 × 20.0/60.0 = 0.0167 mol dm−3, pH 12.2) ✔.
Additional higher level · outcomes 3.1.9 to 3.1.17
9The pOH scale 3.1.9 HL only
The syllabus statement: The pOH scale describes the [OH–] of a solution. pOH = –log10[OH–]; [OH–] = 10–pOH. The skill: interconvert [H+], [OH–], pH and pOH values.
The logarithm that simplifies [H+] works just as well for [OH−]. Defining pOH = −log10[OH−] and taking −log10 of both sides of Kw = [H+][OH−] gives a very convenient result:
Four quantities, [H+], [OH−], pH and pOH, are therefore linked by two definitions and one equation, and knowing any one gives the other three. For a base, it is usually fastest to go [OH−] → pOH → pH.
| Solution | [H+] / mol dm−3 | pH | [OH−] / mol dm−3 | pOH |
|---|---|---|---|---|
| 0.010 mol dm−3 HCl | 1.0 × 10−2 | 2.00 | 1.0 × 10−12 | 12.00 |
| pure water | 1.0 × 10−7 | 7.00 | 1.0 × 10−7 | 7.00 |
| 0.010 mol dm−3 KOH | 1.0 × 10−12 | 12.00 | 1.0 × 10−2 | 2.00 |
Question. An ammonia solution has [OH−] = 1 × 10−4 mol dm−3. Calculate its pH at 298 K.
| pOH | −log10(1 × 10−4) = 4.0 |
| pH | 14.0 − 4.0 = 10.0 |
| Check | An alkaline solution: pH above 7. The distractor 4.0 is the pOH, left unconverted. |
10Ka, Kb, pKa and pKb 3.1.10 HL only
The syllabus statement: The strengths of weak acids and bases are described by their Ka, Kb, pKa or pKb values. The skill: interpret the relative strengths of acids and bases from these data. Quadratic equations are not expected in calculations.
The partial ionization of a weak acid is an equilibrium (Reactivity 2.3), so its extent can be expressed by an equilibrium constant, the acid dissociation constant, Ka. Water is the solvent, present in large excess, and is omitted from the expression.
B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq) Kb = [BH+][OH−] / [B]
A larger Ka means a greater extent of ionization and a stronger acid. As with pH, the numbers span many powers of ten, so they are often expressed as pKa = −log10Ka and pKb = −log10Kb. Because of the minus sign, the relationship reverses: the smaller the pKa, the stronger the acid. Like all equilibrium constants, Ka and Kb depend only on temperature; they do not change with concentration.
| Acid | Ka | pKa | Strength |
|---|---|---|---|
| chloroethanoic acid, ClCH2COOH | 1.3 × 10−3 | 2.87 | strongest of these |
| methanoic acid, HCOOH | 1.8 × 10−4 | 3.75 | ↓ weaker |
| ethanoic acid, CH3COOH | 1.7 × 10−5 | 4.76 | |
| hydrocyanic acid, HCN | 6.2 × 10−10 | 9.21 |
The pH of a weak acid
For a weak acid of initial concentration c, the ionization produces equal amounts of H+ and A−, so [H+] = [A−] = x. If the acid is weak enough, x is so small compared with c that [HA] at equilibrium ≈ c. Then:
This approximation is what lets the guide promise that no quadratic is needed. It assumes (1) the ionization of water contributes negligibly to [H+] and (2) the fraction of acid ionized is small (a few per cent at most). Both are reasonable for acids with Ka below about 10−3 at ordinary concentrations. The same reasoning for a weak base gives [OH−] ≈ √(Kb × c).
Question. (a) Calculate the pH of 0.10 mol dm−3 of a weak acid with Ka = 1.0 × 10−5. (b) Calculate the pH of 1.00 × 10−2 mol dm−3 ammonia, pKb = 4.75, at 298 K.
| (a) [H+] | √(1.0 × 10−5 × 0.10) = √(1.0 × 10−6) = 1.0 × 10−3 mol dm−3 |
| (a) pH | 3.0. Check the approximation: 10−3/0.10 = 1 % ionized, small. |
| (b) Kb | 10−4.75 = 1.78 × 10−5 |
| (b) [OH−] | √(1.78 × 10−5 × 1.00 × 10−2) = 4.22 × 10−4 mol dm−3 |
| (b) pH | pOH = 3.37; pH = 14.00 − 3.37 = 10.6 |
| Check | A weak acid: pH above that of a strong acid of the same concentration (1.0). A weak base: pH below that of 0.010 mol dm−3 NaOH (12.0). |
Exam focus · what the published papers show
On a recent multiple-choice item only 43% of candidates could correctly determine the pH of a solution from its concentration and Ka value; many appear to have taken the Ka value as the [H+]. In structured questions typical challenges were in obtaining the correct value of [H3O+] from the given pH, forgetting the power of 2 in the numerator of the Ka expression – and arithmetic errors, and the pH of ammonia proved to be challenging with many confusing Ka and Kb. Other candidates did not recognize that since it is a weak base, [NH3] at equilibrium is approximately equal to starting concentration.
11The relationship Ka × Kb = Kw 3.1.11 HL only
The syllabus statement: For a conjugate acid–base pair, the relationship Ka × Kb = Kw can be derived from the expressions for Ka and Kb. The skill: solve problems involving these values.
Consider ethanoic acid and its conjugate base, the ethanoate ion. Write each expression and multiply:
The acid and conjugate-base concentrations cancel, leaving the ionic product of water. Taking −log10 of both sides gives the equivalent form pKa + pKb = pKw = 14.00 at 298 K. The chemistry behind the arithmetic is the inverse relationship already met in 3.1.6: a larger Ka forces a smaller Kb for the conjugate base. The strongest acids have the weakest conjugate bases.
Question. The pKa of ethanoic acid is 4.76 at 298 K. Calculate Kb for the ethanoate ion.
| Ka | 10−4.76 = 1.74 × 10−5 |
| Kb | Kw/Ka = 1.00 × 10−14 ÷ 1.74 × 10−5 = 5.8 × 10−10 |
| Alternative | pKb = 14.00 − 4.76 = 9.24; Kb = 10−9.24 = 5.8 × 10−10 |
Exam focus · what the published papers show
A report describes the item on this relationship as a good discriminatory question in which 57% of candidates were able to identify relationship between acid and base dissociation constants in a conjugate acid– base pair. The common wrong answers multiplied the p-values (pKa × pKb); logarithms add when the quantities multiply.
pH, pOH, Ka and Kb — the whole set of relationships
Five quantities, all fixed by any one of them plus the concentration. The model also solves the weak-acid equilibrium exactly, so you can see where the usual approximation starts to fail.
Where this solution sits on the pH scale, with the ordinary landmarks for comparison. The scale is logarithmic: one unit is a factor of ten in [H+].
12The pH of salt solutions 3.1.12 HL only
The syllabus statement: The pH of a salt solution depends on the relative strengths of the parent acid and base. The skill: construct equations for the hydrolysis of ions in a salt, and predict the effect of each ion on the pH of the salt solution, with examples including NH4+, RCOO−, CO32− and HCO3−. The acidity of hydrated transition element ions and Al3+(aq) is not required.
Solutions of salts are not all neutral. Ammonium chloride solution is acidic, sodium ethanoate solution is alkaline, and sodium chloride solution is neutral. The explanation lies in the conjugate relationships of 3.1.11: an ion that is the conjugate of a weak acid or base is itself a significant base or acid, and reacts with water. This reaction is called hydrolysis.
- Anion of a weak acid (CH3COO−, HCOO−, CO32−, HCO3−) accepts a proton from water, releasing OH−: the solution becomes alkaline.CH3COO−(aq) + H2O(l) ⇌ CH3COOH(aq) + OH−(aq)
CO32−(aq) + H2O(l) ⇌ HCO3−(aq) + OH−(aq) - Cation of a weak base (NH4+, RNH3+) donates a proton to water: the solution becomes acidic.NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
- Ions of strong acids and strong bases (Cl−, Br−, NO3−, SO42−, Na+, K+) are such weak conjugates that they do not hydrolyse: no effect on pH.
| Parent acid | Parent base | Example | Solution |
|---|---|---|---|
| strong | strong | NaCl, KNO3, KBr | neutral, pH 7 |
| strong | weak | NH4Cl, NH4Br | acidic, pH < 7 |
| weak | strong | CH3COONa, HCOOK, Na2CO3, NaHCO3 | alkaline, pH > 7 |
| weak | weak | CH3COONH4, HCOONH4 | depends on Ka of the cation versus Kb of the anion |
The carbonate ion is a stronger base than the hydrogencarbonate ion (HCO3− is a weaker acid than H2CO3, so its conjugate base CO32− is stronger), which is why sodium carbonate solution is more alkaline than sodium hydrogencarbonate solution of the same concentration. Hydrogencarbonate is amphiprotic, but its basic behaviour dominates in water, so NaHCO3(aq) is mildly alkaline.
Exam focus · what the published papers show
A recent report found that about half the students recognised that a salt solution would be basic, though most justified this by stating the relative strength of the acid and base involved, rather than being able to gain the second mark by giving an equation for the hydrolysis. Another observed that the pH of solutions of aqueous chlorides was not generally well known with only a small number of candidates gaining full marks. The complete answer names the ion that hydrolyses, writes its equation with water, and identifies whether H3O+ or OH− is released.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
(c) State the Ka expression for ethanoic acid. [1] (d) Calculate the Kb of the conjugate base of ethanoic acid, given pKa(CH3COOH) = 4.76 and Kw = 1.0 × 10−14. [1]
(i) State why NH3 is a Lewis base. [1] (ii) Calculate the pH of a 1.00 × 10−2 mol dm−3 aqueous solution of ammonia. pKb = 4.75 at 298 K. [3]
(e) Determine the concentration of methanoic acid in a solution of pH = 4.12, given pKa(HCOOH) = 3.75. [2] (f) Identify if aqueous solutions of the following salts are acidic, basic, or neutral: sodium methanoate; ammonium chloride; sodium nitrate. [2]
Solutions and mark-scheme guidance · Set R3C
R3C.1 C
pOH = −log(1 × 10−4) = 4.0; pH = 14.0 − 4.0 = 10.0.
R3C.2 A
Ka × Kb = Kw (equivalently pKa + pKb = pKw). Products of constants become sums of p-values, never products.
R3C.3 C
The weakest acid (smallest Ka) has the strongest conjugate base: HCN, Ka = 6.2 × 10−10. The report found strength of conjugate bases appeared to be another good discriminator.
R3C.4 A
Convert everything to one scale. pKa: HClO 7.4; HIO3 0.8; HF 3.25 (Ka 5.6 × 10−4); CH3CH2COOH 4.89 (Ka 1.3 × 10−5). Increasing acidity means decreasing pKa: HClO < CH3CH2COOH < HF < HIO3.
R3C.5 B
[H+] = √(1 × 10−5 × 0.1) = 1 × 10−3; pH = 3. The most common wrong answer, 5, takes Ka as [H+].
R3C.6 C
NH4Cl acidic (NH4+ hydrolyses) < NaCl neutral < CH3COONa alkaline < Na2CO3 more alkaline (CO32− is a stronger base than CH3COO−).
R3C.7 C
NH4Br acidic (strong acid, weak base) < HCOONH4 close to neutral (weak–weak) < KBr neutral < HCOOK alkaline. The order puts the weak–weak salt just below KBr; with Ka(NH4+) ≈ 5.6 × 10−10 larger than Kb(HCOO−) ≈ 5.6 × 10−11, ammonium methanoate is very slightly acidic.
R3C.8 B
Conductivity depends on the concentration of ions, which increases with base strength (smaller pKb): ethanol (pKb 15.5, essentially no ions) < phenylamine < methylamine.
R3C.9 [2]
(c) Ka = [CH3COO−][H3O+]/[CH3COOH] ✔ (H+ accepted for H3O+). (d) Ka = 10−4.76 = 1.7 × 10−5; Kb = 1.0 × 10−14 ÷ 1.7 × 10−5 = 5.8 × 10−10 ✔ (answers 5.7–5.9 × 10−10 accepted).
R3C.10 [4]
(i) Donates a lone (non-bonding) pair of electrons ✔. (ii) Kb = 10−4.75 = 1.78 × 10−5 ✔; [OH−] = √(1.00 × 10−2 × 1.78 × 10−5) = 4.22 × 10−4 ✔; pOH = 3.37, pH = 10.6 ✔. Award [3] for the correct final answer.
R3C.11 [4]
(e) [H+] = [HCOO−] = 10−4.12; Ka = 10−3.75 = (10−4.12)2/[HCOOH] ✔; [HCOOH] = 3.24 × 10−5 mol dm−3 ✔. (f) Sodium methanoate basic; ammonium chloride acidic; sodium nitrate neutral ✔✔ ([2] for three correct, [1] for two).
13pH curves for all four combinations 3.1.13 HL only
The syllabus statement: pH curves of different combinations of strong and weak monoprotic acids and bases have characteristic shapes and features. The skill: interpret the general shapes of pH curves for all four combinations, including the intercept with the pH axis, the equivalence point, the buffer region, and the points where pH = pKa or pOH = pKb.
The curve for a strong acid and strong base (3.1.8) is the reference. Replacing either partner with a weak one changes the curve in ways that each follow from a piece of chemistry already developed.
- Starting pH. A weak acid is only partly ionized, so the intercept is higher (about 2.9 for 0.100 mol dm−3 ethanoic acid, against 1.0 for HCl).
- pH at equivalence. At equivalence the flask contains the salt, and its pH is that of 3.1.12: 7 for strong–strong; above 7 for weak acid–strong base (the anion hydrolyses, about 8.7 here); below 7 for strong acid–weak base (the cation hydrolyses, about 5.3 here); close to 7 for weak–weak, depending on the relative K values.
- Size of the vertical section. The steep section is long for strong–strong, shorter when one partner is weak, and almost absent for weak–weak, which is why a weak–weak titration has no sharp end point.
- Buffer region. With a weak acid, the early part of the curve rises only slowly because the flask contains both the weak acid and its conjugate base: a buffer (3.1.16).
The half-equivalence point
Halfway to the equivalence point, exactly half the weak acid has been converted to its conjugate base, so [HA] = [A−]. Substituting in the Ka expression:
Reading the pH at half the equivalence volume is therefore an experimental method for determining pKa. The same argument for a weak base titrated with a strong acid shows that at half-equivalence [B] = [BH+], so pOH = pKb and pH = 14 − pKb.
Exam focus · what the published papers show
Reports note that candidates struggled with a WB/ WA curve, showing buffering regions or large pH drops at equivalence, and that in sketches many did not identify the buffer region or that the pH at equivalence point is greater than 7. On a multiple-choice item over 75 % could correctly identify the nature of the acid and base that would produce the given titration pH curve. A full-credit sketch for a weak acid with a strong base shows: an S-shape, a buffer region below pH 7, equivalence above pH 7, and pKa read at half the equivalence volume.
14Acid–base indicators 3.1.14 HL only
The syllabus statement: Acid–base indicators are weak acids, where the components of the conjugate acid–base pair have different colours. The pH of the end point of an indicator, where it changes colour, approximately corresponds to its pKa value. The skill: construct equilibria expressions to show why the colour of an indicator changes with pH, using the generalized formula HInd.
An acid–base indicator is a weak acid whose un-ionized form and conjugate base absorb different wavelengths of visible light, so they have different colours.
colour A colour B
Only a few drops are used, so the indicator does not affect the pH of the solution; instead, the solution’s [H+] controls the position of the indicator’s equilibrium. In acid, the high [H+] pushes the equilibrium to the left (Le Châtelier), and colour A of HInd dominates. In alkali, OH− removes H+, the equilibrium shifts right, and colour B of Ind− dominates.
Rearranging Ka(HInd) shows exactly when the change happens:
When [H+] = Ka(HInd), that is when pH = pKa(HInd), the two forms are present in equal amounts and the solution shows the intermediate colour. This is the end point of the indicator. The eye sees a colour change over roughly one pH unit either side of the pKa, when one form outnumbers the other by about ten to one, which is why indicators are quoted with a pH range.
Universal indicator is a mixture of several indicators with overlapping ranges, so its colour changes continuously across the whole pH scale. That makes it useful for estimating pH but useless for titrations, where a single sharp change is needed.
Exam focus · what the published papers show
A report found that many candidates could explain the behaviour of indicators, but there were also some poor answers that did not acknowledge the importance of equilibrium in the action of an indicator. On a multiple-choice item relating an indicator’s colour to pH and pKa, 55% of candidates obtained the correct answer, and it was described as one of the most challenging questions on the paper. Another report records answers that stated a single colour without stating the complete colour change.
15Choosing an indicator for a titration 3.1.15 HL only
The syllabus statement: An appropriate indicator for a titration has an end point range that coincides with the pH at the equivalence point. The skill: identify an appropriate indicator for a titration from the identity of the salt and the pH range of the indicator, and distinguish between the terms “end point” and “equivalence point”.
The equivalence point is where the amounts of acid and base added are in the stoichiometric ratio of the equation; it is a property of the reaction. The end point is where the indicator changes colour; it is a property of the indicator. A titration is accurate when the chosen indicator’s end point falls within the vertical section of the curve, so that the end point and equivalence point occur at the same volume, to within one drop.
The choice is made in two steps. First, predict the pH at equivalence from the salt formed (3.1.12). Second, choose the indicator whose colour-change range lies within the steep part of the curve around that pH.
| Titration | Salt at equivalence | pH at equivalence | Suitable indicator |
|---|---|---|---|
| strong acid + strong base | neutral (e.g. NaCl) | 7 | any whose range lies within ≈ 3–11: methyl orange, bromothymol blue or phenolphthalein |
| weak acid + strong base | alkaline (e.g. CH3COONa) | > 7 | phenolphthalein (8.3–10.0) |
| strong acid + weak base | acidic (e.g. NH4Cl) | < 7 | methyl orange (3.1–4.4) or methyl red |
| weak acid + weak base | near 7 | ≈ 7 | none suitable: no sharp vertical section |
Exam focus · what the published papers show
In one session indicator choice was generally well done although a number of candidates chose bromothymol blue as a suitable indicator for weak base with a strong acid. Another noted that few candidates identified a suitable acid-base indicator, where the whole of the range fitted within the rapid pH change of both curves. The whole range, not only the pKa, must lie inside the steep section.
16Buffer solutions 3.1.16 HL only
The syllabus statement: A buffer solution is one that resists change in pH on the addition of small amounts of acid or alkali. The skill: describe the composition of acidic and basic buffers and explain their actions.
Blood is held at pH 7.35–7.45; a change of a few tenths of a unit is life-threatening, yet metabolism constantly releases acids into it. Adding 1 cm3 of 1 mol dm−3 HCl to a litre of pure water changes the pH from 7 to 3. The same acid added to a litre of blood changes its pH by only a few hundredths. Blood is a buffer.
Composition
A buffer contains substantial and comparable amounts of a weak acid and its conjugate base, or of a weak base and its conjugate acid.
- Acidic buffer (pH < 7): a weak acid and a salt of its conjugate base, e.g. CH3COOH and CH3COONa.
- Basic buffer (pH > 7): a weak base and a salt of its conjugate acid, e.g. NH3 and NH4Cl.
A buffer can also be made by partial neutralization: excess weak acid with some strong base (for example 2 mol CH3COOH with 1 mol NaOH gives 1 mol CH3COOH and 1 mol CH3COO−), or excess weak base with some strong acid. The weak component must be in excess; complete neutralization leaves only the salt, which is not a buffer.
How a buffer works
When a small amount of strong acid is added, the H+ ions are removed by the conjugate base, which is present in large amount:
When a small amount of strong base is added, the OH− ions are removed by the weak acid:
In each case a strong acid or base is replaced by a weak one, and because both components are present in large amounts, their ratio changes only slightly. For a basic buffer the same logic applies: added H+ reacts with NH3 to form NH4+; added OH− reacts with NH4+ to form NH3 and water.
A buffer made from a strong acid and its salt (HCl with NaCl) cannot work: Cl− is too weak a base to remove added H+. The system must be a weak conjugate pair, so that both components are present in significant amounts and each can react. Buffers have a limited capacity: once one component is used up, the pH changes sharply, as it does at the end of the buffer region on a titration curve.
Exam focus · what the published papers show
For an ammonia buffer, the published marking notes credit: equilibrium shifts to right/H+ reacts with NH3 ✔ and «as large excess» ratio [NH3]:[NH4+] «and hence pH» almost unchanged ✔, and also accept strong acid/H+ converted to a weak acid/NH4+. Two ideas are needed: an equation or statement of what the added ion reacts with, and why the pH hardly changes.
Across sessions the definition of a buffer is very well known, but the explanation of the action of buffers proved to be more challenging with only the stronger candidates giving a complete response in terms of protonation of the conjugate base and increased dissociation of the acid. Answers were poorly answered because equations were not used to explain buffer action or the dissociation equations for the base and acid were given rather than their reactions with H+ or OH-. On combinations, over 70% of candidates could identify the pair of solutions that would form a buffer, though some were unaware that the weak component must be in excess.
17The pH of a buffer solution 3.1.17 HL only
The syllabus statement: the pH of a buffer depends on both the pKa or pKb of its acid or base and the ratio of the concentration of acid or base to the concentration of the conjugate base or acid. The skill: solve problems involving the composition and pH of a buffer solution, using the equilibrium constant, including the effect of dilution.
In a buffer the weak acid is barely ionized, and its ionization is further suppressed by the large amount of conjugate base present. So, to a good approximation, [HA] at equilibrium equals the amount of acid put in and [A−] equals the amount of salt put in. Rearranging Ka gives:
The two forms are the same equation; the second is often called the Henderson–Hasselbalch equation, and either may be used. They show the two factors named in the guide: the pKa fixes the pH region in which the buffer works, and the ratio fine-tunes it. With equal concentrations of acid and conjugate base, log 1 = 0 and pH = pKa. A buffer is most effective within about one pH unit of its pKa, so a buffer for a target pH is designed by choosing an acid with pKa close to that pH.
Dilution
Diluting a buffer with water reduces [HA] and [A−] by the same factor, so their ratio, and therefore the pH, is unchanged. What dilution does reduce is the buffer capacity: with fewer moles of each component in a given volume, a smaller addition of acid or alkali will exhaust one of them.
Question. 20.0 cm3 of 0.100 mol dm−3 NaOH is added to 50.0 cm3 of 0.100 mol dm−3 ethanoic acid (pKa = 4.76). Calculate the pH of the buffer formed.
| Amounts | n(CH3COOH) = 5.00 × 10−3 mol; n(NaOH) = 2.00 × 10−3 mol |
| After reaction | CH3COOH left: 3.00 × 10−3 mol; CH3COO− formed: 2.00 × 10−3 mol (same volume, so the ratio of amounts is the ratio of concentrations) |
| pH | 4.76 + log10(2.00/3.00) = 4.76 − 0.18 = 4.58 |
| Check | More acid than conjugate base, so pH is below pKa: consistent. |
Question. A buffer contains 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 dm3 (pKa 4.76). Calculate the pH after adding 0.010 mol HCl, and compare with adding the same HCl to 1.00 dm3 of water.
| Before | ratio 1 : 1, pH = 4.76 |
| After | H+ + CH3COO− → CH3COOH: acid 0.110 mol, base 0.090 mol; pH = 4.76 + log(0.090/0.110) = 4.67 |
| Water | [H+] = 0.010 mol dm−3: pH falls from 7.00 to 2.00 |
Buffer or water? Add a little acid or alkali and compare
Set the weak acid, its conjugate base and the amount of strong acid or base added. The model finds the new ratio from the neutralization, then the pH from pKa + log([A−]/[HA]), and compares the change with the same addition to pure water.
Top marker: the buffer before and after the addition. Bottom marker: the same addition to pure water (pH 7.00). Negative additions are strong base (OH−).
Exam focus · what the published papers show
Reports record that some candidates incorrectly inverted [conjugate base] and [acid] in the Henderson-Hasselbalch equation, that the calculation of the ratio of a conjugate acid-base pair to create a buffer with a specific pH was poorly done, and that on one item less than a third of the candidates were able to predict the pH of the buffer correctly even though many realized that the volume of NaOH must be less than the volume of acid. Check the direction of the answer: excess acid means pH below pKa.
Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination; items marked “Practice” were written for these notes.
Ethanoic acid is titrated with potassium hydroxide. On the pH curve, point B lies on the flat part of the curve before the equivalence point and point C at the equivalence point. (i) Identify the major species, other than water and potassium ions, at points B and C. [2] (ii) State a suitable indicator for this titration. [1] (iii) Suggest, giving a reason, which point on the curve is considered a buffer region. [1]
(i) Sketch the titration curve of methanoic acid with sodium hydroxide, showing how you would determine the pKa of methanoic acid. [2] (ii) Identify an indicator that could be used for this titration. [1]
An aqueous solution containing high concentrations of both NH3 and NH4+ acts as a buffer as a result of the equilibrium NH3(aq) + H+(aq) ⇌ NH4+(aq). Referring to this equilibrium, outline why adding a small volume of strong acid would leave the pH of the buffer almost unchanged. [2]
Justify whether a 1.0 dm3 solution made from 0.10 mol NH3 and 0.20 mol HCl will form a buffer solution. [1]
A buffer is needed at pH 5.00. (a) Explain why ethanoic acid (pKa 4.76) is a better choice than methanoic acid (pKa 3.75). (b) Calculate the ratio [CH3COO−]/[CH3COOH] required. (c) State the effect on the pH of diluting this buffer tenfold. [3]
Solutions and mark-scheme guidance · Set R3D
R3D.1 C
The curve is for a weak base titrated with a strong acid: the steep section is centred on pH 5, below 7, as expected for an acidic salt. D (1.5) is the final pH of the excess acid, not the equivalence point.
R3D.2 B
A weak base with a strong acid: the start is below the pH of a strong base, the early section falls gradually (buffer region), and the vertical section and equivalence lie below pH 7.
R3D.3 C
Equivalence at 50 cm3; at half-equivalence, 25 cm3, pH ≈ 9.2, so pOH = pKb = 14.0 − 9.2 = 4.8. The distractor 9.2 is the pH itself.
R3D.4 B
The buffer region is the flat section before equivalence, where both the weak acid and its conjugate base are present in significant amounts; B is at half-equivalence.
R3D.5 A
pH 7 is two units above the pKa of 5.1, so [In−] ≫ [HIn] and colour B is seen. At pH 3 the acid form dominates (colour A). The report found this one of the most challenging questions on the paper, with 55 % correct.
R3D.6 D
The curve starts at pH 12 (strong base in the flask), and the vertical section lies roughly between pH 11.5 and 6, so the equivalence point is above 7: a weak acid added to a strong base. Phenolphthalein, pKa 9.6, changes colour inside the steep section; the other three change below it.
R3D.7 D
Mixing equal volumes: 0.02 mol NH3 with 0.01 mol HCl gives 0.01 mol NH3 and 0.01 mol NH4+, a basic buffer. C neutralizes completely; A and B contain no weak conjugate pair. The report notes over 70 % correct, with the commonest error unaware that the weak component must be in excess.
R3D.8 D
Half-neutralization forms a buffer: 10.0 cm3 of NaOH converts half the acid, so [CH3COOH] = [CH3COO−] and pH = pKa ≈ 4.8. 40.0 cm3 would be twice the equivalence volume.
R3D.9 [4]
(i) B: CH3COOH and CH3COO− ✔; C: CH3COO− ✔. (ii) Phenolphthalein ✔ (phenol red or bromothymol blue also accepted). (iii) B, with a reason: little change in pH on small additions of base, or half the equivalence volume, or similar amounts of weak acid and conjugate base ✔.
R3D.10 [3]
(i) Increasing S-shaped curve with a buffer region below pH 7 and equivalence above pH 7 ✔; pKa = pH at half-equivalence (half the volume needed for neutralization) ✔. (ii) Phenolphthalein or phenol red ✔.
R3D.11 [2]
Added H+ reacts with NH3, shifting the equilibrium to the right ✔; because both species are present in large excess, the ratio [NH3] : [NH4+], and hence the pH, is almost unchanged ✔.
R3D.12 [1]
No: the HCl is in excess, so all the NH3 is neutralized; the solution contains NH4Cl and excess HCl, with no weak base left ✔.
R3D.13 [3]
(a) The best buffer has pKa within about one unit of the target pH; 4.76 is much closer to 5.00 than 3.75 is ✔. (b) log ratio = 5.00 − 4.76 = 0.24, ratio = 100.24 = 1.7 ✔. (c) No change: both concentrations fall by the same factor, so the ratio is unchanged (the capacity decreases) ✔.
Review · Reactivity 3.1
18Misconceptions, the examiner’s view, and the question types
- “A proton is a hydrogen atom.” Why it is wrong: H+ has no electron; a hydrogen atom has one. Correct model: acid–base reactions transfer H+ only. Consequence: “hydrogen” in place of “proton” loses the mark.
- “Strong means concentrated.” Strength is extent of ionization; concentration is amount per volume. A dilute strong acid and a concentrated weak acid both exist.
- “A weak acid needs less alkali to neutralize it.” Equal amounts need equal volumes; the equilibrium is pulled over as H+ is removed.
- “pH 7 always means neutral.” Neutral means [H+] = [OH−]; pH 7 is neutral only at 298 K. Hot pure water has pH below 7 and is neutral.
- “All salt solutions are neutral.” Ions from weak acids or weak bases hydrolyse, making the solution alkaline or acidic.
- “A smaller pKa means a weaker acid.” The opposite: smaller pKa, larger Ka, stronger acid.
- “A buffer keeps the pH constant whatever is added.” It resists change on adding small amounts; its capacity is finite.
- “Diluting a buffer changes its pH.” The ratio [A−]/[HA] is unchanged, so the pH is too.
- “Amphiprotic and amphoteric mean the same.” Amphiprotic requires proton donation and acceptance; Al2O3 is amphoteric but not amphiprotic.
Evidence base: the IB Diploma chemistry subject reports quoted in this chapter.
Answered well: identifying conjugate pairs in a given equation; recognizing the combination that describes a strong Brønsted–Lowry acid; calculating the pH of a strong base from its concentration (77 % on one item); interconverting Ka and pKa (nearly 80 % on one item); the definition of a buffer; identifying the type of titration from a curve (over 75 %) and a buffer mixture from a list (over 70 %).
Found difficult: (1) writing conjugate bases with the correct charge (HCO3− → CO32−; OH− → O2−); (2) distinguishing amphiprotic from amphoteric (42 % on one item); (3) the pH of alkalis, where pOH was given as pH or strength was taken to mean pH 13; (4) weak-acid calculations, where Ka was taken as [H+] (43 % on one item) and Ka and Kb were confused; (5) explaining buffer action with equations; (6) the Henderson–Hasselbalch ratio inverted; (7) sketching pH curves with the correct equivalence volume and pH; (8) choosing an indicator for a weak base.
What successful answers did: followed the proton to assign every role; wrote single arrows for strong and equilibrium arrows for weak; converted between pH, pOH, [H+] and [OH−] explicitly; checked every calculated pH against whether the solution should be acidic or alkaline; explained buffers with an equation for the reaction of each added ion; and gave each distinguishing test with its expected result.
Eight question types cover the sub-topic.
| If the question asks… | …then |
|---|---|
| Identify the acid, base or conjugate pair | Follow the proton; pairs differ by one H+ and one unit of charge, on opposite sides. |
| Show amphiprotic behaviour | Two equations: with H3O+ (accepts) and with OH− (donates). |
| pH, pOH, [H+], [OH−] | pH = −log[H+]; Kw = 10−14; pH + pOH = 14 at 298 K. |
| Strong or weak? Distinguish them | Extent of ionization; tests: pH, conductivity, rate with Mg or CaCO3, each with its result. |
| HL · pH of a weak acid or base | [H+] = √(Kac) or [OH−] = √(Kbc); KaKb = Kw. |
| HL · pH of a salt | Name the parents; write the hydrolysis equation; state which ion is released. |
| HL · pH curve and indicator | Start pH, equivalence pH from the salt, buffer region, pH = pKa at half-equivalence; indicator range inside the steep part. |
| HL · Buffer composition, action, pH | Weak conjugate pair in comparable amounts; equations for added H+ and OH−; pH = pKa + log([A−]/[HA]); dilution: no change. |
19Quick check
- Identify the acid, base and both conjugate pairs in HNO2(aq) + H2O(l) ⇌ NO2−(aq) + H3O+(aq).
- Write the conjugate base of HPO42− and the conjugate acid of CH3NH2.
- Calculate the pH of 0.0050 mol dm−3 HNO3(aq).
- Calculate [OH−] in a solution of pH 9.40 at 298 K.
- Explain why 0.1 mol dm−3 ethanoic acid has a higher pH than 0.1 mol dm−3 hydrochloric acid.
- Write the equation for the reaction of potassium hydrogencarbonate with hydrochloric acid.
- HL: Calculate the pH of 0.20 mol dm−3 methanoic acid, pKa 3.75.
- HL: Predict, with an equation, whether sodium carbonate solution is acidic, neutral or alkaline.
- HL: Choose an indicator for titrating aqueous ammonia with hydrochloric acid, and explain.
- HL: Calculate the pH of a solution containing 0.050 mol dm−3 NH3 and 0.10 mol dm−3 NH4Cl (pKb(NH3) = 4.75).
- Acid HNO2, base H2O; pairs HNO2/NO2− and H3O+/H2O.
- PO43−; CH3NH3+.
- pH = −log(0.0050) = 2.30.
- pOH = 14.00 − 9.40 = 4.60; [OH−] = 10−4.60 = 2.5 × 10−5 mol dm−3.
- Ethanoic acid is weak: only partly ionized, so [H+] is much smaller than 0.1 mol dm−3 and the pH is higher.
- KHCO3(s) + HCl(aq) → KCl(aq) + H2O(l) + CO2(g).
- Ka = 1.78 × 10−4; [H+] = √(1.78 × 10−4 × 0.20) = 5.96 × 10−3; pH = 2.22.
- Alkaline: CO32−(aq) + H2O(l) ⇌ HCO3−(aq) + OH−(aq); carbonate is the anion of a weak acid.
- Methyl orange (or methyl red): the salt NH4Cl is acidic, so equivalence is below pH 7 and the steep section lies in the acidic range.
- pOH = 4.75 + log(0.10/0.050) = 5.05; pH = 8.95.
20Summary and knowledge organiser
Essential knowledge
- Brønsted–Lowry acid = proton donor, base = proton acceptor (needs a lone pair); alkali = soluble base giving OH−(aq); H+(aq) ≡ H3O+(aq).
- Conjugate pairs differ by one H+; amphiprotic species donate and accept protons (H2O, HCO3−, H2PO4−).
- pH = −log[H+]; one pH unit = tenfold [H+]; Kw = [H+][OH−] = 10−14 at 298 K; Kw rises with temperature.
- Strong = fully ionized (HCl, HBr, HI, HNO3, H2SO4; group 1 hydroxides); weak = partly ionized; equilibria lie towards the weaker conjugate.
- Acid + hydroxide/oxide → salt + water; + carbonate/hydrogencarbonate → salt + water + CO2; + NH3 → ammonium salt.
- Strong–strong pH curve: start pH from [acid], vertical section, pH 7 at equivalence.
- HL pOH = −log[OH−]; pH + pOH = 14; Ka, Kb, pKa, pKb; KaKb = Kw; [H+] ≈ √(Kac).
- HL Salts: anion of weak acid → alkaline; cation of weak base → acidic; hydrolysis equations.
- HL Four pH curves; pH = pKa (or pOH = pKb) at half-equivalence; indicators HInd ⇌ H+ + Ind−, end point at pH ≈ pKa(HInd).
- HL Buffers: weak acid + conjugate base or weak base + conjugate acid; pH = pKa + log([A−]/[HA]); dilution leaves pH unchanged.
Examination checklist
- Say “proton” or “H+”, never “hydrogen”.
- Single arrow for strong, equilibrium arrow for weak.
- Check the sign: acids below pH 7, alkalis above.
- Give every distinguishing test with its expected result for each substance.
- HL Write Ka with the square of [H+]; never take Ka as [H+].
- HL Explain buffers with the equation for each added ion.
Knowledge organiser
| Outcome | Key facts and relationships | Must-remember distinctions and common errors |
|---|---|---|
| BL acids and bases 3.1.1 | Donor/acceptor; H+(aq) ≡ H3O+(aq); base needs a lone pair. | Base ≠ alkali; proton ≠ hydrogen atom. |
| Conjugate pairs 3.1.2 | Differ by one H+ and one charge. | HCO3− → CO32−; H3O+/OH− is not a pair. |
| Amphiprotic 3.1.3 | H2O, HCO3−, H2PO4−, HPO42−. | Amphoteric is broader (Al2O3). |
| pH and Kw 3.1.4–3.1.5 | pH = −log[H+]; Kw = 10−14 (298 K). | Neutral = [H+] = [OH−]; pH 7 only at 298 K. |
| Strong/weak 3.1.6 | Extent of ionization; weaker conjugate favoured. | Strong ≠ concentrated; same volume of alkali. |
| Neutralization 3.1.7 | Salt + water (+ CO2); NH3 gives ammonium salt. | Never H2CO3 as a product. |
| pH curves 3.1.8, 3.1.13 | Start, steep section, equivalence, plateau; buffer region; pH = pKa at ½ eq. | Equivalence pH from the salt: 7, >7 or <7. |
| Ka, Kb 3.1.9–3.1.11 HL | pOH; √(Kac); KaKb = Kw; pKa + pKb = 14. | Smaller pKa = stronger acid. |
| Salts 3.1.12 HL | Hydrolysis of NH4+, RCOO−, CO32−, HCO3−. | Justify with an equation. |
| Indicators 3.1.14–3.1.15 HL | HInd ⇌ H+ + Ind−; end point ≈ pKa. | End point ≠ equivalence point; whole range inside steep section. |
| Buffers 3.1.16–3.1.17 HL | Weak conjugate pair; pH = pKa + log([A−]/[HA]). | Weak component in excess; dilution: same pH, less capacity. |