Related topics: Amount of Substance · Period 3 Oxides
Where this sits
You have used acids and alkalis since GCSE, but this A-level topic is almost entirely new: it makes acidity quantitative. The whole topic rests on one idea (acidity is set by the concentration of hydrogen ions), measured on the logarithmic pH scale. From there come weak-acid equilibria (the chemical equilibria of AS applied to acids), pH curves, and buffers. It is calculation-heavy, and a habit worth building early is to work out what species are present after any reaction before reaching for a formula.
Brønsted–Lowry acids & bases
The Brønsted–Lowry model defines acids and bases by what they do with protons (H+ ions):
Key definitions
A Brønsted–Lowry acid is a proton (H+) donor.
A Brønsted–Lowry base is a proton (H+) acceptor.
An acid–base reaction is therefore a transfer of protons. When an acid donates its proton it becomes a conjugate base; when a base accepts one it becomes a conjugate acid. The two make a conjugate acid–base pair, differing by a single H+.
CH3COOH + H2O ⇌ CH3COO− + H3O+
Some species can do either job depending on the reaction: the hydrogencarbonate ion HCO3− can donate a proton (to become CO32−) or accept one (to become H2CO3).
Exam questions
Q1
[1 mark]
Which species cannot function as a Brønsted–Lowry acid? Tick (✓) one box.
CH3COO− has no proton to donate (it is the conjugate base of ethanoic acid), so it cannot act as an acid. HCO3−, H3O+ and NH4+ can all donate an H+.
Q2
[1 mark]
Which species can behave as a Brønsted–Lowry acid in aqueous solution? Tick (✓) one box.
HCO3− can donate a proton to form CO32−. SO42− has no easily donated H; BF3 is an electron-pair acceptor (a Lewis acid, not Brønsted–Lowry); NH3 is a base.
Source: AQA A-Level Chemistry past papers.
The pH scale & strong acids
Hydrogen-ion concentrations span an enormous range, so acidity is measured on a logarithmic scale:
pH = −log10[H+]
Reversing it gives [H+] = 10−pH. Because the scale is logarithmic, a fall of one pH unit means ten times the hydrogen-ion concentration.
A strong acid fully dissociates, so its hydrogen-ion concentration equals its concentration, but a diprotic acid (such as H2SO4) releases two H+ per molecule, so [H+] is twice the acid concentration.
Worked example: pH of a strong acid, and back again
(a) Calculate the pH of 0.0250 mol dm−3 sulfuric acid (assume full dissociation).
Step 1. Find [H+] from the concentration. H2SO4 is diprotic, so:
[H+] = 2 × 0.0250 = 0.0500 mol dm−3
Step 2. Take the log.
pH = −log10(0.0500) = 1.30
(b) A solution of hydrochloric acid has pH = 2.60. Find its concentration.
Step 1. Reverse the log.
[H+] = 10−2.60 = 2.51 × 10−3 mol dm−3
Step 2. Relate [H+] to the acid. HCl is monoprotic and fully dissociated, so [HCl] = [H+] = 2.51 × 10−3 mol dm−3.
Diluting a strong acid follows directly from the log scale: every ten-fold dilution cuts [H+] to a tenth, so the pH rises by exactly one unit. A two-fold dilution raises it by only log102 ≈ 0.30. Halving the concentration never halves (or doubles) the pH, because pH is logarithmic, not linear.
Interactive: pH from [H+], and back again
Exam questions
Q3
[1 mark]
A strong acid H2X dissociates fully: H2X(aq) → 2H+(aq) + X2−(aq). What is the pH of a 0.020 mol dm−3 solution? Tick (✓) one box.
Two H+ per molecule: [H+] = 2 × 0.020 = 0.040 mol dm−3; pH = −log(0.040) = 1.40. Forgetting the ×2 gives 1.70.
Q4
[1 mark]
Which statement about pH is correct? Tick (✓) one box.
Kw increases with temperature, so pure water has a higher [H+] (and pH below 7) above 298 K, while remaining neutral. Nitric acid at 2.0 mol dm−3 has pH = −log 2.0 = −0.30; diprotic H2SO4 has a lower pH than HCl at the same concentration.
Source: AQA A-Level Chemistry past papers.
The ionic product of water & strong bases
Water itself dissociates very slightly, and the equilibrium constant for this gives the ionic product of water:
H2O(l) ⇌ H+(aq) + OH−(aq)
Kw = [H+][OH−] = 1.00 × 10−14 mol2 dm−6 (at 298 K)
Kw comes from the ordinary equilibrium constant for this dissociation: [H2O] is (almost) constant (it is enormous compared with [H+] and [OH−], and the equilibrium lies far to the left), so it is incorporated into Kw (Kw = Kc[H2O]) rather than written separately. In pure water every dissociating molecule gives one H+ and one OH−, so [H+] = [OH−] = √Kw.
For a strong base, find [OH−] from the concentration (×2 for a Group 2 hydroxide), then use Kw to get [H+] = Kw/[OH−] and hence the pH.
Worked example: pH of a strong base
Calculate the pH of 0.150 mol dm−3 NaOH at 298 K (Kw = 1.00 × 10−14 mol2 dm−6).
Step 1. Write down [OH−]. NaOH is a strong base with one OH− per formula unit:
[OH−] = 0.150 mol dm−3
Step 2. Convert to [H+] with Kw.
[H+] = Kw/[OH−] = 1.00 × 10−140.150 = 6.67 × 10−14 mol dm−3
Step 3. Take the log.
pH = −log10(6.67 × 10−14) = 13.18
Interactive: strong bases, routed through Kw
Precision points
- The dissociation of water is endothermic, so Kw increases with temperature. Pure water above 25 °C therefore has a pH below 7, but it is still neutral, because [H+] = [OH−].
- For Group 2 hydroxides (and any diprotic species), remember the factor of 2 in [OH−].
- Warming an alkali changes its pH (through Kw) but not its concentration, so the volume of acid needed to neutralise it in a titration is unchanged. pH and amount of substance are different questions.
| Temperature / °C | Kw / mol2 dm−6 | pH of pure water |
|---|---|---|
| 10 | 2.93 × 10−15 | 7.27 |
| 20 | 6.81 × 10−15 | 7.08 |
| 25 | 1.00 × 10−14 | 7.00 |
| 30 | 1.47 × 10−14 | 6.92 |
| 50 | 5.48 × 10−14 | 6.63 |
Exam questions
Q5
[1 mark]
What is the concentration of NaOH(aq), in mol dm−3, that has pH = 14.30? (Kw = 1.00 × 10−14 mol2 dm−6.) Tick (✓) one box.
[H+] = 10−14.30 = 5.01 × 10−15; [OH−] = Kw/[H+] = 1.00 × 10−145.01 × 10−15 = 2.00 mol dm−3. NaOH gives one OH− per formula, so [NaOH] = 2.00.
Q6
[1 mark]
Water dissociates slightly: H2O(l) ⇌ H+(aq) + OH−(aq). Explain why [H2O] is not shown in the Kw expression.
Show answer
[H2O] is (almost) constant: it is very large compared with [H+] and [OH−], so it is incorporated into Kw (Kw = Kc[H2O]). 1 mark
Q7
[1 mark]
Kw = 6.40 × 10−15 mol2 dm−6 at 18 °C and 1.00 × 10−14 mol2 dm−6 at 25 °C. Which statement is correct? Tick (✓) one box.
In pure water [OH−] = √Kw = √(6.40 × 10−15) = 8.00 × 10−8. Kw rising with temperature means the dissociation is endothermic; the pH at 18 °C (7.10) differs from 25 °C (7.00); and water is neutral at every temperature: it never becomes more or less acidic, because [H+] = [OH−] throughout.
Q8
[3 marks]
Calculate the pH of a 0.0100 mol dm−3 solution of strontium hydroxide at 10 °C (assume complete dissociation). At 10 °C, Kw = 2.93 × 10−15 mol2 dm−6.
Show answer
[OH−] = 2 × 0.0100 = 0.0200 mol dm−3. 1 mark
[H+] = Kw/[OH−] = 2.93 × 10−150.0200 = 1.47 × 10−13. 1 mark
pH = −log(1.47 × 10−13) = 12.83. 1 mark
Omitting the ×2 for Sr(OH)2 gives 12.53.
Q9
[3 marks]
At 50 °C, Kw = 5.48 × 10−14 mol2 dm−6. Calculate the pH of pure water at 50 °C (to 2 dp), and explain why the water is still neutral.
Show answer
In pure water [H+] = [OH−], so [H+] = √Kw = √(5.48 × 10−14) = 2.34 × 10−7. 1 mark
pH = −log(2.34 × 10−7) = 6.63. 1 mark
It is neutral because [H+] = [OH−] (each water molecule that dissociates gives one of each). 1 mark
Q10
[2 marks]
The pH of a barium hydroxide solution is lower at 50 °C than at 10 °C. At 50 °C, a 25 cm3 sample of this solution was neutralised by 22.45 cm3 of hydrochloric acid. Deduce the volume of this hydrochloric acid needed to neutralise another 25 cm3 sample at 10 °C, and explain your answer.
Show answer
The same volume: 22.45 cm3. 1 mark
The amount (concentration) of OH− in the sample is unchanged. Temperature changes Kw, and so the pH, but not the amount of base present. 1 mark
Source: AQA A-Level Chemistry past papers.
Weak acids and the acid dissociation constant
A weak acid only partially dissociates in water, so most of it stays as undissociated molecules:
hydrochloric acid: strong
Fully dissociated: every molecule has split into ions, so [H+] equals the acid concentration.
ethanoic acid: weak
Only partially dissociated: almost all molecules stay intact, so [H+] is far below the acid concentration.
The position of that equilibrium is measured by the acid dissociation constant, Ka:
Ka = [H+][A−] / [HA]
and pKa = −log10Ka (a smaller pKa means a stronger weak acid; reversing it, Ka = 10−pKa). To find the pH, make two standard approximations: the acid’s dissociation gives equal amounts of H+ and A− ([H+] = [A−]), and so little dissociates that [HA] ≈ the concentration you started with. These reduce the expression to:
[H+] = √(Ka[HA])
These approximations hold when the acid is genuinely weak and not too dilute. For a stronger or very dilute acid the fraction that dissociates is no longer negligible, so [HA] falls measurably below the starting concentration and the shortcut loses accuracy.
Worked example: pH of a weak acid from pKa
Calculate the pH of 0.0500 mol dm−3 methanoic acid (pKa = 3.75 at 298 K), to 2 decimal places.
Step 1. Convert pKa to Ka.
Ka = 10−3.75 = 1.78 × 10−4 mol dm−3
Step 2. Apply the weak-acid shortcut.
[H+] = √(Ka[HA]) = √(1.78 × 10−4 × 0.0500) = 2.98 × 10−3 mol dm−3
Step 3. Take the log, and quote 2 decimal places.
pH = −log10(2.98 × 10−3) = 2.53
Diluting a weak acid adds one step: work out the new concentration first (moles are unchanged, the volume grows), then apply [H+] = √(Ka[HA]) to that diluted concentration. Because [H+] depends on the square root of the concentration, a weak acid’s pH creeps up on dilution far more slowly than a strong acid’s: the equilibrium partly re-dissociates to replace the H+ that dilution removes.
Worked example: pH of a diluted weak acid
25.0 cm3 of 0.400 mol dm−3 ethanoic acid is diluted with water to 200.0 cm3. Calculate the pH of the diluted solution (Ka = 1.74 × 10−5 mol dm−3).
Step 1. Find the new concentration. The volume grows eight-fold (25.0 → 200.0 cm3):
[CH3COOH] = 0.400 × 25.0200.0 = 0.0500 mol dm−3
Step 2. Apply the weak-acid shortcut to the diluted concentration.
[H+] = √(1.74 × 10−5 × 0.0500) = 9.33 × 10−4 mol dm−3
Step 3. Take the log.
pH = −log10(9.33 × 10−4) = 3.03
An eight-fold dilution raised the pH only from 2.58 to 3.03, less than half a unit, where a strong acid would rise by log108 ≈ 0.90.
Precision points
- Defining a weak acid: say it partially (slightly) dissociates/ionises in aqueous solution. Mark schemes do not accept “the reaction is reversible” on its own, and naming the wrong ions loses the mark.
- Square brackets are essential in a Ka expression. Round brackets are not credited. Write the actual species: Ka = [CH3COO−][H+]/[CH3COOH].
- When a question says “give your answer to 2 decimal places”, a correct pH to 1 decimal place loses the final mark: the decimal places are the mark.
Interactive: weak-acid pH and Ka
Exam questions
Q11
[2 marks]
State the meaning of the term weak acid, and give the expression for Ka for propanoic acid, CH3CH2COOH.
Show answer
A weak acid only partially (slightly) dissociates / ionises in aqueous solution. 1 mark
Ka = [CH3CH2COO−][H+] / [CH3CH2COOH]. 1 mark
Square brackets are essential.
Q12
[3 marks]
Calculate the pH of a 0.150 mol dm−3 solution of ethanoic acid at 25 °C (to 2 dp). For ethanoic acid, Ka = 1.74 × 10−5 mol dm−3.
Show answer
[H+] = √(Ka[HA]) = √(1.74 × 10−5 × 0.150). 1 mark
[H+] = 1.62 × 10−3 mol dm−3. 1 mark
pH = −log(1.62 × 10−3) = 2.79. 1 mark
Q13
[1 mark]
A 0.100 mol dm−3 solution of a weak acid has pH = 2.50. What is its Ka, in mol dm−3? Tick (✓) one box.
[H+] = 10−2.50 = 3.16 × 10−3. Ka = [H+]2/[HA] = (3.16 × 10−3)2/0.100 = 1.00 × 10−4.
Q14
[4 marks]
Calculate the pH of a 0.100 mol dm−3 propanoic acid solution, to 2 decimal places. For propanoic acid, pKa = 4.87.
Show answer
Ka = 10−pKa = 10−4.87 = 1.35 × 10−5 mol dm−3. 1 mark
[H+] = √(Ka[HA]), i.e. [H+]2 = Ka[CH3CH2COOH]. 1 mark
[H+] = √(1.35 × 10−5 × 0.100) = 1.16 × 10−3 mol dm−3. 1 mark
pH = −log(1.16 × 10−3) = 2.94. 1 mark
Q15
[4 marks]
A student dilutes 25.0 cm3 of 0.500 mol dm−3 propanoic acid by adding water until the total volume is 100.0 cm3. Calculate the pH of this diluted solution, to 2 decimal places. For propanoic acid, Ka = 1.35 × 10−5 mol dm−3.
Show answer
[C2H5COOH] after dilution = 0.500 × 25.0/100.0 = 0.125 mol dm−3. 1 mark
[H+] = √(Ka[HA]) = √(1.35 × 10−5 × 0.125). 1 mark
[H+] = 1.30 × 10−3 mol dm−3. 1 mark
pH = −log(1.30 × 10−3) = 2.89. 1 mark
The trap is skipping the dilution step and using 0.500. The new concentration must come first.
Source: AQA A-Level Chemistry past papers.
Half-neutralisation shortcut
When exactly half a weak acid has been neutralised, half of it remains as HA and half has become A−, so [HA] = [A−]. Substituting into the Ka expression leaves [H+] = Ka, i.e. pH = pKa, so the pH at the half-equivalence point of a titration curve reads Ka off directly.
pH curves, indicators & Required Practical 9
Following the pH through a titration gives a pH curve. Its shape depends on whether each reactant is strong or weak, and the key feature is the steep, near-vertical section around the equivalence point.
| Titration | Vertical section (approx.) | Suitable indicator |
|---|---|---|
| Strong acid–strong base | pH ~3–11 (long) | methyl orange or phenolphthalein |
| Strong acid–weak base | pH ~3–7 | methyl orange |
| Weak acid–strong base | pH ~7–11 | phenolphthalein |
| Weak acid–weak base | no sharp vertical | none suitable |
An indicator is suitable only if its colour-change range falls entirely within the vertical section of the curve, so that it changes sharply at the equivalence point. A weak acid–weak base titration has no steep section, so no indicator gives a sharp end point. Universal indicator is never suitable for any titration: it changes colour gradually across the whole pH range, so there is no single sharp colour change to mark the end point.
strong acid + strong base
Long vertical section: methyl orange or phenolphthalein both change inside it.
strong acid + weak base
Vertical section ends low: methyl orange fits; phenolphthalein misses it.
weak acid + strong base
Vertical section sits high: phenolphthalein fits; methyl orange changes far too early.
weak acid + weak base
No vertical section: the pH drifts through the equivalence point, so no sharp end point.
Exam questions walk the weak acid–strong base curve feature by feature, so learn its anatomy. The starting pH is the weak acid’s own pH. Calculate it with [H+] = √(Ka[HA]). As base is added, the flask holds a shrinking amount of HA alongside the A− being formed (a buffer), so the pH rises only gradually across the middle of the curve (the buffer region). At half-neutralisation, [HA] = [A−] and pH = pKa. Then comes the steep jump at the equivalence point, and finally a flattening tail as excess base accumulates.
The pH at the equivalence point is set by the salt
At the equivalence point the flask holds only the salt (and water), and its ions decide the pH:
- Strong acid–strong base → a neutral salt (e.g. NaCl) → equivalence at pH 7.
- Weak acid–strong base → the salt contains a conjugate base (e.g. CH3COO−, which accepts protons from water) → equivalence above pH 7.
- Strong acid–weak base → the salt contains a conjugate acid (e.g. NH4+, a proton donor) → equivalence below pH 7.
So mixing exactly equal amounts of KOH and ethanoic acid gives not a buffer but a solution of potassium ethanoate, pH above 7.
Interactive: build a pH curve, then test an indicator
acid
base
test
Required practical 9: pH changes in titrations
You measure how pH changes when a weak acid reacts with a strong base and when a strong acid reacts with a weak base, using a pH meter (calibrated with buffer solutions of known pH, and with the probe washed with distilled water between solutions so no residue carries over). Add the alkali dropwise near the equivalence point, where a small addition causes a large pH change, so the steep section is captured accurately.
Everything that separates a good curve from a useless one:
- Calibrate with three buffers, not one: pH 4.00, 7.00 and 9.20. Three points give a correction line and check the meter is still responding linearly; one could only shift the readings.
- Rinse the probe with deionised water between every solution, and never wipe the bulb dry. Carry-over is a systematic error and the glass bulb is fragile.
- Switch to 0.20 cm3 portions near the equivalence point. The reason to write is that the pH changes very rapidly there for a very small addition. 2.0 cm3 portions jump straight over the vertical section.
- Stir after every addition before reading, or the probe measures a local pocket rather than the bulk solution.
- Join the points with a smooth curve, never straight lines, and read the equivalence point from the middle of the vertical section.
This is Required Practical 9. The full apparatus, method and safety are collected in the required practicals guide.
Exam questions
Q16
[1 mark]
A 0.10 mol dm−3 aqueous solution of an acid is added slowly to 25 cm3 of a 0.10 mol dm−3 aqueous solution of a base. Which acid–base pair has the highest pH at the equivalence point? Tick (✓) one box.
At equivalence only the salt remains, so compare the salts. Sodium ethanoate (weak acid–strong base) contains the basic ethanoate ion → the highest pH. HCl/NaOH gives neutral NaCl (pH 7); NH4Cl (strong acid–weak base) is acidic; the weak–weak pair is close to neutral.
Q17
[1 mark]
Identify the indicator most suitable for the titration of propanoic acid with sodium hydroxide. Bromocresol green (3.8–5.4); bromothymol blue (6.0–7.6); thymol blue (8.0–9.6).
Show answer
Thymol blue: a weak acid–strong base titration has its vertical section at higher pH (~7–11), and only thymol blue’s range (8.0–9.6) lies within it. 1 mark
Q18a
[1 mark]
A student produces a pH curve by adding sodium hydroxide solution to aqueous propanoic acid. Suggest why the student adds the sodium hydroxide solution dropwise around the equivalence point.
Show answer
There is a large (rapid) change in pH for a small addition of alkali near the equivalence point. Dropwise additions are needed to capture it. 1 mark
Q18b
[3 marks]
Give an expression for Ka for propanoic acid (CH3CH2COOH). Use this expression to show that pH = pKa when half of the propanoic acid has reacted with sodium hydroxide.
Show answer
Ka = [CH3CH2COO−][H+] / [CH3CH2COOH]. 1 mark
At half-neutralisation, [CH3CH2COOH] = [CH3CH2COO−], so the two cancel. 1 mark
That leaves Ka = [H+], and taking −log of both sides gives pKa = pH. 1 mark
Q18c
[2 marks]
Methyl orange (pH range 3.1–4.4) and universal indicator are not suitable indicators for the titration of propanoic acid with sodium hydroxide. State the reason why each indicator is not suitable.
Show answer
Methyl orange’s range does not fall within the steep section. It would change colour before the equivalence point. 1 mark
Universal indicator changes colour gradually across a range of colours, so there is no one sharp colour change at the end point. 1 mark
Q19
[2 marks]
A student plans to titrate butanoic acid with ethylamine. Explain why this titration could not be followed using an indicator.
Show answer
This is a weak acid–weak base titration. 1 mark
The pH changes only gradually at the equivalence point (no steep/vertical section), so there is no sharp colour change for an indicator to show. 1 mark
Q20
[1 mark]
Hydrochloric acid is titrated with sodium hydroxide solution. State why bromocresol green (3.8–5.4), phenol red (6.8–8.4) and thymolphthalein (9.3–10.5) are all suitable indicators for this titration.
Show answer
All three have their colour-change range within the steep (vertical) part of the strong acid–strong base curve, roughly pH 3 to 11. 1 mark
Source: AQA A-Level Chemistry past papers.
Buffer solutions
A buffer solution resists changes in pH when a small amount of acid or base is added, or when it is diluted. An acidic buffer is a mixture of a weak acid and the salt of that weak acid (a basic buffer uses a weak base and its salt).
There are two ways to make an acidic buffer, and exam questions use both. Either mix the weak acid with a solution of its salt directly (ethanoic acid + sodium ethanoate), or partially neutralise an excess of the weak acid with a strong base. The base converts some HA into A−, leaving a mixture of both. Either way, the buffer works because it holds a large reservoir of both the weak acid HA and its conjugate base A−:
How an acidic buffer resists pH change
- Add a small amount of acid (H+): it reacts with the conjugate base (H+ + A− → HA), removing the added H+.
- Add a small amount of base (OH−): it reacts with the weak acid (OH− + HA → A− + H2O), removing the added OH−.
- Because both HA and A− are present in large amounts, the ratio [HA]/[A−] barely changes, so [H+] (and the pH) stays almost constant.
- Dilution lowers [HA] and [A−] by the same factor, so their ratio (and therefore the pH) does not change.
A basic buffer (for example ammonia with ammonium chloride) works the same way in reverse, using the weak base NH3 and its conjugate acid NH4+. Added acid reacts with the base (NH3 + H+ → NH4+) and added base reacts with the ammonium ion (NH4+ + OH− → NH3 + H2O), so the ratio of NH3 to NH4+ (and the pH) barely changes.
Buffers matter wherever a steady pH is needed: blood is held near pH 7.4 by a carbonic acid–hydrogencarbonate buffer (H2CO3/HCO3−), and buffers are used in shampoos, foods, and many industrial and biochemical processes.
To calculate the pH, rearrange the Ka expression for the ratio of acid to salt:
[H+] = Ka × [HA] / [A−]
Because the acid and its salt share one solution, the volume cancels in the ratio, so moles work just as well as concentrations. In a partial-neutralisation question, do the reaction bookkeeping first: every mole of added OH− converts one mole of HA into A−.
Worked example: buffer pH by partial neutralisation
20.0 cm3 of 0.100 mol dm−3 NaOH is added to 25.0 cm3 of 0.100 mol dm−3 propanoic acid (Ka = 1.35 × 10−5 mol dm−3). Calculate the pH of the solution formed.
Step 1. Find the moles before the reaction.
n(HA) = 0.0250 × 0.100 = 2.50 × 10−3 mol; n(OH−) = 0.0200 × 0.100 = 2.00 × 10−3 mol
Step 2. React them: what is present afterwards? The OH− converts an equal amount of HA into A−:
n(HA) left = 2.50 × 10−3 − 2.00 × 10−3 = 5.0 × 10−4 mol; n(A−) = 2.00 × 10−3 mol
Step 3. Apply the buffer equation with the mole ratio (the volume cancels):
[H+] = 1.35 × 10−5 × 5.0 × 10−42.00 × 10−3 = 3.38 × 10−6 mol dm−3
Step 4. Take the log.
pH = −log10(3.38 × 10−6) = 5.47
The hardest buffer questions run the calculation in reverse: they state the target pH and ask for the mass of salt to dissolve in the acid. The chain is the same one backwards: pH → [H+] → rearrange for [A−] → moles → mass.
Worked example: the mass of salt for a target pH
What mass of sodium ethanoate (CH3COONa, Mr = 82.0) must dissolve in 250 cm3 of 0.200 mol dm−3 ethanoic acid to give a buffer of pH = 4.30? (Ka = 1.74 × 10−5 mol dm−3; assume the volume is unchanged.)
Step 1. Convert the target pH to [H+].
[H+] = 10−4.30 = 5.01 × 10−5 mol dm−3
Step 2. Rearrange the Ka expression for [A−].
[A−] = Ka[HA][H+] = 1.74 × 10−5 × 0.2005.01 × 10−5 = 6.94 × 10−2 mol dm−3
Step 3. Convert to moles in the actual volume.
n(CH3COO−) = 6.94 × 10−2 × 0.250 = 1.74 × 10−2 mol
Step 4. Convert moles to mass.
mass = 1.74 × 10−2 × 82.0 = 1.42 g
Interactive: buffer pH
Exam questions
Q21
[1 mark]
Equal volumes of pairs of solutions are mixed. Which pair forms a buffer solution? Tick (✓) one box.
Ammonia (a weak base) with ammonium chloride (its salt) is a basic buffer. The others are two bases, two weak acids, and a strong acid with a strong alkali. None is a weak-species/salt pair.
Q22
[1 mark]
Equal volumes of two solutions, each with the same concentration, are mixed together at 298 K. Which two solutions, when mixed, form a solution with a pH > 7? Tick (✓) one box.
Equal amounts of KOH and CH3COOH neutralise completely, leaving only potassium ethanoate, the salt of a weak acid, whose basic ethanoate ion gives pH > 7. HCOOH/HCOOK is an acidic buffer (pH < 7); NH3 + HCl gives acidic NH4Cl; NH4Cl/KCl contains the weakly acidic NH4+ ion.
Q23
[3 marks]
State how a buffer solution can be made from solutions of potassium hydroxide and ethanoic acid, give an equation for the reaction, and state how the buffer resists a change in pH when a small amount of acid is added.
Show answer
Add enough KOH to partially neutralise the ethanoic acid, leaving a mixture of ethanoic acid and ethanoate ions (use the acid in excess). 1 mark
KOH + CH3COOH → CH3COOK + H2O. 1 mark
Added H+ reacts with the ethanoate ions (CH3COO− + H+ → CH3COOH), so [H+] stays almost constant. 1 mark
Q24
[5 marks]
A buffer is made by adding 2.00 g of sodium hydroxide to 500 cm3 of 1.00 mol dm−3 ethanoic acid at 25 °C. Calculate its pH (to 2 dp). For ethanoic acid, Ka = 1.74 × 10−5 mol dm−3.
Show answer
n(NaOH) = 2.0040.0 = 0.0500 mol; n(CH3COOH) at start = 0.500 × 1.00 = 0.500 mol. 1 mark
After reaction: n(CH3COOH) = 0.500 − 0.0500 = 0.450 mol. 1 mark
n(CH3COO−) = 0.0500 mol. 1 mark
[H+] = Ka × n(HA)/n(A−) = 1.74 × 10−5 × (0.450/0.0500) = 1.57 × 10−4. 1 mark
pH = −log(1.57 × 10−4) = 3.80. 1 mark
The volume cancels in the ratio, so it is not needed.
Q25
[5 marks]
A mixture of methanoic acid and sodium methanoate in aqueous solution acts as an acidic buffer. Calculate the mass, in g, of sodium methanoate (HCOONa, Mr = 68.0) that must be added to 25.0 cm3 of 0.100 mol dm−3 methanoic acid to produce a buffer solution with pH = 4.05 at 298 K. For methanoic acid, pKa = 3.75. Assume the volume of the solution remains constant.
Show answer
Ka = 10−3.75 = 1.78 × 10−4 mol dm−3. 1 mark
[H+] = 10−4.05 = 8.91 × 10−5 mol dm−3. 1 mark
[HCOO−] = Ka[HCOOH]/[H+] = 1.78 × 10−4 × 0.1008.91 × 10−5 = 0.200 mol dm−3. 1 mark
n(HCOONa) = 0.200 × 0.0250 = 5.00 × 10−3 mol. 1 mark
mass = 5.00 × 10−3 × 68.0 = 0.34 g. 1 mark
Q26
[6 marks]
A buffer solution with a pH of 4.50 is made by dissolving x g of sodium propanoate (C2H5COONa) in a solution of propanoic acid. The final volume of buffer solution is 500 cm3 and the final concentration of the propanoic acid is 0.250 mol dm−3. Calculate x. For propanoic acid, Ka = 1.35 × 10−5 mol dm−3.
Show answer
[H+] = 10−4.50 = 3.16 × 10−5 mol dm−3. 1 mark
Rearrange: [C2H5COO−] = Ka[C2H5COOH]/[H+]. 1 mark
[C2H5COO−] = 1.35 × 10−5 × 0.2503.16 × 10−5 = 0.1068 mol dm−3. 1 mark
Mr(C2H5COONa) = 96.0. 1 mark
n = 0.1068 × 0.500 = 0.0534 mol. 1 mark
x = 0.0534 × 96.0 = 5.13 g. 1 mark
Accept 5.09–5.14 g. Six marks, four of them before any chemistry “happens”: the chain is pH → [H+] → [A−] → moles → mass.
Source: AQA A-Level Chemistry past papers.
The species-present workflow
Nearly every acid–base calculation is really the same task: decide what is present after any reaction, then choose the matching method.
Decide the species, then the method
- Strong acid alone → pH = −log[H+] (remember ×2 for a diprotic acid).
- Strong base alone → [H+] = Kw/[OH−] (×2 for a Group 2 hydroxide).
- Weak acid alone → [H+] = √(Ka[HA]).
- Strong acid + strong base → find the excess H+ or OH−, divide by the total volume, then the pH.
- Weak acid partly neutralised → a buffer: [H+] = Ka[HA]/[A−].
- Weak acid exactly neutralised → only the salt remains: no buffer, pH above 7.
And check the temperature: away from 25 °C, use the Kw value given for that temperature. The method does not change, the number does.
Precision points
- Examiner reports single out working the species present after reaction before any formula, the factor of 2 for diprotic acids and hydroxide, and quoting pH to the right number of decimal places.
- In a buffer explanation, name the species that reacts (H+ with A−; OH− with HA), not just “the equilibrium shifts”.
Capstone quiz: six past-paper questions
Six real AQA multiple-choice questions on this topic, in the style that opens Paper 3. Pick one answer each. The reasoning appears once you commit.
Q1
[1 mark]
Which can not function as a Brønsted–Lowry acid?
A Brønsted–Lowry acid donates a proton, so it needs an H it can release. CH3COO− is the conjugate base of ethanoic acid. The acidic H has already gone, and the remaining hydrogens sit on the CH3 group, which does not ionise in water. HCO3− is the tempting answer because it is negatively charged and usually met as a base, but it is amphoteric and can still donate its H to give CO32−.
Q2
[1 mark]
Equal volumes of pairs of solutions are mixed. Which pair forms a buffer solution?
A buffer needs a weak acid or weak base together with a salt of its conjugate. NH3 with NH4Cl is a weak base plus its conjugate acid, so it can mop up both added acid and added alkali. The trap is the two weak acids: ethanoic and methanoic acid are both weak, but neither supplies the conjugate base of the other, so there is no reservoir to react with added H+.
Q3
[1 mark]
A strong acid H2X dissociates in aqueous solution: H2X(aq) → 2H+(aq) + X2−(aq). What is the pH of a 0.020 mol dm−3 solution of this acid?
The equation releases two protons per molecule, so [H+] = 2 × 0.020 = 0.040 mol dm−3 and pH = −log(0.040) = 1.40. The 1.70 option is what you get from −log(0.020), forgetting the factor of 2, the single most common slip on diprotic strong acids.
Q4
[1 mark]
A 0.100 mol dm−3 solution of a weak acid has pH = 2.50. What is the value of Ka for this acid, in mol dm−3?
Work backwards: [H+] = 10−2.50 = 3.16 × 10−3 mol dm−3, then Ka = [H+]2[HA] = (3.16 × 10−3)20.100 = 1.00 × 10−4. The 3.16 × 10−3 option is tempting because it is the first number your calculator shows, but that is [H+], not Ka.
Q5
[1 mark]
A 0.10 mol dm−3 aqueous solution of an acid is added slowly to 25 cm3 of a 0.10 mol dm−3 aqueous solution of a base. Which acid–base pair has the highest pH at the equivalence point?
At the equivalence point only the salt is left, so the real question is which salt solution is most alkaline. Weak acid + strong base gives sodium ethanoate, whose ethanoate ion takes protons from water and pushes the pH above 7 (typically 8–9). HCl and NaOH tempts because NaOH is the strongest base on offer, but that pair gives a neutral salt and an equivalence point of exactly pH 7. Of the NH3 pairs, HCl gives ammonium chloride, which is acidic; CH3COOH gives ammonium ethanoate, which is close to neutral because ethanoic acid and ammonia are almost equally weak. Neither beats sodium ethanoate.
Q6
[1 mark]
Values of the ionic product of water (Kw) at different temperatures are given. Kw = 6.40 × 10−15 mol2 dm−6 at 18 °C; Kw = 1.00 × 10−14 mol2 dm−6 at 25 °C. Which statement is correct?
In pure water [H+] = [OH−], so [OH−] = √Kw = √(6.40 × 10−15) = 8.00 × 10−8 mol dm−3. The pH option is the trap: people assume pure water is always pH 7, but a smaller Kw at 18 °C means a smaller [H+] and a pH of about 7.10. The same data kills the exothermic claim: Kw rises with temperature, so dissociation must be endothermic.
Source: AQA A-Level Chemistry past papers.
Capstone questions
Q27
[1 mark]
Which change causes the pH of 10 cm3 of 1.0 mol dm−3 NaOH to be halved at 298 K? (Kw = 1.0 × 10−14.) Tick (✓) one box.
The NaOH starts at pH 14, so “halved” means pH 7 (neutral). Only adding an exactly equal amount of HCl (10 cm3 of 1.0 mol dm−3) neutralises completely. Dilutions barely move a strong base’s pH (doubling the volume takes it to ~13.7; even a thousand-fold dilution only reaches ~11), and 5 cm3 of HCl leaves excess OH− (pH ≈ 13.5).
Q28
[5 marks]
36.25 cm3 of 0.200 mol dm−3 sodium hydroxide are added to 25.00 cm3 of 0.150 mol dm−3 hydrochloric acid. Calculate the pH of the final solution at 25 °C (Kw = 1.00 × 10−14).
Show answer
n(OH−) = 0.03625 × 0.200 = 7.25 × 10−3 mol; n(H+) = 0.02500 × 0.150 = 3.75 × 10−3 mol. 1 mark
Excess OH− = 7.25 × 10−3 − 3.75 × 10−3 = 3.50 × 10−3 mol. 1 mark
Total volume = 61.25 cm3, so [OH−] = 3.50 × 10−30.06125 = 5.71 × 10−2 mol dm−3. 1 mark
[H+] = Kw/[OH−] = 1.00 × 10−145.71 × 10−2 = 1.75 × 10−13. 1 mark
pH = −log(1.75 × 10−13) = 12.76. 1 mark
Q29
[5 marks]
35.0 cm3 of 0.150 mol dm−3 aqueous sodium hydroxide are mixed with 20.0 cm3 of a 0.100 mol dm−3 solution of hydrochloric acid. The temperature of the solution formed is 40 °C, where Kw = 2.92 × 10−14 mol2 dm−6. Calculate the pH of the solution formed, to 2 decimal places.
Show answer
n(OH−) = 0.0350 × 0.150 = 5.25 × 10−3 mol; n(H+) = 0.0200 × 0.100 = 2.00 × 10−3 mol. 1 mark
Excess OH− = 3.25 × 10−3 mol. 1 mark
Total volume = 55.0 cm3, so [OH−] = 3.25 × 10−30.0550 = 5.91 × 10−2 mol dm−3. 1 mark
[H+] = 2.92 × 10−145.91 × 10−2 = 4.94 × 10−13. 1 mark
pH = −log(4.94 × 10−13) = 12.31. 1 mark
Same workflow as Q28. The only difference is using the Kw value for 40 °C.
Q30a
[2 marks]
Calcium hydroxide is almost insoluble in water, but reacts with dilute hydrochloric acid: Ca(OH)2(s) + 2HCl(aq) → CaCl2(aq) + 2H2O(l). A student adds 100 cm3 of 0.100 mol dm−3 hydrochloric acid to 0.600 g of solid calcium hydroxide (Mr = 74.1). Show, by calculation, that the calcium hydroxide is in excess.
Show answer
n(HCl) = 0.100 × 0.100 = 0.0100 mol; n(Ca(OH)2) = 0.60074.1 = 8.10 × 10−3 mol. 1 mark
0.0100 mol of HCl needs only 0.0100 divided by 2 = 5.00 × 10−3 mol of Ca(OH)2, and 8.10 × 10−3 > 5.00 × 10−3, so the calcium hydroxide is in excess. 1 mark
State what the 5.00 × 10−3 is (“the amount of Ca(OH)2 needed”), not just “0.01/2”.
Q30b
[5 marks]
The final mixture contains a saturated solution of Ca(OH)2 at 293 K. At 293 K the solubility of Ca(OH)2 in this solution is 0.400 g dm−3 and Kw = 6.80 × 10−15 mol2 dm−6. Calculate the pH of this solution, to 2 decimal places.
Show answer
Solubility in mol dm−3 = 0.40074.1 = 5.40 × 10−3. 1 mark
[OH−] = 2 × 5.40 × 10−3 = 1.08 × 10−2 mol dm−3. 1 mark
[H+] = Kw/[OH−] = 6.80 × 10−151.08 × 10−2 = 6.30 × 10−13. 1 mark
pH = −log[H+]. 1 mark
pH = −log(6.30 × 10−13) = 12.20. 1 mark
Three separate traps in one question: g dm−3 → mol dm−3 first, then ×2 for the hydroxide, then the non-standard Kw.
Source: AQA A-Level Chemistry past papers.
3.1.12 Acids and bases: Quick-reference summary
- Brønsted–Lowry: acid = proton donor, base = proton acceptor; a conjugate pair differs by one H+.
- pH = −log10[H+] and [H+] = 10−pH. Strong acids fully dissociate (a diprotic acid gives 2 H+ per molecule).
- Kw = [H+][OH−] (1.00 × 10−14 at 298 K). Strong base: [H+] = Kw/[OH−]. Kw rises with temperature, so neutral water is below pH 7 above 25 °C, still neutral because [H+] = [OH−].
- Weak acid: Ka = [H+][A−]/[HA]; pKa = −log Ka; [H+] = √(Ka[HA]).
- pH curves: the indicator must change colour within the steep (vertical) section; a weak acid–weak base titration has no sharp jump, so no indicator is suitable (and universal indicator never is). At half-neutralisation of a weak acid, pH = pKa.
- Equivalence pH is set by the salt: strong–strong → 7; weak acid–strong base → above 7; strong acid–weak base → below 7.
- Buffer: a weak acid + its salt (acidic) resists pH change: added H+ reacts with A−, added OH− reacts with HA; dilution leaves the ratio (and pH) unchanged. pH from Ka = [H+][A−]/[HA]; for a target pH, run it backwards (pH → [H+] → [A−] → moles → mass).