Edexcel International A Level Chemistry: Acid-base equilibria: titration curves and buffers (YCH11)

This guide teaches the second half of Topic 14: Acid-base Equilibria from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers outcomes 14.15 to 14.23, including Core Practical 11. This content is in Unit 4: Rates, Equilibria and Further Organic Chemistry (IA2). It assumes outcomes 14.1 to 14.14 (pH, Ka and Kw).

Every topic is on the course hub and the printable Edexcel Chemistry checklist. Then use the revision notes and practice questions for this unit, or a free diagnostic.

The outcomes covered

Outcome	What you must be able to do
14.15	Draw and interpret titration curves for strong and weak, monoprotic and diprotic acids and bases
14.16	Choose a suitable indicator from a titration curve and indicator data
14.17	Say what a buffer solution is
14.18	Explain how a buffer works
14.19	Calculate the pH of a buffer
14.20	Calculate the concentrations needed to make a buffer of a given pH
14.21	Use a titration curve to show buffer action and find Ka at half-neutralisation
14.22	Explain why buffers matter in cells, in blood (H₂CO₃/HCO₃⁻) and in foods
14.23	Core Practical 11: finding the Ka value for a weak acid

Unless a question says otherwise, work at 298 K, where Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

14.15 Titration curves

A titration curve plots pH against the volume added from the burette. Compare four features: the starting pH, the steep vertical section centred on the equivalence point (where acid and base have reacted in the ratio of the equation), the pH at equivalence, and the final pH, which approaches that of the burette solution.

Acid + base (base added)	Start pH	Equivalence pH	Vertical section (approx.)
Strong + strong (HCl + NaOH)	about 1	7	pH 3 to 11
Weak + strong (CH₃COOH + NaOH)	about 3	above 7 (anion is a weak base)	pH 7 to 11
Strong + weak (HCl + NH₃)	about 1	below 7 (NH₄⁺ is a weak acid)	pH 3 to 7
Weak + weak (CH₃COOH + NH₃)	about 3	close to 7	no clear vertical section

If acid is added to base, the curve is reflected: it starts high and falls. To find the equivalence volume, read the volume at the middle of the vertical section. To sketch a curve, calculate the equivalence volume first, then mark the start pH, the half-equivalence point and the final pH.

Worked example 1: sketching a weak acid-strong base curve

30.0 cm³ of 0.0600 mol dm⁻³ butanoic acid (use pKa = 4.82) is titrated with 0.0900 mol dm⁻³ NaOH.

moles acid = 0.0300 × 0.0600 = 1.80 × 10⁻³ mol
equivalence volume = 1.80 × 10⁻³ / 0.0900 = 20.0 cm³
start: Ka = 10^-4.82 = 1.51 × 10⁻⁵
       [H⁺] = √(Ka × c) = √(1.51 × 10⁻⁵ × 0.0600) = 9.53 × 10⁻⁴, pH = 3.02
10.0 cm³: [acid] = [salt], so pH = pKa = 4.82
26.0 cm³: excess OH⁻ = 0.0060 × 0.0900 = 5.40 × 10⁻⁴ mol in 56.0 cm³
       [OH⁻] = 9.64 × 10⁻³, [H⁺] = Kw/[OH⁻], pH = 11.98

The equivalence point is at 20.0 cm³ and above pH 7, because the butanoate ion is a weak base.

Diprotic acids and bases

A diprotic acid, H₂X, reacts with two moles of NaOH per mole. If its two pKa values are well apart, the curve has two vertical sections, and the second equivalence volume is twice the first. Half-way to the first equivalence point, pH = pKa₁; half-way between the two, pH = pKa₂. Sulfuric acid is usually treated as a strong diprotic acid: one vertical section, but twice the NaOH needed by the same amount of HCl.

Sodium carbonate is a diprotic base. With hydrochloric acid it reacts in two steps:

CO₃²⁻ + H⁺ → HCO₃⁻        (first equivalence, pH about 8 to 9)
HCO₃⁻ + H⁺ → H₂O + CO₂    (second equivalence, pH about 4)

Worked example 2

20.0 cm³ of 0.0750 mol dm⁻³ Na₂CO₃ is titrated with 0.120 mol dm⁻³ HCl.

moles CO₃²⁻ = 0.0200 × 0.0750 = 1.50 × 10⁻³ mol
first equivalence:  1.50 × 10⁻³ / 0.120 = 12.5 cm³
second equivalence: twice the moles of HCl, so 25.0 cm³

14.16 Choosing an indicator

An indicator is a weak acid, HIn, whose two forms have different colours. The end point (colour change) should match the equivalence point, so choose an indicator whose whole range lies inside the vertical section.

Typical ranges (in a question, always use the values you are given):

Indicator	Colour in acid	Colour in alkali	Range (pH)
Methyl orange	red	yellow	3.1 to 4.4
Methyl red	red	yellow	4.4 to 6.2
Bromothymol blue	yellow	blue	6.0 to 7.6
Phenolphthalein	colourless	pink	8.2 to 10.0

• Strong acid + strong base: any of the four.
• Weak acid + strong base: phenolphthalein (methyl orange would change in the buffer region).
• Strong acid + weak base: methyl orange or methyl red.
• Weak acid + weak base: none gives a sharp end point; use a pH meter.
• Sodium carbonate + HCl: phenolphthalein for the first end point, methyl orange for the second.

14.17 and 14.18 Buffer solutions

A buffer solution is one whose pH changes very little when a small amount of acid or alkali is added to it. It does not hold the pH constant, and a large addition overwhelms it.

Acidic buffers contain a weak acid and its conjugate base (usually supplied as a salt), for example propanoic acid with sodium propanoate. Basic buffers contain a weak base and its conjugate acid, for example ammonia with ammonium chloride.

How an acidic buffer works

HA ⇌ H⁺ + A⁻

The salt is fully ionised and the weak acid barely dissociated, so both A⁻ and HA are present in large amounts.

• Small amount of acid added: the extra H⁺ ions react with A⁻ to form HA. The equilibrium shifts to the left. Because [HA] and [A⁻] are both large, their ratio changes only a little, so [H⁺] and pH change only a little.
• Small amount of alkali added: OH⁻ + HA → A⁻ + H₂O. The ratio [HA]/[A⁻] again changes only slightly.

In an NH₃/NH₄Cl buffer, NH₃ + H⁺ → NH₄⁺ and NH₄⁺ + OH⁻ → NH₃ + H₂O.

14.19 Calculating the pH of a buffer

Rearrange the Ka expression:

[H⁺] = Ka × [HA] / [A⁻]        or        pH = pKa − log([HA]/[A⁻])

Assume HA’s dissociation is negligible and all A⁻ comes from the salt. Both share one volume, so moles can replace concentrations. Another route is to add less strong base than the weak acid needs for neutralisation: the leftover acid and the anion formed make the buffer.

Worked example 3: partial neutralisation

45.0 cm³ of 0.120 mol dm⁻³ lactic acid (use Ka = 1.4 × 10⁻⁴ mol dm⁻³) is mixed with 18.0 cm³ of 0.125 mol dm⁻³ NaOH. Find the pH, then the pH after adding 1.00 cm³ of 0.500 mol dm⁻³ HCl.

moles HA at start = 0.0450 × 0.120 = 5.40 × 10⁻³ mol
moles OH⁻ added   = 0.0180 × 0.125 = 2.25 × 10⁻³ mol
HA left = 3.15 × 10⁻³ mol;  A⁻ formed = 2.25 × 10⁻³ mol
[H⁺] = 1.4 × 10⁻⁴ × 3.15/2.25 = 1.96 × 10⁻⁴ mol dm⁻³
pH = 3.71

HCl = 5.00 × 10⁻⁴ mol converts A⁻ to HA: HA = 3.65 × 10⁻³, A⁻ = 1.75 × 10⁻³
[H⁺] = 1.4 × 10⁻⁴ × 3.65/1.75 = 2.92 × 10⁻⁴ mol dm⁻³
pH = 3.53

The pH falls by less than 0.2. The same HCl added to 63.0 cm³ of pure water (64.0 cm³ in total) would give pH 2.11.

14.20 Making a buffer of a chosen pH

Convert the target pH to [H⁺], then rearrange Ka for the unknown concentration.

Worked example 4: mass of salt needed

You need a buffer of pH 5.10 made from 250 cm³ of 0.180 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³). What mass of sodium ethanoate, CH₃COONa (Mr = 82.0), must be dissolved? Assume the volume does not change.

[H⁺] = 10^-5.10 = 7.94 × 10⁻⁶ mol dm⁻³
[A⁻] = Ka × [HA] / [H⁺] = 1.74 × 10⁻⁵ × 0.180 / 7.94 × 10⁻⁶ = 0.394 mol dm⁻³
moles = 0.394 × 0.250 = 0.0986 mol
mass  = 0.0986 × 82.0 = 8.08 g

If equal volumes of acid and salt solutions are mixed, both concentrations halve and the ratio is unchanged, so the salt concentration follows directly from the ratio.

14.21 Buffer action and Ka from a titration curve

On a weak acid-strong base curve, the gently rising section before the vertical section is a buffer region: the flask holds both HA and A⁻, so added NaOH changes the pH only slowly. On a strong acid-weak base curve, the weak base and its conjugate acid buffer in the same way.

At the half-neutralisation point, half the acid has become A⁻, so [HA] = [A⁻] and:

pH = pKa    and    Ka = 10^-pH

Find the equivalence volume from the vertical section, halve it, and read the pH at that volume.

Worked example 5

25.0 cm³ of a weak monoprotic acid is titrated with 0.110 mol dm⁻³ NaOH. The vertical section is centred on 18.40 cm³, and the pH at 9.20 cm³ is 3.75. Find Ka and the concentration of the acid.

half-equivalence volume = 18.40 / 2 = 9.20 cm³, so pKa = 3.75
Ka = 10^-3.75 = 1.78 × 10⁻⁴ mol dm⁻³
moles NaOH = 0.01840 × 0.110 = 2.024 × 10⁻³ mol = moles acid
[acid] = 2.024 × 10⁻³ / 0.0250 = 0.0810 mol dm⁻³

For a weak base titrated with strong acid, the half-way pH equals the pKa of the base’s conjugate acid (NH₄⁺ for ammonia).

14.22 Buffers in living things and in food

Blood is kept within a narrow pH range, close to 7.4. The main buffer is carbonic acid and hydrogencarbonate ions:

H₂CO₃ ⇌ H⁺ + HCO₃⁻

When cells release acids (for example lactic acid during hard exercise), extra H⁺ combines with HCO₃⁻ to form H₂CO₃, which can leave as carbon dioxide through the lungs. Added OH⁻ reacts with H₂CO₃ to form more HCO₃⁻. Enzymes work only in a narrow pH range, so this matters. Cells also contain buffers, such as phosphate systems and proteins.

Foods: bacteria and fungi can change the pH of food as they grow, spoiling it. A buffer (a weak acid with its salt) added to the food resists that pH change and slows deterioration.

14.23 Core Practical 11: finding Ka for a weak acid

One method uses half-neutralisation:

1. Calibrate a pH meter with buffer solutions of known pH.
2. Titrate 25.0 cm³ of the weak acid with NaOH using phenolphthalein. Record the volume.
3. To a second 25.0 cm³ portion of acid, add exactly half that volume of NaOH (or add 25.0 cm³ more acid to the neutralised flask).
4. Measure the pH. Now [HA] = [A⁻], so pKa = pH and Ka = 10^-pH.

Neither concentration needs to be known. Or measure the pH of an acid of known concentration and use Ka = [H⁺]²/[HA] (outcome 14.14). The specification also suggests plotting titration curves with a data logger and analysing vinegar. Errors come from an uncalibrated or unrinsed meter, or reading before the value settles.

Common errors

• Assuming every equivalence point is at pH 7.
• Picking an indicator that starts or finishes changing colour outside the vertical section.
• Using starting moles in a buffer ratio instead of the moles left after reaction.
• Writing [H⁺] = Ka × [A⁻]/[HA] with the ratio upside down.
• Reading the pH at the equivalence point and calling it pKa: it is the pH at half that volume.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 14: Acid-base Equilibria, outcomes 14.15 to 14.23, including Core Practical 11.