Acid-Base Titrations: Study Guide (Chemistry 8.5)

What a pH titration curve tells you

In Topic 4.6 a titration was a way to measure an amount. Here you add a pH probe. As titrant runs in from the burette, you record the pH after each addition and plot pH against volume of titrant added. That plot is a titration curve, and it summarises the whole experiment in one picture.

Every acid-base titration curve has the same four regions. Think of a weak acid HA being titrated with sodium hydroxide:

Region	What has been added	Major species in the flask	How to find the pH
Start	no base	HA (barely ionised)	weak acid equilibrium (Topic 8.3)
Before equivalence	some OH⁻, all used up	HA and A⁻ (a buffer)	Henderson-Hasselbalch, or the Ka expression
Equivalence point	moles OH⁻ = moles HA at start	A⁻ only (plus Na⁺)	A⁻ is a weak base: use Kb
After equivalence	more OH⁻ than HA	A⁻ and excess OH⁻	excess OH⁻ ÷ total volume

The reaction in the flask is the one from Topic 8.4: HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l). It goes essentially to completion, so for each mole of OH⁻ added, one mole of HA becomes one mole of A⁻.

Strong and weak acids compared

Figure 1 shows two titrations with the same amounts. Each flask holds 30.00 mL of an acid at 0.0800 M, and the titrant is 0.100 M NaOH. One acid is HCl (strong). The other is a fictional weak acid, HQ, with Ka = 2.5 × 10⁻⁵ (pKa = 4.60).

Four differences are worth learning from the picture:

1. Start. HCl is fully ionised, so pH = −log(0.0800) = 1.10. HQ is only about 2% ionised, so its pH is higher, 2.85.
2. Buffer region. Once some HQ has turned into Q⁻, the flask holds a conjugate pair, which resists pH change. The weak-acid curve is therefore flat in the middle. HCl forms no buffer.
3. Equivalence pH. HCl gives pH 7.00. HQ gives pH 8.62, because Q⁻ is a weak base.
4. Same equivalence volume. Both need 24.00 mL. Strength changes the shape, not how much base is needed.

The equivalence point: moles, not pH

For an acid or base that gives or takes one proton (monoprotic), the equivalence point is reached when

moles of titrant added = moles of analyte present at the start.

This is true for strong and weak acids and bases alike. If HQ is only 2% ionised at the start, how can all of it react? Each OH⁻ reacts either with H₃O⁺ or directly with an HQ molecule. As H₃O⁺ is removed, the equilibrium HQ ⇌ H⁺ + Q⁻ keeps shifting to the right. By the equivalence point every HQ molecule has been used.

On a curve, the equivalence point is the middle of the steepest part (the point of inflection), not the top of the jump. Read its volume off the x-axis, then use n = M × V.

The half-equivalence point: pH = pKa

Halfway to the equivalence volume, exactly half of the original HA has been turned into A⁻. So [HA] = [A⁻]. Put that into the Ka expression:

Ka = [H₃O⁺][A⁻] / [HA], and with [A⁻] = [HA] this becomes Ka = [H₃O⁺], so pH = pKa.

This is the most useful single reading on a weak-acid curve. Find the equivalence volume, halve it, read the pH there, and you have pKa (and Ka = 10^(−pKa)). In Figure 1, the pH at 12.00 mL is 4.60, the pKa of HQ.

For a weak base B titrated with a strong acid, the same idea applies to the pair BH⁺/B. At half-equivalence [B] = [BH⁺], so the pH equals the pKa of BH⁺. Then pKb of B = 14.00 − pKa (at 25 °C).

What sets the pH at the equivalence point

At the equivalence point the original acid and the added base have both been used up. The pH is fixed by what is left in the flask:

Titration	Major species at equivalence	pH at equivalence (25 °C)
Strong acid + strong base (HCl + NaOH)	Na⁺, Cl⁻, H₂O; neither ion reacts with water	7.00 (neutral)
Weak acid + strong base (HQ + NaOH)	Q⁻ (conjugate base), Na⁺	above 7: Q⁻ + H₂O ⇌ HQ + OH⁻
Weak base + strong acid (B + HCl)	BH⁺ (conjugate acid), Cl⁻	below 7: BH⁺ + H₂O ⇌ B + H₃O⁺

The weak-base curve is the weak-acid curve upside down: it starts high (basic), falls slowly through a buffer region, drops steeply at equivalence, and the equivalence point sits below pH 7.

Worked example 1: pH at four points on a weak acid curve

Question. For the HQ titration in Figure 1 (30.00 mL of 0.0800 M HQ, Ka = 2.5 × 10⁻⁵, titrated with 0.100 M NaOH), find: (a) the equivalence volume; (b) the pH after 18.00 mL; (c) the pH at equivalence; (d) the pH after 30.00 mL.

(a) n(HQ) = 30.00 mL × 0.0800 M = 2.400 mmol. At equivalence n(OH⁻) = 2.400 mmol, so V = 2.400 mmol ÷ 0.100 M = 24.00 mL.

(b) 18.00 mL × 0.100 M = 1.800 mmol OH⁻, all used up.

	HQ	OH⁻	Q⁻
Before reaction (mmol)	2.400	1.800	0
After reaction (mmol)	0.600	0	1.800

Both members of the pair are present, so this is a buffer. Both share the same volume, so the mole ratio equals the concentration ratio: pH = pKa + log([Q⁻]/[HQ]) = 4.60 + log(1.800 / 0.600) = 4.60 + 0.48 = 5.08.

(c) At equivalence all 2.400 mmol of HQ is now Q⁻, in 30.00 + 24.00 = 54.00 mL. [Q⁻] = 2.400 ÷ 54.00 = 0.04444 M. Kb(Q⁻) = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 2.5 × 10⁻⁵ = 4.0 × 10⁻¹⁰. [OH⁻] ≈ √(Kb × [Q⁻]) = √(4.0 × 10⁻¹⁰ × 0.04444) = 4.22 × 10⁻⁶ M, so pOH = 5.38 and pH = 8.62.

(d) 30.00 mL × 0.100 M = 3.000 mmol OH⁻; 3.000 − 2.400 = 0.600 mmol in excess, in 60.00 mL. [OH⁻] = 0.0100 M, pOH = 2.00, pH = 12.00. The small amount of OH⁻ made by Q⁻ is negligible beside the excess.

Check. Each answer sits where it should on Figure 1: 5.08 is a little above pKa (more Q⁻ than HQ), 8.62 is basic, and 12.00 is on the flat top.

Worked example 2: reading a weak base curve

Question. A student titrates 25.00 mL of a solution of a fictional weak base, B, with 0.150 M HCl. The curve falls from pH 11.1, the steepest point is at 20.00 mL, and the pH at 10.00 mL is 9.20. Find (a) the concentration of B; (b) Kb for B; (c) the pH at the equivalence point.

(a) Moles of HCl at equivalence: 20.00 mL × 0.150 M = 3.000 mmol. The reaction B + H₃O⁺ → BH⁺ + H₂O is 1:1, so n(B) = 3.000 mmol and [B] = 3.000 mmol ÷ 25.00 mL = 0.120 M.

(b) 10.00 mL is half of 20.00 mL, so [B] = [BH⁺] there and pH = pKa(BH⁺) = 9.20. Then pKb = 14.00 − 9.20 = 4.80 and Kb = 10^(−4.80) = 1.6 × 10⁻⁵.

(c) At equivalence the flask holds 3.000 mmol BH⁺ in 25.00 + 20.00 = 45.00 mL: [BH⁺] = 0.06667 M. Ka(BH⁺) = 10^(−9.20) = 6.31 × 10⁻¹⁰. [H₃O⁺] ≈ √(6.31 × 10⁻¹⁰ × 0.06667) = 6.49 × 10⁻⁶ M, so pH = 5.19.

Interpretation. The equivalence point is acidic, as it must be for a weak base titrated with a strong acid. The starting pH, 11.1, agrees with a 0.120 M base with this Kb, which is a useful check on the reading.

Polyprotic acids: one jump per proton

A diprotic acid, H₂X, loses its protons one at a time. The first proton is removed (H₂X → HX⁻) before the second (HX⁻ → X²⁻) starts in earnest, as long as the two pKa values are well apart. So the curve shows two steep rises, and the second needs the same volume of titrant as the first, because each H₂X has one of each proton.

How to read a polyprotic curve:

• Number of acidic protons = number of clear jumps, provided each proton is acidic enough to show one. (Phosphoric acid, H₃PO₄, has three acidic protons but shows only two clear jumps with NaOH: its third proton is so weakly acidic that the last jump is lost in the high pH of excess hydroxide.)
• Equal spacing. The second equivalence volume is twice the first (40.00 mL against 20.00 mL in Figure 2). The moles of acid come from the first equivalence volume.
• Each pKa. Halfway to the first equivalence point, pH ≈ pKa₁ (here 4.00 at 10.00 mL). Halfway between the first and second equivalence points, pH ≈ pKa₂ (here 8.00 at 30.00 mL).
• Major species. Before the first jump: H₂X and HX⁻. At the first equivalence point: mainly HX⁻. Between the jumps: HX⁻ and X²⁻. At the second equivalence point: mainly X²⁻. After it: X²⁻ and excess OH⁻.

You need to reason about which species are present in large or small amounts at any point. You are not expected to calculate the concentration of every species along a polyprotic curve; that is outside the course. Full pH calculations are for monoprotic titrations, as in the worked examples above.

Worked example 3: identifying species on a diprotic curve

Question. Use Figure 2. (a) How many moles of H₂X were titrated? (b) Name the two most abundant acid-containing species at 30.00 mL and say how their amounts compare. (c) At 20.00 mL, which acid-containing species dominates, and how do the amounts of H₂X and X²⁻ compare?

(a) First equivalence at 20.00 mL: 20.00 mL × 0.100 M = 2.00 mmol of H₂X. (Using 40.00 mL would double-count: that volume removes both protons.)

(b) 30.00 mL is halfway between 20.00 and 40.00 mL. All the H₂X has become HX⁻, and half of that HX⁻ has become X²⁻. So HX⁻ and X²⁻ are present in equal amounts, which is why the pH there equals pKa₂.

(c) At the first equivalence point HX⁻ dominates. A little HX⁻ gives up a proton (making X²⁻) and a little accepts one (making H₂X). So H₂X and X²⁻ are both present only in small, roughly equal amounts, which is why the pH there, about 6.0, lies roughly midway between pKa₁ and pKa₂. Neither is a major species. This qualitative answer is all the course asks for.

Common misconceptions

• “The equivalence point is always at pH 7.” Only for strong acid with strong base. Weak acid gives a basic equivalence point; weak base gives an acidic one.
• “A weak acid needs less base because it is only partly ionised.” The equilibrium shifts as OH⁻ removes acid, so every HA molecule is eventually used. Equal moles of acid need equal moles of base.
• “The half-equivalence point is where pH is half of the equivalence pH.” It is where the volume is half the equivalence volume. The pH there equals pKa.
• “pH = pKa at half-equivalence works for HCl too.” A strong acid has no undissociated HA, so there is no pair to balance.
• “The equivalence point is where the curve flattens at the top.” It is the middle of the steep rise.
• “For a diprotic acid, use the second equivalence volume to find the moles of acid with a 1:1 ratio.” That volume removed two protons per molecule. Use the first equivalence volume, or halve the second.
• “Adding water to the flask before titrating changes the equivalence volume.” It changes the starting pH a little, but the moles of acid, and so the equivalence volume, stay the same.

Where this leads

Next, Topic 8.6, Molecular Structure of Acids and Bases, explains why some acids are strong and others weak, which is what gives the curves in Figure 1 their different shapes. Topic 8.7 shows how to choose an indicator whose colour change matches the equivalence pH, and Topics 8.8 to 8.10 study the buffer region in detail. Look back at Acid-Base Reactions and Buffers for the mixing calculations used in Worked example 1. Now try the practice questions, then use the revision notes and the checklist.

Reading about titrations or using a simulation does not replace the hands-on laboratory work the course requires.