E. Chemical Systems and Equilibrium · Acid-Base and Solubility Equilibria

Titration Curves II: Weak Acids and Weak Bases

E2.5 E3.8 E3.7 E2.4About 75 minutes

Learning goals

I can describe the shape of a weak acid-strong base titration curve and a weak base-strong acid titration curve.
I can explain why the equivalence point is above 7 for a weak acid and below 7 for a weak base.
I can calculate the pH at the start, in the buffer region, at half-equivalence, at equivalence and after equivalence.
I can use a titration curve to find an acid's concentration and its Ka.

Words to know

half-equivalence point
The point where half the weak acid (or weak base) has been neutralized. There, [HA] = [A⁻] and pH = pKa.

buffer region
The fairly flat part of a weak acid or weak base titration curve where both partners of the conjugate pair are present.

equivalence-point pH
The pH when exactly enough titrant has been added. It depends on the salt formed: 7 for strong-strong, above 7 for weak acid-strong base, below 7 for weak base-strong acid.

Same moles, different curve

Titrate 25.00 mL of 0.100 mol/L acetic acid with 0.100 mol/L NaOH, and the equivalence point still comes at 25.00 mL: there are just as many moles of acid as in 0.100 mol/L HCl. But the curve looks very different. It starts higher, rises quickly, flattens out in the middle, and the equivalence point is not at pH 7. Every one of those features can be explained with what you already know: weak acid calculations, buffers and salt hydrolysis. In fact, a weak acid titration is a tour of this whole unit.

Five points on a weak acid-strong base curve

For 25.00 mL of 0.100 mol/L CH₃COOH (Ka = 1.8 × 10⁻⁵) titrated with 0.100 mol/L NaOH, the reaction is CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O, and it goes to completion. As always, do the stoichiometry first, then decide what kind of solution is left:

Start (0 mL): only the weak acid. Use the weak-acid calculation (Ka and an ICE table): pH = 2.87.
Buffer region (between 0 mL and equivalence): some CH₃COOH has become CH₃COO⁻, so both partners are present. It is a buffer: use pH = pKa + log(n(A⁻) ÷ n(HA)). The curve is fairly flat here.
Half-equivalence (12.50 mL): exactly half the acid has been converted, so n(A⁻) = n(HA) and pH = pKa = 4.74. This is the easiest way to measure Ka from a curve.
Equivalence (25.00 mL): all the acid has become acetate. Acetate is a weak base, so this is a basic salt solution: pH = 8.72.
After equivalence: excess OH⁻ controls the pH, just as with HCl. The curves for strong and weak acids are the same here.

Titration of 25.00 mL of 0.100 mol/L acetic acid with 0.100 mol/L NaOH. The flat buffer region is centred on the half-equivalence point, where pH = pKa. The equivalence point is basic.

Worked example

Example 1: In the buffer region

Find the pH after 10.00 mL of 0.100 mol/L NaOH is added to 25.00 mL of 0.100 mol/L CH₃COOH.

n(CH₃COOH) = 2.500 × 10⁻³ mol; n(OH⁻) = 1.000 × 10⁻³ mol.
Reaction to completion: CH₃COOH left = 1.500 × 10⁻³ mol; CH₃COO⁻ formed = 1.000 × 10⁻³ mol.
Both partners are present, so it is a buffer. The volumes cancel in the ratio, so use moles.
pH = 4.74 + log(1.000 ÷ 1.500) = 4.74 − 0.18 = 4.57.

pH = 4.57.

Worked example

Example 2: At the equivalence point

Find the pH at the equivalence point of the same titration.

At 25.00 mL, all 2.500 × 10⁻³ mol of acid has become CH₃COO⁻, in a total volume of 50.00 mL.
[CH₃COO⁻] = 2.500 × 10⁻³ ÷ 0.05000 = 0.0500 mol/L.
Acetate is a weak base: Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.
[OH⁻] = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ mol/L. pOH = 5.28.
pH = 14.00 − 5.28 = 8.72.

pH = 8.72. Remember to use the total volume, which halves the concentration of the salt here.

Worked example

Example 3: After the equivalence point

Find the pH after 35.00 mL of NaOH.

Excess OH⁻ = 3.500 × 10⁻³ − 2.500 × 10⁻³ = 1.000 × 10⁻³ mol, in 60.00 mL.
The acetate ion also makes a little OH⁻, but far less than the excess strong base, so ignore it.
[OH⁻] = 1.000 × 10⁻³ ÷ 0.06000 = 0.01667 mol/L. pOH = 1.78; pH = 12.22.

pH = 12.22, the same as for HCl at this volume.

Strong acid (dashed) and weak acid (solid), same concentration, same base. The weak acid starts higher, has a flatter buffer region and a shorter jump, and its equivalence point is basic. Both reach equivalence at the same volume.

Weak base titrated with strong acid

Now titrate 25.00 mL of 0.100 mol/L NH₃ with 0.100 mol/L HCl. Everything is reversed:

Start: a weak base, pH 11.13 (from Kb).
Buffer region: NH₃ and NH₄⁺ are both present. Use pH = pKa(NH₄⁺) + log(n(NH₃) ÷ n(NH₄⁺)), with pKa(NH₄⁺) = 9.25.
Half-equivalence (12.50 mL): pH = pKa(NH₄⁺) = 9.25 (equivalently, pOH = pKb(NH₃)).
Equivalence (25.00 mL): only NH₄Cl, an acidic salt. [NH₄⁺] = 0.0500 mol/L, [H₃O⁺] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol/L, pH = 5.28.
After equivalence: excess H₃O⁺ controls the pH.

Summary of equivalence-point pH (25 °C): strong acid + strong base: 7; weak acid + strong base: above 7; weak base + strong acid: below 7. Weak acid + weak base titrations give curves with no sharp jump at all, so they are not used for analysis.

Titration of 25.00 mL of 0.100 mol/L NH₃ with 0.100 mol/L HCl. The curve falls; its buffer region is centred on pH 9.25 and its equivalence point is acidic.

Worked example

Example 4: Finding concentration and Ka from a curve

A curve for 20.00 mL of an unknown monoprotic acid titrated with 0.1000 mol/L NaOH shows the equivalence point at 16.00 mL and pH 4.20 at 8.00 mL. Find the acid's concentration and Ka, and suggest its identity using the Ka table.

n(NaOH) at equivalence = 0.1000 × 0.01600 = 1.600 × 10⁻³ mol = n(acid).
c(acid) = 1.600 × 10⁻³ ÷ 0.02000 = 0.0800 mol/L.
8.00 mL is half of 16.00 mL, so it is the half-equivalence point: pKa = 4.20.
Ka = 10⁻⁴·²⁰ = 6.3 × 10⁻⁵.

c = 0.0800 mol/L, Ka = 6.3 × 10⁻⁵, which matches benzoic acid in the table.

Just for fun

Chemists love the half-equivalence point because it hands over pKa for free: one pH reading, no ICE table, no quadratic. It's the closest thing chemistry has to a cheat code, and it's completely legal.

Hands-on activity

Analyse a titration like a lab chemist

You need: graph paper, a ruler, a pencil and a calculator.

A student titrated 20.00 mL of an unknown acid with 0.1000 mol/L NaOH and recorded these pH-meter readings:

V (mL): 0, 2.00, 4.00, 6.00, 8.00, 10.00, 12.00, 14.00, 15.00, 15.50, 16.00, 16.50, 17.00, 18.00, 20.00, 24.00
pH: 2.65, 3.38, 3.73, 3.98, 4.20, 4.42, 4.68, 5.05, 5.38, 5.69, 8.42, 11.14, 11.43, 11.72, 12.00, 12.26

Plot pH (vertical) against volume (horizontal) and draw a smooth curve.
Find the equivalence volume: the middle of the steepest part.
Calculate the acid's concentration.
Find the half-equivalence volume and read the pH there. Calculate Ka.
Is the equivalence point acidic, neutral or basic? What does that tell you?
Check: use your Ka and concentration to calculate the starting pH. Does it match 2.65?

What to look for: equivalence at about 16.00 mL, so c = 0.0800 mol/L. Half-equivalence at 8.00 mL, pH 4.20, so Ka ≈ 6.3 × 10⁻⁵ (benzoic acid). The equivalence point (pH about 8.4) is basic, confirming a weak acid. Starting pH from Ka: x = √(6.3 × 10⁻⁵ × 0.0800) = 2.2 × 10⁻³, pH = 2.65 ✓. This is the same analysis you would do with data from a real lab or a simulation.

Your turn

For the acetic acid titration in this lesson, find the pH after 5.00 mL of NaOH.

Find the pH after 20.00 mL of NaOH.

Why is the pH at the equivalence point of a weak acid-strong base titration greater than 7?

For the NH₃/HCl titration in this lesson, find the pH after 10.00 mL of HCl.

A weak acid's half-equivalence point is at pH 3.74. What is its Ka? Which acid in the table could it be?

Which titration has an equivalence point below pH 7: (a) HNO₃ + KOH (b) HF + NaOH (c) CH₃NH₂ + HCl?

More practice: warm-up

At the half-equivalence point of a weak acid titration, pH = ?

What type of solution exists in the flask between the start and the equivalence point?

Is the equivalence point of NH₃ titrated with HCl acidic, basic or neutral?

Does a weak acid need more, less or the same volume of base to reach equivalence as the same amount of a strong acid?

Where is the curve flattest in a weak acid-strong base titration?

More practice: on your own

20.00 mL of 0.150 mol/L formic acid (Ka = 1.8 × 10⁻⁴) is titrated with 0.100 mol/L NaOH. Find the equivalence volume and the pH at half-equivalence.

Find the pH at the equivalence point of the formic acid titration.

Find the starting pH of the formic acid titration.

For the NH₃/HCl titration in this lesson, find the pH after 30.00 mL of HCl.

Two acids of equal concentration are titrated with the same NaOH. Acid X's curve starts at pH 1.0 and has an equivalence pH of 7.0; acid Y starts at pH 3.0 and has an equivalence pH of 8.9. Classify each, and explain which has the larger jump at equivalence.

Challenge

Show algebraically why pH = pKa at the half-equivalence point.

Phosphoric acid is triprotic. Describe the main features you would expect on its titration curve with NaOH, and explain why only two equivalence points are usually seen clearly.

25.00 mL of 0.100 mol/L CH₃COOH is titrated with 0.100 mol/L NaOH. At what volume of NaOH is the pH equal to 5.00?

Check your understanding

Quiz

What is the pH at the half-equivalence point when a weak acid with Ka = 1.0 × 10⁻⁵ is titrated with NaOH?

Why is the equivalence point of NH₃ titrated with HCl below pH 7?

In the buffer region of a weak acid-strong base titration, the flask contains

Compared with 0.100 mol/L HCl, a titration curve of 0.100 mol/L CH₃COOH with the same NaOH has

A 25.00 mL sample of a weak acid reaches equivalence at 18.00 mL of 0.120 mol/L NaOH. The pH is 4.50 after 9.00 mL. What is Ka?

After the equivalence point in a weak acid-strong base titration, the pH is controlled mainly by

For parents and teachers

Big idea (E2.5, E3.8): a weak acid titration passes through four kinds of solution: a weak acid, a buffer, a basic salt at equivalence, and excess strong base. At half-equivalence, pH = pKa. Weak base titrations are the mirror image, with an acidic equivalence point.

How to help: Ask your teen to point to a spot on the curve and tell you which of the four kinds of solution is in the flask and which equation they would use. The data-analysis activity is exactly what a first-year university lab report asks for; ask them to explain how they found the equivalence volume.

Common difficulties:

Using the strong-acid method in the buffer region, or the buffer equation at the start (where there is no A⁻ yet).
At equivalence, forgetting that the salt concentration uses the total volume.
Assuming every equivalence point is at pH 7.
Confusing the half-equivalence point (pH = pKa) with the equivalence point.