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Titration curves

What a titration curve shows

A titration curve plots the pH of the solution in the flask against the volume of titrant added.
Its shape tells you the strengths of the acid and base involved, and where the equivalence point lies.

The equivalence point

The equivalence point is where the acid and base have been mixed in exactly the stoichiometric ratio — all the acid has been neutralised, with no excess of either.
Its pH is not always 7. It depends on what salt has been formed:

Titration	Salt formed	pH at equivalence
strong acid + strong base	neutral salt	7
weak acid + strong base	salt of a weak acid → basic	above 7
strong acid + weak base	salt of a weak base → acidic	below 7

Weak acid + weak base titrations are excluded from this standard, because the curve has no sharp vertical section and no usable end point.

Strong acid with strong base

Starts low — a strong acid is fully dissociated, so a 0.1 mol L⁻¹ solution begins at pH 1.
The pH rises very slowly at first, because the acid is in large excess.
Near the equivalence point it rises almost vertically through several pH units.
Equivalence at pH 7, since $\text{NaCl}$ has no ion that reacts with water.
Then it levels off as excess base dominates.

Weak acid with strong base

Four features distinguish this curve, and questions ask about each.

The initial pH is higher — a weak acid is only partly dissociated, so $[\text{H}_3\text{O}^+]$ starts much lower than for a strong acid of the same concentration.
A buffer region appears in the early-to-middle part, where the pH changes only slowly. Here both $\text{HA}$ and $\text{A}^-$ are present in quantity, so the solution is a buffer.
The half-equivalence point sits in the middle of that region. Exactly half the acid has been converted, so $[\text{HA}] = [\text{A}^-]$ and

$\text{pH} = \text{p}K_a$

The equivalence point is above pH 7, because the salt formed contains the conjugate base of a weak acid, which reacts with water:

$\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$

Reading the pKa off the graph

Find the equivalence volume from the steepest part of the curve.
Halve it to get the half-equivalence volume.
Read the pH at that volume. That value is the $\text{p}K_a$ .
This is the quickest way to identify an unknown weak acid, and it requires no calculation at all.

Why the vertical section matters

The vertical section is where a single drop of titrant changes the pH by several units.
It is long for a strong acid + strong base, shorter for a weak acid + strong base, and essentially absent for weak + weak.
A titration is only usable if there is a vertical section, because that is what makes the end point sharp — which is why the standard excludes weak + weak.

Tips and Tricks

The equivalence point is not always pH 7. Decide what salt has been formed and ask whether either of its ions reacts with water. A weak acid titrated with a strong base always gives an equivalence point above 7.

Do not confuse the equivalence point with the end point. The equivalence point is a fact about the stoichiometry; the end point is where the indicator changes colour. A good indicator makes them coincide.

At half-equivalence, $\text{pH} = \text{p}K_a$ — and only there. Reading the pKa anywhere else on the curve is wrong.

When asked to sketch a curve, mark the initial pH, the equivalence volume and pH, and the shape of the vertical section. An unlabelled S-shape rarely scores.

If asked why the pH rises rapidly near equivalence, say that the remaining acid is nearly used up, so each further drop of base removes a large fraction of what is left, and once the acid is gone the excess base takes over immediately.

Worked ExampleA full weak acid titration

$25.0$ mL of $0.120$ mol L⁻¹ ethanoic acid is titrated with $0.150$ mol L⁻¹ sodium hydroxide. $K_a(\text{CH}_3\text{COOH}) = 1.74 \times 10^{-5}$ .

Calculate (a) the initial pH, (b) the volume of NaOH needed to reach equivalence, (c) the pH at half-equivalence, and (d) the pH at the equivalence point.

Part (a) — the initial pH

Before any base is added this is simply a weak acid solution:

$[\text{H}_3\text{O}^+] = \sqrt{K_a c} = \sqrt{1.74 \times 10^{-5} \times 0.120} = \sqrt{2.088 \times 10^{-6}}$

$= 1.445 \times 10^{-3} \text{ mol L}^{-1}$

Part (b) — the equivalence volume

At equivalence, moles of base = moles of acid.

$n(\text{CH}_3\text{COOH}) = cV = 0.120 \times 0.0250 = 3.00 \times 10^{-3} \text{ mol}$

$V(\text{NaOH}) = \frac{n}{c} = \frac{3.00 \times 10^{-3}}{0.150} = 0.0200 \text{ L}$

Part (c) — the pH at half-equivalence

Half-equivalence is at 10.0 mL, where exactly half the ethanoic acid has been converted to ethanoate.

So $[\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]$ , the ratio is 1, and

$[\text{H}_3\text{O}^+] = K_a \times 1 = 1.74 \times 10^{-5}$

Part (d) — the pH at the equivalence point

Step 1 — Identify what is in the flask. All the ethanoic acid has been converted to ethanoate ion. The solution is now simply a solution of sodium ethanoate — a weak base.

Step 2 — Find its concentration, allowing for dilution. The total volume is now $25.0 + 20.0 = 45.0$ mL.

$[\text{CH}_3\text{COO}^-] = \frac{3.00 \times 10^{-3}}{0.0450} = 0.0667 \text{ mol L}^{-1}$

This dilution step is essential — the base has diluted the original solution nearly twofold.

Step 3 — Derive $K_b$ for the ethanoate ion.

$K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.747 \times 10^{-10}$

Step 4 — Treat it as a weak base.

$[\text{OH}^-] = \sqrt{K_b c} = \sqrt{5.747 \times 10^{-10} \times 0.0667} = \sqrt{3.831 \times 10^{-11}}$

$= 6.190 \times 10^{-6} \text{ mol L}^{-1}$

Step 5 — Convert to pH.

$\text{pOH} = -\log(6.190 \times 10^{-6}) = 5.208$

$\text{pH} = 14 - 5.208 = 8.79$