Indicators and Titration Curves

Lessons

Senior 6 · Chemistry

Indicators and Titration Curves

3 lessons

Lesson 2 of 3

Titration Curves

25 min

Watching the pH during a titration

In a winery, a food-testing laboratory or a water-quality laboratory, an acid–base titration is often run with a pH electrode in the flask instead of an indicator. The pH is read after every addition from the burette, and the readings are plotted against the volume added. The graph, a titration curve (or pH curve), shows far more than an indicator can: the exact equivalence point, whether the acid is strong or weak, and even its $K_{a}$ .

In this lesson you will build titration curves from the pH calculations of Unit 8, and learn to read every part of one.

Definition

Titration curve (pH curve): a graph of the pH of the solution in the flask against the volume of titrant added. Example: the S-shaped curve for HCl titrated with NaOH.

Equivalence point: the volume at which the amount of titrant added is exactly the amount needed by the equation. On the curve it is the middle of the steep (vertical) section. Example: 25.0 cm³ of 0.100 mol dm⁻³ NaOH for 25.0 cm³ of 0.100 mol dm⁻³ HCl.

Vertical section (steep section): the part of the curve near the equivalence point where a drop or two of titrant changes the pH by several units.

Half-equivalence point: the volume that is half the equivalence volume. For a weak acid titrated with a strong base, $[\mathrm{HA}] = [\mathrm{A^{-}}]$ here, so $\mathrm{pH} = \mathrm{p}K_a$ .

Buffer region: the flat part of a weak acid (or weak base) curve before equivalence, where the flask contains a buffer of the weak acid and its salt.

The experiment

Figure 2.1 shows the set-up. A known volume of acid, here 25.0 cm³, is pipetted into a beaker with a stirrer bar. A pH electrode, calibrated with buffers as in Unit 8, dips into it. The alkali is added from the burette, in 1–2 cm³ portions at first and then drop by drop close to the expected equivalence point, where the pH changes fastest. The pH is recorded after each addition.

Figure 2.1: Apparatus for recording a titration curve.

Strong acid with strong base

For 25.0 cm³ of 0.100 mol dm⁻³ HCl titrated with 0.100 mol dm⁻³ NaOH, the reaction is $\mathrm{H_3O^{+}(aq) + OH^{-}(aq) \rightarrow 2H_2O(l)}$ . Until the equivalence point, the pH comes from the excess acid; after it, from the excess alkali. Because pH is a logarithm, the pH barely moves while there is still plenty of acid: half the acid has to be used up just to raise the pH by 0.48. Near 25.0 cm³ the excess becomes tiny and the pH shoots up (solid line in Figure 2.2).

Weak acid with strong base

For 0.100 mol dm⁻³ ethanoic acid ( $\mathrm{p}K_a = 4.76$ ), four things are different (dashed line in Figure 2.2):

The start is higher, pH 2.88 instead of 1.00, because a weak acid is only slightly ionised.
There is a buffer region. After the first few cm³, the flask holds both $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COO^{-}}$ , and the curve flattens.
At half-equivalence, pH = p $K_{a}$ . At 12.5 cm³ half the acid has become ethanoate, so $\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]} = 1$ and the Henderson–Hasselbalch equation gives $\mathrm{pH} = \mathrm{p}K_a + \log_{10}1 = \mathrm{p}K_a$ .
The equivalence point is alkaline, pH 8.73, because the flask now contains sodium ethanoate, a salt that hydrolyses: $\mathrm{CH_3COO^{-} + H_2O \rightleftharpoons CH_3COOH + OH^{-}}$ .

After equivalence, both curves follow the same line: the pH is set by the excess NaOH, whatever the acid was.

Figure 2.2: Strong and weak acid titrated with a strong base (values calculated by Ganzaa for 0.100 mol dm⁻³ solutions at 25 °C).

Titrations with a weak base

Figure 2.3 shows the same two acids titrated with 0.100 mol dm⁻³ aqueous ammonia, a weak base ( $\mathrm{p}K_a(\mathrm{NH_4^{+}}) = 9.25$ ). With HCl, the equivalence point is acidic (pH 5.28) because ammonium chloride hydrolyses, $\mathrm{NH_4^{+} + H_2O \rightleftharpoons NH_3 + H_3O^{+}}$ , and the vertical section is short (pH 3.7 to 6.9). With ethanoic acid and ammonia there is no vertical section: the pH drifts upward through 7.00 at equivalence and no sharp change marks it.

Figure 2.3: Strong and weak acid titrated with a weak base.

The four types compared

Acid + base (0.100 mol dm⁻³, 25.0 cm³ of acid)	Starting pH	pH at equivalence	pH from 0.10 cm³ before to 0.10 cm³ after equivalence
strong + strong (HCl + NaOH)	1.00	7.00	3.70 → 10.30
weak + strong ( $\mathrm{CH_3COOH}$ + NaOH)	2.88	8.73	7.16 → 10.30
strong + weak (HCl + $\mathrm{NH_3}$ )	1.00	5.28	3.70 → 6.85
weak + weak ( $\mathrm{CH_3COOH}$ + $\mathrm{NH_3}$ )	2.88	7.00	6.85 → 7.16

0.10 cm³ is about two drops from a burette. The bigger the jump across those few drops, the easier the end point is to detect. The size of the jump also depends on concentration: with 0.0100 mol dm⁻³ HCl and NaOH the same two drops either side only take the pH from 4.70 to 9.30.

Worked example

Problem. 25.0 cm³ of 0.100 mol dm⁻³ HCl is titrated with 0.100 mol dm⁻³ NaOH. Calculate the pH after (a) 0.00 cm³, (b) 24.90 cm³, (c) 25.00 cm³ and (d) 25.10 cm³ of NaOH have been added.

Solution. $n(\mathrm{HCl}) = 0.0250 \times 0.100 = 2.50 \times 10^{-3}\,\mathrm{mol}$ .

(a) $[\mathrm{H_3O^{+}}] = 0.100$ , pH = 1.00.

(b) $n(\mathrm{NaOH}) = 2.490 \times 10^{-3}\,\mathrm{mol}$ , so $1.0 \times 10^{-5}\,\mathrm{mol}$ of acid is left in 49.90 cm³:

$[\mathrm{H_3O^{+}}] = \frac{1.0 \times 10^{-5}}{0.04990} = 2.0 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}, \quad \mathrm{pH} = 3.70$

(d) $1.0 \times 10^{-5}\,\mathrm{mol}$ of $\mathrm{OH^{-}}$ is in excess, in 50.10 cm³:

$[\mathrm{OH^{-}}] = \frac{1.0 \times 10^{-5}}{0.05010} = 2.0 \times 10^{-4}, \quad \mathrm{pOH} = 3.70, \quad \mathrm{pH} = 14.00 - 3.70 = 10.30$

Answer: (a) 1.00; (b) 3.70; (c) 7.00; (d) 10.30. Just 0.20 cm³ of alkali, about four drops, raises the pH by 6.6 units.

Worked example

Problem. 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid ( $K_a = 1.74 \times 10^{-5}$ , $\mathrm{p}K_a = 4.76$ ) is titrated with 0.100 mol dm⁻³ NaOH. Calculate the pH (a) at the start, (b) after 12.50 cm³, (c) at the equivalence point and (d) after 30.00 cm³.

Solution.

(a) Weak acid: $[\mathrm{H_3O^{+}}] = \sqrt{K_a c} = \sqrt{1.74 \times 10^{-5} \times 0.100} = 1.32 \times 10^{-3}$ ; pH = 2.88.

(b) Half the acid has been converted to ethanoate, so $[\mathrm{HA}] = [\mathrm{A^{-}}]$ and pH = p $K_{a}$ = 4.76.

(c) At 25.00 cm³ all the acid is $\mathrm{CH_3COO^{-}}$ : $\frac{2.50 \times 10^{-3}}{0.0500} = 0.0500\,\mathrm{mol\,dm^{-3}}$ . For the base,

$K_b = \frac{K_w}{K_a} = \frac{1.00 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}$

$[\mathrm{OH^{-}}] = \sqrt{K_b c} = \sqrt{5.75 \times 10^{-10} \times 0.0500} = 5.36 \times 10^{-6}, \quad \mathrm{pOH} = 5.27, \quad \mathrm{pH} = 8.73$

(d) Excess $\mathrm{OH^{-}}$ = $0.100 \times 0.00500 = 5.00 \times 10^{-4}\,\mathrm{mol}$ in 55.0 cm³: $[\mathrm{OH^{-}}] = 9.09 \times 10^{-3}$ , pOH = 2.04, pH = 11.96. (The ethanoate ion's own effect is negligible beside this excess.)

Answer: (a) 2.88; (b) 4.76; (c) 8.73; (d) 11.96.

Worked example

Problem. 20.0 cm³ of a solution of a weak monoprotic acid HX is titrated with 0.150 mol dm⁻³ NaOH. The curve shows the equivalence point at 16.0 cm³, and the pH after 8.0 cm³ is 3.86. Find the concentration of HX, its $K_{a}$ , and the pH you would expect before any alkali is added.

Solution. At equivalence, $n(\mathrm{HX}) = n(\mathrm{NaOH}) = 0.150 \times 0.0160 = 2.40 \times 10^{-3}\,\mathrm{mol}$ , so

$c(\mathrm{HX}) = \frac{2.40 \times 10^{-3}}{0.0200} = 0.120\,\mathrm{mol\,dm^{-3}}$

8.0 cm³ is the half-equivalence volume, so $\mathrm{p}K_a = 3.86$ and $K_a = 10^{-3.86} = 1.38 \times 10^{-4}$ .

$[\mathrm{H_3O^{+}}] = \sqrt{1.38 \times 10^{-4} \times 0.120} = 4.07 \times 10^{-3}, \quad \mathrm{pH} = 2.39$

Answer: 0.120 mol dm⁻³; $K_a = 1.38 \times 10^{-4}\,\mathrm{mol\,dm^{-3}}$ ; starting pH about 2.39. Reading p $K_{a}$ at half-equivalence is how chemists measure the $K_{a}$ of a new acid.

Warning

Common mistakes with titration curves

Assuming the equivalence point is always at pH 7. Only a strong acid with a strong base gives a neutral salt solution; a weak acid with a strong base gives an alkaline one, and a strong acid with a weak base an acidic one.
Reading p $K_{a}$ at the equivalence point. It is read at half the equivalence volume.
Thinking a weaker acid needs less alkali. Equal volumes of 0.100 mol dm⁻³ HCl and 0.100 mol dm⁻³ ethanoic acid need exactly the same volume of NaOH; only the shape of the curve differs.
Forgetting to use the total volume (acid + titrant) when you calculate a concentration after mixing.

Tip

To sketch any titration curve, fix four points first: the starting pH, the half-equivalence pH (= p $K_{a}$ for a weak acid), the equivalence volume and pH (from the salt formed), and the final pH (from the excess titrant). Then join them with the right shape: flat, then steep, then flat.

Chemistry Student: pH curves, titrations. Watch for how the starting pH, the pH at equivalence and the length of the vertical section change for each of the four combinations of strong and weak acid and base.

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Practice

Sketch the curve for 25.0 cm³ of 0.100 mol dm⁻³ NaOH titrated with 0.100 mol dm⁻³ HCl (acid added to the alkali). Label the starting pH, the equivalence point and the final pH.
20.0 cm³ of 0.150 mol dm⁻³ HCl is titrated with 0.100 mol dm⁻³ NaOH. Find the equivalence volume, and the pH after 10.0 cm³ and after 32.0 cm³ of NaOH.
20.0 cm³ of 0.200 mol dm⁻³ methanoic acid ( $\mathrm{p}K_a = 3.75$ ) is titrated with 0.200 mol dm⁻³ NaOH. Calculate the starting pH, the pH at half-equivalence and the pH at equivalence.
Explain why the pH at the equivalence point of HCl titrated with aqueous ammonia is below 7. Calculate it for a final $\mathrm{NH_4Cl}$ concentration of 0.0500 mol dm⁻³ ( $\mathrm{p}K_a(\mathrm{NH_4^{+}}) = 9.25$ ).
From a titration curve, a student reads the half-equivalence pH of an unknown acid as 4.2 and the pH at equivalence as 8.5. Which of these acids is it most likely to be: benzoic acid ( $\mathrm{p}K_a = 4.20$ ), lactic acid ( $\mathrm{p}K_a = 3.86$ ) or hydrochloric acid? Explain.
Compare the curves for 0.100 mol dm⁻³ and 0.0100 mol dm⁻³ HCl titrated with NaOH of the same concentration. Which parts change, and why does it matter?
Evaluate this claim: "Because ethanoic acid is weak, it takes less sodium hydroxide to neutralise 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid than 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid."

Show answersHide answers

Starts at pH 13.00, falls slowly, then drops steeply from about 10.3 to 3.7 around 25.0 cm³ with the equivalence point at pH 7.00, then levels off towards about 1.5. It is the mirror image of the curve in Figure 2.2.
$V = \frac{0.150 \times 20.0}{0.100} = 30.0\,\mathrm{cm^3}$ . After 10.0 cm³: excess acid $= 3.00 \times 10^{-3} - 1.00 \times 10^{-3} = 2.00 \times 10^{-3}\,\mathrm{mol}$ in 30.0 cm³, $[\mathrm{H_3O^{+}}] = 0.0667$ , pH = 1.18. After 32.0 cm³: excess $\mathrm{OH^{-}} = 2.0 \times 10^{-4}\,\mathrm{mol}$ in 52.0 cm³, $[\mathrm{OH^{-}}] = 3.85 \times 10^{-3}$ , pOH = 2.41, pH = 11.59.
Start: $\sqrt{10^{-3.75} \times 0.200} = 5.96 \times 10^{-3}$ , pH = 2.22. Half-equivalence: pH = 3.75. Equivalence: $[\mathrm{HCOO^{-}}] = 0.100$ , $K_b = 10^{-10.25}$ , $[\mathrm{OH^{-}}] = \sqrt{10^{-10.25} \times 0.100} = 2.37 \times 10^{-6}$ , pOH = 5.625, so pH = 8.375, about 8.4.
The salt $\mathrm{NH_4Cl}$ is present, and $\mathrm{NH_4^{+}}$ is a weak acid: $\mathrm{NH_4^{+} + H_2O \rightleftharpoons NH_3 + H_3O^{+}}$ . $[\mathrm{H_3O^{+}}] = \sqrt{10^{-9.25} \times 0.0500} = 5.30 \times 10^{-6}$ , pH = 5.28.
Benzoic acid. pH 4.2 at half-equivalence means $\mathrm{p}K_a = 4.2$ , which matches benzoic acid; lactic acid would give 3.86; HCl would give an equivalence pH of 7, not 8.5, and has no buffer region.
The starting pH rises from 1.00 to 2.00 and the final pH falls by one unit, so the vertical section shrinks (about 3.7–10.3 becomes about 4.7–9.3 for ±0.10 cm³). The equivalence volume and pH 7.00 do not change. A shorter jump means fewer indicators change sharply within it, so the end point is harder to detect.
The claim is wrong. Both solutions contain $2.50 \times 10^{-3}\,\mathrm{mol}$ of acid, and both react 1 : 1 with NaOH, so both need 25.0 cm³ of 0.100 mol dm⁻³ NaOH. As $\mathrm{OH^{-}}$ removes $\mathrm{H_3O^{+}}$ , the weak acid's equilibrium shifts right and keeps supplying more. "Weak" changes the pH values on the curve, not the amount of alkali needed.

Sources and credits

Image: PH reading of diluted wine sample during a titration assay.JPG by Agne27, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: pH Curves, Titrations (A-Level IB Chemistry), Chemistry Student, embedded from YouTube.
The text, apparatus diagram, titration curves (calculated from the charge balance at 25 °C with $\mathrm{p}K_a = 4.76$ for ethanoic acid and 9.25 for the ammonium ion), table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Watching the pH during a titration

2The experiment

3Strong acid with strong base

4Weak acid with strong base

5Titrations with a weak base

6The four types compared

Watching the pH during a titration

The experiment

Strong acid with strong base

Weak acid with strong base

Titrations with a weak base

The four types compared