Acid-base indicators and titration curves — Chemistry

Before this

This leads to

Ka, Kb and pH calculations · coming soon
Oxidation and reduction

A colour change at pH 4 that finds pH 7

Here is a puzzle from the titration lectures. You titrate hydrochloric acid with sodium hydroxide solution. At the equivalence point, the flask holds only sodium chloride solution, so its pH is 7 (at 25 °C). Yet the lecture on the technique said that methyl orange works for this titration. Methyl orange turns orange at a pH of about 4, far from 7.

So the indicator changes colour while the solution is still clearly acidic. Surely the burette reading must then be too small? Commit to a guess before you read on.

Predict — how far off is the end point?

25.0 cm³ of 0.100 mol/dm³ hydrochloric acid is titrated with 0.100 mol/dm³ sodium hydroxide, so the equivalence point is at 25.00 cm³. Methyl orange finishes changing colour at about pH 4.4. How early will the end point come?

This figure zooms in on the titration of 25.0 cm³ of 0.100 mol/dm³ hydrochloric acid with 0.100 mol/dm³ sodium hydroxide, between 24.0 and 26.0 cm³.

One drop from a burette is about 0.05 cm³. One drop before the equivalence point, the pH is 4.00. One drop after it, the pH is 10.00. Methyl orange finishes its change at pH 4.4, which the curve reaches at about 24.98 cm³: less than one drop early.

So the indicator does not have to change at the pH of the equivalence point. It only has to change somewhere on the near-vertical part of the curve. This lecture explains where that steep part comes from, why its position changes when the acid or the base is weak, and how an indicator works in the first place. It also keeps two promises. The titration lecture said that a later lecture would explain why a weak acid moves the jump up and a weak base moves it down. The buffers lecture said that the flat middle part of some titration curves is a buffer forming in the flask.

An indicator is a weak acid with two colours

An indicator is a weak acid, HIn, whose molecule has one colour while its conjugate base, In⁻, has another.

HIn(aq) ⇌ H⁺(aq) + In⁻(aq)

Use Le Chatelier's principle. In an acidic solution, [H⁺] is high, so the position of equilibrium lies to the left. Most of the indicator is HIn, and you see the first colour. In an alkaline solution, OH⁻ removes H⁺, so the position lies to the right. Most of it is In⁻, and you see the second colour. (For some indicators the two forms carry different charges, but the idea is exactly the same.)

Now use the buffers lecture. The indicator is a weak acid, so it has its own acid dissociation constant, written K_In. The same steps that gave the Henderson–Hasselbalch equation give:

pH = p K_{In} + lo g_{10} \frac{[ In ^{-} ]}{[ HIn ]}

When pH = pK_In, the log is zero, so the two forms are equal and you see the in-between colour. When the pH is one unit higher, $[In^{-}] \div [HIn] = 10$ . Your eye then sees almost only the In⁻ colour. When it is one unit lower, the ratio is $\frac{1}{10}$ , and you see almost only the HIn colour.

That is it. That is the whole idea. An indicator changes colour over about two pH units, centred on its pK_In. Outside that range, the solution can go on changing, and the indicator shows nothing more.

Indicator	Colour of HIn	Colour of In⁻	pK_In	Changes between pH
methyl orange	red	yellow	3.7	3.2 and 4.4
methyl red	red	yellow	5.1	4.2 and 6.3
phenolphthalein	colourless	pink	9.3	8.2 and 10.0

(pK_In values and the methyl red range from Chemguide. Tables differ by about 0.1 in the ranges. The ranges are not exactly pK_In ± 1, because the eye sees some colours more easily than others: a faint pink in a colourless solution is easy to notice.)

Stop and think

A few drops of methyl orange are added to a solution, and it turns yellow. Is the solution alkaline?

Show the answer

Not necessarily. Methyl orange is yellow at every pH above about 4.4. A solution of pH 5 is acidic, and methyl orange is yellow in it. So is a solution of pH 13. An indicator tells you only which side of its own range the pH is on. That is why litmus, which changes near pH 7, is used to test whether something is acidic or alkaline, while methyl orange is used for titrations.

Why the curve has a cliff

A titration curve (or pH curve) shows the pH in the flask against the volume added from the burette. In this lecture, the acid is always in the flask and the base is added, so every curve rises.

Worked Example 1 — The pH just before and just after

25.0 cm³ of 0.100 mol/dm³ HCl is titrated with 0.100 mol/dm³ NaOH. Find the pH after 24.95 cm³ and after 25.05 cm³ of NaOH, one drop either side of the equivalence point.

Think first. Before the equivalence point, some HCl is left over. After it, some NaOH is left over. Find the leftover moles, then divide by the total volume.

Moles of HCl at the start: $0.100 \times \frac{25.0}{1000} = 2.50 \times 1 0^{- 3}$ mol.

At 24.95 cm³: moles of NaOH added $= 0.100 \times \frac{24.95}{1000} = 2.495 \times 1 0^{- 3}$ mol. Acid left over $= 5.0 \times 1 0^{- 6}$ mol, in $25.0 + 24.95 = 49.95$ cm³. So $[H^{+}] = 5.0 \times 1 0^{- 6} \div 0.04995 \approx 1.0 \times 1 0^{- 4}$ mol/dm³, and pH ≈ 4.00.

At 25.05 cm³: NaOH left over $= 5.0 \times 1 0^{- 6}$ mol in 50.05 cm³, so $[OH^{-}] \approx 1.0 \times 1 0^{- 4}$ mol/dm³. Then $[H^{+}] = K_{w} \div [OH^{-}] \approx 1.0 \times 1 0^{- 10}$ mol/dm³, and pH ≈ 10.00.

Check. Two drops moved the pH by 6 units. ✓ The two answers are symmetrical about pH 7, because the leftover amounts are equal. ✓

Why such a cliff? Because pH is a logarithm. Each time the leftover acid falls to a tenth, the pH rises by only one unit. Early in the titration, the leftover acid falls slowly: after 20 cm³ of NaOH, a fifth of the acid is still there, and the pH is only 1.95. Near the end, the leftover acid falls by a factor of ten with every tiny addition. From 1.0 cm³ left (pH ≈ 2.7) to 0.1 cm³ left (pH ≈ 3.7) to one drop left (pH 4.00), the steps get shorter and shorter in volume. Then the last trace of acid goes, and one more drop of alkali takes the pH straight to 10.

A weak acid: a buffer forms in the flask

Now titrate 25.0 cm³ of 0.100 mol/dm³ ethanoic acid, a weak acid, with the same sodium hydroxide. The curve changes in three places.

Ethanoic acid (solid) and hydrochloric acid (dashed), each 25.0 cm³ of 0.100 mol/dm³, with 0.100 mol/dm³ NaOH. Curves calculated from the equilibrium constants (K_a = 1.74 × 10⁻⁵ mol/dm³). After the equivalence point, the two curves are the same.

It starts higher. Ethanoic acid is only slightly dissociated, so its starting pH is about 2.88, not 1.00. (You met this number as a preview in the pH lecture.)

It has a flat middle. As soon as some NaOH is added, it turns some ethanoic acid into ethanoate ions: CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l). The flask now holds a weak acid and its conjugate base. That is exactly a buffer, made by part-neutralising a weak acid, as the buffers lecture described. So the pH changes slowly here.

Halfway to the equivalence point, at 12.50 cm³, half the acid has become ethanoate. Then [CH₃COO⁻] = [CH₃COOH], the log term in the Henderson–Hasselbalch equation is zero, and pH = pK_a = 4.76. This half-neutralisation point gives a way to measure pK_a with a pH meter.

Its equivalence point is alkaline, at about pH 8.73. At the equivalence point, the flask holds sodium ethanoate solution. The ethanoate ion is a base, the conjugate base of a weak acid. It takes a proton from a few water molecules:

CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)

This makes a little OH⁻, so the solution is slightly alkaline. That is the reason the titration lecture promised: the salt of a weak acid and a strong base is not neutral. Calculating the 8.73 exactly needs a Pre-University method. The direction, alkaline, follows from the conjugate base alone.

The steep part is now shorter and higher. One drop before the equivalence point the pH is about 7.46, and one drop after it is 10.00. Phenolphthalein, 8.2 to 10.0, changes right on it. Methyl orange is useless here. It changes between pH 3.2 and 4.4, which this curve passes at about 0.5 to 7.6 cm³, long before the equivalence point.

Worked Example 2 — Measuring pKa from a curve

A student titrates 25.0 cm³ of an unknown weak acid HX with NaOH, using a pH meter. The steep part is centred at 18.40 cm³. At 9.20 cm³, the meter reads 3.85. Find pK_a and K_a for HX.

Think first. 9.20 cm³ is exactly half of 18.40 cm³. So this is the half-neutralisation point, where [X⁻] = [HX].

At that point, pH = pK_a, so pK_a = 3.85. Then $K_{a} = 1 0^{- 3.85} \approx 1.4 \times 1 0^{- 4}$ mol/dm³.

Check. pK_a 3.85 is smaller than ethanoic acid's 4.76, so HX is a stronger acid, with a K_a about eight times larger ( $1.4 \times 1 0^{- 4} \div 1.74 \times 1 0^{- 5} \approx 8.1$ ). ✓ The concentrations of HX and NaOH are not needed: only the halfway volume matters. ✓

A weak base, and two weak partners

Now keep hydrochloric acid in the flask and add 0.100 mol/dm³ ammonia, a weak base, instead of sodium hydroxide. Before the equivalence point, nothing changes: the leftover HCl controls the pH, as before. At the equivalence point, the flask holds ammonium chloride. The ammonium ion is a weak acid, the conjugate acid of ammonia:

NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq)

So the equivalence point is acidic, at about pH 5.28. After it, the extra ammonia and the ammonium ions make a buffer, so the pH rises only slowly towards about 9.25, the pK_a of the ammonium ion. The steep part runs from about pH 4.00 one drop before to 6.55 one drop after. Methyl orange and methyl red both change inside it. Phenolphthalein would not start to turn pink until about 27.2 cm³, more than 2 cm³ too late.

Finally, ethanoic acid with ammonia: both partners weak. There is no cliff at all. One drop before the equivalence point the pH is about 6.93, and one drop after it is about 7.08. An indicator would change gradually over several cm³, so no simple indicator gives a sharp end point. Chemists use a pH meter instead, and take the equivalence point from the middle of the steepest part of the curve.

Choosing an indicator

Choose an indicator whose whole colour-change range lies on the steep part of the titration curve. Then the end point falls within a drop of the equivalence point.

The steep part sits on the side of the strong partner: alkaline for a weak acid with a strong base (phenolphthalein), acidic for a strong acid with a weak base (methyl orange or methyl red). Both work for strong with strong. None works for weak with weak.

Test the rule on all four pairs below.

Graph builder — titration curves and indicators

25.0 cm³ of 0.100 mol/dm³ acid is in the flask. 0.100 mol/dm³ base is added from the burette. Choose the acid, the base and an indicator. Then move the burette. The − and + buttons add or remove one drop (0.05 cm³).

Acid:

Base:

Indicator:

Volume of base added: 20.00 cm³

This table shows, for each pair, the pH one drop before and one drop after the equivalence point (25.00 cm³), and the volumes over which each indicator changes colour.

Acid + base	pH at 24.95 → 25.05 cm³	Equivalence pH	Methyl orange changes	Phenolphthalein changes
HCl + NaOH	4.00 → 10.00	7.00	24.69–24.98 cm³ ✓	25.00–25.05 cm³ ✓
ethanoic acid + NaOH	7.46 → 10.00	8.73	0.51–7.59 cm³ ✗	24.99–25.05 cm³ ✓
HCl + ammonia	4.00 → 6.55	5.28	24.69–24.98 cm³ ✓	starts at 27.2 cm³ ✗
ethanoic acid + ammonia	6.93 → 7.08	7.00	0.51–7.59 cm³ ✗	starts at 27.2 cm³ ✗

An indicator works when its whole change happens within a drop or two of 25.00 cm³. Methyl red (4.2–6.3) works for HCl with either base: its whole change happens within 0.03 cm³ of 25.00 cm³.

You should have found that the same indicator can be excellent in one titration and useless in another. The question is never "which indicator is best?" It is "where is this curve steep, and which indicator changes there?"

Worked Example 3 — Ammonia solution in a cleaning product

A student wants to find the concentration of ammonia in a household cleaner by titrating it with standard hydrochloric acid. Which indicator should she use: methyl orange, methyl red or phenolphthalein?

Think first. Ammonia is a weak base and hydrochloric acid is strong. Where is the steep part?

At the equivalence point, the flask holds ammonium chloride. The ammonium ion is a weak acid, so the equivalence point is acidic, about pH 5.3 for 0.100 mol/dm³ solutions. The steep part runs from about pH 4 to about 6.5. Methyl orange (3.2–4.4) or methyl red (4.2–6.3) changes on it. Phenolphthalein (8.2–10.0) does not.

Check. The steep part is on the side of the strong partner, the acid, so it is acidic. ✓ This does not depend on which solution is in the burette. If the acid is added to the ammonia, the curve falls instead of rising, but its steep part is in the same place. ✓

Three traps

Trap 1 — "The equivalence point is always at pH 7"

Only for a strong acid with a strong base, and only at 25 °C. With a weak acid, the salt's conjugate base makes the equivalence point alkaline (about 8.7 for ethanoic acid). With a weak base, the conjugate acid makes it acidic (about 5.3 for ammonia).

Trap 2 — "The indicator must change at the pH of the equivalence point"

It must change on the steep part of the curve, which can be several pH units wide. Methyl orange changes near pH 4 and still finds the HCl–NaOH equivalence point within a drop. What matters is the volume at which the colour changes, not the pH.

Trap 3 — Reading pKa from the wrong point

pH = pK_a at half the equivalence volume, where half the acid has been neutralised. It is not the pH at the equivalence point, and it is not halfway up the pH axis.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Bromothymol blue has pK_In 7.0 and changes between pH 6.0 and 7.6 (yellow to blue). Using the numbers in this lecture, would it be a good indicator for (a) HCl with NaOH, (b) HCl with ammonia?

Hint 1
Where is the steep part of each curve? Use the pH one drop before and one drop after the equivalence point.

Hint 2
(a) 4.00 → 10.00. (b) 4.00 → 6.55. Is the whole range 6.0–7.6 inside each?

Full answer
(a) Yes. The steep part runs from pH 4.00 to 10.00 within two drops, and 6.0–7.6 lies inside it. In fact the whole change happens within about 0.01 cm³ of 25.00 cm³. (b) No. The steep part only reaches about pH 6.55, then the ammonia buffer makes the pH rise slowly. Bromothymol blue would finish changing at about 25.6 cm³, more than ten drops late. Check: the equivalence point for (b) is acidic (pH ≈ 5.3), so an indicator that changes at 7 is on the wrong side. ✓
25.0 cm³ of 0.100 mol/dm³ HCl has 20.0 cm³ of 0.100 mol/dm³ NaOH added. Find the pH.

Hint 1
Find the moles of HCl at the start and the moles of NaOH added.

Hint 2
The leftover H⁺ is in a total volume of 45.0 cm³.

Full answer
HCl: $2.50 \times 1 0^{- 3}$ mol. NaOH: $0.100 \times \frac{20.0}{1000} = 2.00 \times 1 0^{- 3}$ mol. Leftover H⁺: $5.0 \times 1 0^{- 4}$ mol. Volume: 45.0 cm³ = 0.0450 dm³. $[H^{+}] = 5.0 \times 1 0^{- 4} \div 0.0450 \approx 1.11 \times 1 0^{- 2}$ mol/dm³, so pH $= - lo g_{10} (1.11 \times 1 0^{- 2}) \approx 1.95$ . Check: 80% of the acid has been neutralised, yet the pH has only risen from 1.00 to 1.95, which is why the start of the curve is so flat. ✓
A weak acid is titrated with NaOH. The equivalence point is at 22.40 cm³. At 11.20 cm³, the pH is 4.20. (a) Find pK_a and K_a. (b) Is the acid stronger or weaker than ethanoic acid? (c) Which indicator would you use, and why?

Hint 1
Compare 11.20 cm³ with 22.40 cm³.

Hint 2
At the half-neutralisation point, pH = pK_a. Then $K_{a} = 1 0^{- p K_{a}}$ .

Full answer
(a) 11.20 cm³ is half of 22.40 cm³, so pK_a = 4.20 and $K_{a} = 1 0^{- 4.20} \approx 6.3 \times 1 0^{- 5}$ mol/dm³. (b) Stronger: its pK_a is smaller than 4.76 (K_a larger than 1.74 × 10⁻⁵). (c) Phenolphthalein. It is a weak acid with a strong base, so the salt's conjugate base makes the equivalence point alkaline, and the steep part lies in the alkaline region. Check: methyl orange (3.2–4.4) would change between roughly 2 and 14 cm³, around the half-neutralisation point, long before 22.40 cm³. ✓
A stretch. A student titrates ethanoic acid with ammonia solution, using a pH meter, and records pH 6.93 at 24.95 cm³ and pH 7.08 at 25.05 cm³. (a) Explain why no simple indicator would give a sharp end point. (b) How can she still find the equivalence volume from her readings? (c) Why is the pH at the equivalence point close to 7 here, even though neither partner is strong?

Hint 1
(a) Compare the pH change over two drops with the width of an indicator's range.

Hint 2
(c) Compare the pK_a of ethanoic acid (4.76) with the pK_a of the ammonium ion (9.25). How far is each from 7?

Full answer
(a) Two drops change the pH by only 0.15. An indicator needs about 1.5 to 2 pH units to change colour completely, so its change would be spread over several cm³, and the end point would be vague. (b) Plot pH against volume and take the equivalence point at the middle of the steepest part, where the pH changes fastest per drop. (c) At the equivalence point, the flask holds ammonium ethanoate. The ethanoate ion makes a little OH⁻, and the ammonium ion makes a little H⁺. The two pK_a values, 4.76 and 9.25, are almost the same distance from 7 (2.24 and 2.25), so the two effects nearly cancel. Check: the curve passes pH 7.00 at 25.00 cm³. ✓

The bigger idea

The first indicators came from plants. In 1664, Robert Boyle described how syrup of violets, a blue syrup used as a medicine, turned red with acids and green with alkalis. He used such colour changes to sort substances into acids and alkalis. (Source: R. Boyle, Experiments and Considerations Touching Colours, 1664.) Today a pH meter gives a number instead of a colour. But for a single titration, a well-chosen indicator is still fast, cheap and reliable.

Look back at one titration curve and see how much of this topic it holds. The cliff comes from pH being a logarithm. The flat middle is a buffer, and its halfway point gives pK_a. The position of the equivalence point comes from conjugate acids and bases, from the first lecture on acid–base theory. And the indicator itself is one more weak acid in equilibrium, with its own pK_In. That completes the Upper Secondary lectures on acids and bases. The Pre-University lectures return to weak acids, and calculate the pH values that this lecture could only quote.

Key concepts in this lesson

An indicator is a weak acid HIn whose molecule and conjugate base In⁻ have different colours.
An indicator changes colour over about two pH units, centred on its pK_In.
A titration curve rises steeply near the equivalence point because pH is a logarithm, so the last traces of acid make the largest pH changes.
A good indicator changes colour entirely on the steep part of the curve, so the end point falls within a drop of the equivalence point.
Titrating a weak acid with a strong base forms a buffer in the flask, and at the half-neutralisation point pH = pK_a.
The equivalence point is alkaline for a weak acid with a strong base, and acidic for a strong acid with a weak base, because of the salt's conjugate base or conjugate acid.