Every titration so far has quietly assumed you could see the end point. Time to look at how that actually works. The surprise is that an indicator is not some special dye — it is a weak acid, obeying exactly the same rules as ethanoic acid, with one extra feature: its two forms are different colours.
📘 What you need to know
An indicator is a weak acid whose undissociated form HIn and dissociated form In− have different colours.
HIn(aq) ↔ H+(aq) + In−(aq), with colour 1 on the left and colour 2 on the right.
In acid, the extra H+ pushes the equilibrium left, so you see colour 1.
In alkali, OH− removes H+ and pulls it right, so you see colour 2.
The colour you see depends on the ratio [HIn] : [In−], not on a single fixed pH.
At the end point the two are equal, which makes Ka = [H+] and therefore pKa = pH.
The colour change spreads over roughly pKa ± 1, so about two pH units.
Some indicators are weak bases instead: BOH ↔ B+ + OH−, with the colours the other way round.
An indicator is just an equilibrium
Write HIn for the whole indicator molecule, where “In” stands for the rest of it. Dissolve it in water and it behaves like any weak acid.
Now everything follows from Le Chatelier. Drop the indicator into acid and the solution is already full of H+, so the equilibrium is pushed back to the left and nearly all of it stays as HIn. Drop it into alkali and the OH− mops up H+, so the equilibrium shifts right and nearly all of it becomes In−.
The colours shown are methyl orange. The middle beaker is not a separate substance — it is simply what a roughly even mixture of red and yellow forms looks like.
Why the end point gives you pKa
The end point is defined as the moment when the two forms are present in equal amounts, because that is when the colour looks half changed. Put [HIn] = [In−] into the Ka expression and they cancel.
A very useful cancellation
Ka = [H+][In−] ÷ [HIn]
when [In−] = [HIn], Ka = [H+]
so pKa = pH at the end point
You have met this cancellation before. It is exactly what happened at the half-equivalence point of a weak acid titration, for exactly the same reason: the two members of a conjugate pair were present in equal amounts.
Why the change takes two pH units
Your eye is not very good at spotting a minority colour. As a rule of thumb, one form has to outnumber the other by about 10 to 1 before you see it as a pure colour.
A ratio of 10 : 1 corresponds to being one pH unit away from pKa, and 1 : 10 to being one unit the other side. So the colour drifts through its transition across roughly pKa − 1 to pKa + 1.
This is why an indicator has a range rather than a single switching pH. The published range in the data booklet is exactly this two-unit window.
Indicator
Colour in acid
Colour in alkali
pKa
pH range
methyl orange
red
yellow
3.7
3.1 – 4.4
bromophenol blue
yellow
blue
4.2
3.0 – 4.6
methyl red
red
yellow
5.1
4.4 – 6.2
bromothymol blue
yellow
blue
7.0
6.0 – 7.6
phenolphthalein
colourless
pink
9.6
8.3 – 10.0
Check the pattern. Every range is centred on the pKa and is close to two units wide. The real ranges are not perfectly symmetrical, because some colours are easier for the eye to pick out than others — phenolphthalein’s pink shows up against colourless very quickly.
Indicators that are weak bases
Not every indicator is an acid. Some are weak bases, and they work by the same logic with the colours swapped.
A weak base indicator
BOH(aq) ↔ B+(aq) + OH−(aq)
colour 1 ↔ colour 2
Here adding alkali pushes the equilibrium left, so colour 1 appears in alkaline conditions, and colour 2 shows up in acid. That is the opposite way round from an HIn indicator, so read the equation before predicting anything.
Worked examples
WORKED EXAMPLE
An indicator has Ka = 6.3 × 10−5. Estimate the pH range over which it changes colour.
Step 1: find pKapKa = −log10(6.3 × 10−5) = 4.20Step 2: that is the centre of the range
At pH 4.20 the two forms are present in equal amounts.
Step 3: add and subtract one unit4.20 − 1 = 3.20 and 4.20 + 1 = 5.20It changes colour over about pH 3.2 to 5.2say “about” — real ranges shift a little depending on how visible the colours are
WORKED EXAMPLE
Use Le Chatelier’s principle to explain why methyl orange is red in a strongly acidic solution.
Step 1: write the equilibriumHIn(aq) ↔ H+(aq) + In−(aq), red on the left and yellow on the rightStep 2: say what the acid adds
A strongly acidic solution has a high concentration of H+, which is a product of this equilibrium.
Step 3: apply the principle
The system opposes that increase by shifting to the left, using up H+ and converting In− back into HIn.
Almost all of it becomes HIn, the red form, so the solution looks redname the direction of the shift and the form that results — both usually carry a mark
WORKED EXAMPLE
An indicator has Ka = 1.0 × 10−5. Deduce which colour it shows in a solution of pH 3.00.
Step 1: find [H+] in the solution[H+] = 10−3.00 = 1.0 × 10−3 mol dm−3Step 2: rearrange Ka to get the ratio[In−] ÷ [HIn] = Ka ÷ [H+] = 1.0 × 10−5 ÷ 1.0 × 10−3= 0.010, which is 1 : 100Step 3: decide what dominates
HIn outnumbers In− a hundred to one, far past the ten to one needed.
It shows colour 1, the acid colourthe pH is 2 units below the pKₐ of 5.00, so it is well outside the range
💡 Exam tip
Always write the HIn equilibrium before explaining anything. Nearly every mark scheme starts there.
Use pKa ± 1 to estimate a range, and say “approximately” — the booklet values are not perfectly symmetrical.
Do not say the indicator “reacts with the acid”. It shifts its own equilibrium position.
The indicator must be added in a few drops only. Add too much and it starts changing the pH it is supposed to measure.
Remember which colour goes with which form: HIn is the acid colour, In− the alkaline one.
For a weak base indicator the colours are reversed, so check the equation you are given.
⚠ Common mix-up
Thinking an indicator switches at one exact pH. It drifts across roughly two units.
Confusing the end point with the equivalence point. The end point belongs to the indicator; the equivalence point belongs to the chemistry.
Saying the indicator changes colour “because it is neutralised”. It is an equilibrium shift, not a neutralisation.
Getting the colours backwards. HIn is what survives in acid, so HIn carries the acid colour.
Forgetting that In− is a conjugate base. That is why alkali removes H+ and shifts the equilibrium right.
Assuming any indicator suits any titration. Matching them up is the whole of the next page.
Up next: Choosing an Acid–Base Indicator. You now know each indicator has a two-unit window. The next page lines those windows up against the curves from earlier and shows which pairings actually work.
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