Acid-Base Titration Simulation

Titration Setup

0.0 cm³
Volume Added
0.0 cm³
Current pH
13.0
Equivalence Point
25.0 cm³
Indicator Colour
Pink
Balanced Equation
HCl + NaOH → NaCl + H₂O

Titration Curve (pH vs Volume)

Key Concepts:
- The equivalence point is where moles of acid = moles of base
- The endpoint is where the indicator changes colour
- For strong acid/strong base, equivalence point is at pH 7
- The steep rise in pH near the equivalence point is characteristic of titrations

What is a titration?

A titration is a method for finding the unknown concentration of a solution by reacting it with a solution of known concentration, added a little at a time until the reaction is exactly complete.

In an acid–base titration the reaction is neutralisation:

acid + alkali → salt + water

Two terms are constantly confused, and examiners test the difference:

How to use this simulation

  1. Choose your acid — HCl or H₂SO₄, both 0.1 mol/dm³ — and your alkali, NaOH or KOH, 25 cm³ of 0.1 mol/dm³.
  2. Pick an indicator. Try all three; the difference between them is the point of this exercise.
  3. Add acid with the slider, or use Add Drop (0.5 cm³) when you get close to the colour change — exactly as you would slow down near the endpoint at the bench.
  4. Watch Current pH and the titration curve as you go. Notice how little happens for a long time, then how violently pH swings over one or two drops.
  5. Export Data (CSV) gives you the pH-volume pairs to plot yourself.

Try this first. Run HCl against NaOH and note the volume at equivalence. Now switch to H₂SO₄ and predict the new volume before you add anything. If your prediction was 25 cm³, look at the balanced equation again — the answer is in the mole ratio, and it is one of the most common places marks are lost.

Choosing the right indicator

An indicator is only suitable if it changes colour within the steep, near-vertical part of the pH curve. Outside that range, the colour changes gradually and you cannot identify a sharp endpoint.

IndicatorIn acidIn alkaliBest used for
PhenolphthaleinColourlessPinkStrong acid–strong alkali; weak acid–strong alkali
Methyl orangeRedYellowStrong acid–strong alkali; strong acid–weak alkali
Universal indicatorFull colour range, changing graduallyNot suitable

Why universal indicator is wrong for titrations — and this is worth a mark on its own — is that it changes colour gradually across a wide pH range. There is no single drop at which it flips, so you cannot judge an endpoint. Select it in the simulation and watch: the colour drifts rather than snapping. Universal indicator is for estimating pH, not for titrating.

Since this simulation uses strong acids with strong alkalis, the equivalence point sits at pH 7 and both phenolphthalein and methyl orange work.

The real apparatus, and why each detail matters

The rinsing rules, which come up repeatedly:

Concordant results

One titration is never enough. The standard procedure:

  1. Do a quick rough (trial) titration to find the approximate endpoint.
  2. Repeat carefully, adding dropwise as you approach that volume.
  3. Continue until you have results that are concordant — agreeing within 0.10 cm³.
  4. Calculate the mean titre from the concordant results only. The rough titration is excluded, and so is any anomalous result.

Being asked why the rough titration is left out is common: it is less accurate, because the acid was added quickly rather than dropwise near the endpoint.

The calculation

Every acid–base titration calculation is the same three steps.

Worked example — 25.0 cm³ of NaOH needed 25.0 cm³ of 0.100 mol/dm³ HCl. Find the concentration of the NaOH.

HCl + NaOH → NaCl + H₂O

  1. Moles of the known substance: n = c × V ÷ 1000 = 0.100 × 25.0 ÷ 1000 = 2.50 × 10⁻³ mol HCl
  2. Use the mole ratio. It is 1 : 1, so moles of NaOH = 2.50 × 10⁻³ mol
  3. Concentration of the unknown: c = n ÷ (V ÷ 1000) = 2.50 × 10⁻³ ÷ 0.0250 = 0.100 mol/dm³

To convert to grams per dm³, multiply by the relative formula mass: 0.100 × 40 = 4.0 g/dm³ for NaOH.

The sulfuric acid case. H₂SO₄ is diprotic — each molecule provides two H⁺ ions:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The ratio is now 1 : 2, so neutralising the same 2.50 × 10⁻³ mol of NaOH needs only 1.25 × 10⁻³ mol of H₂SO₄ — half the volume, 12.5 cm³. Step 2 is where this is either caught or lost.

Common exam mistakes

Exam-style questions

1. State the difference between the endpoint and the equivalence point. [2 marks]

The equivalence point is where the moles of acid and alkali are exactly in the ratio of the balanced equation (1). The endpoint is where the indicator changes colour, which is the observed approximation to it (1).

2. A student rinsed the conical flask with sodium hydroxide solution before starting. Explain the effect on the titre. [3 marks]

Extra moles of alkali are present in the flask (1), so more acid is needed to neutralise it (1), giving a titre that is too high and a calculated concentration that is too high (1).

3. 25.0 cm³ of KOH required 22.40 cm³ of 0.150 mol/dm³ HCl. Calculate the concentration of the KOH. [3 marks]

Moles HCl = 0.150 × 22.40 ÷ 1000 = 3.36 × 10⁻³ mol (1). Ratio HCl : KOH is 1 : 1, so moles KOH = 3.36 × 10⁻³ mol (1). c = 3.36 × 10⁻³ ÷ 0.0250 = 0.134 mol/dm³ (1).

4. Explain why universal indicator is unsuitable for a titration. [2 marks]

It changes colour gradually over a wide pH range (1), so there is no sharp colour change at the endpoint and the exact volume cannot be identified (1).

5. A student obtained titres of 24.10, 23.65, 23.70 and 23.60 cm³. Calculate the mean titre they should use. [2 marks]

The concordant results — those within 0.10 cm³ of each other — are 23.65, 23.70 and 23.60 (1). 24.10 is the rough/anomalous result and is excluded. Mean = (23.65 + 23.70 + 23.60) ÷ 3 = 23.65 cm³ (1).