Topic 14 · page 9 of 18

Reading a titration curve

A titration curve plots pH against the volume added. Its start, its steep section and its end all come from the amounts of acid and base, and you can calculate any point on it.

14 min read14.15

Read first: Strong and weak acids, Water and strong bases

What the curve shows 14.15

In a titration you add one solution from a burette to a measured volume of another in a flask. If you record the pH after each addition and plot pH against the volume added, you get a titration curve.

The first curve to learn is the simplest: 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid in the flask, with 0.100 mol dm⁻³ sodium hydroxide added from the burette.

A titration curve for 25.0 cm³ of 0.100 mol per cubic decimetre hydrochloric acid with 0.100 mol per cubic decimetre sodium hydroxide. The pH starts at 1.00, rises slowly, then jumps from about pH 3 to pH 11 around 25.0 cm³. The equivalence point is at 25.0 cm³ and pH 7.00. The curve then levels off near pH 12.5.
Strong acid with strong base. Computed from the amounts present, 0.100 mol per cubic decimetre each.

The curve has four parts you can name.

  1. The start. The pH is set by the solution in the flask alone: 0.100 mol dm⁻³ HCl has pH 1.00.
  2. The slow rise. The acid is being used up, but a lot remains, so the pH creeps up.
  3. The vertical section. Close to equivalence a single drop changes the pH by several units. This is the part an indicator needs to fall inside.
  4. The finish. After equivalence the pH is set by the excess titrant, and flattens towards the pH of the burette solution.

Calculating points on the curve 14.15

Every point comes from the moles left over. Find the moles in excess, divide by the total volume, and then use the strong acid or base rules from earlier pages.

Worked exampleBefore equivalence

Calculate the pH after 20.0 cm³ of 0.100 mol dm⁻³ NaOH is added to 25.0 cm³ of 0.100 mol dm⁻³ HCl.

  1. Moles of H⁺ at the start = 0.100 × 0.0250 = 2.50 × 10⁻³ mol
  2. Moles of OH⁻ added = 0.100 × 0.0200 = 2.00 × 10⁻³ mol
  3. Moles of H⁺ left = 5.0 × 10⁻⁴ mol. Total volume = 45.0 cm³ = 0.0450 dm³.
  4. [H⁺] = 5.0 × 10⁻⁴ ÷ 0.0450 = 0.0111 mol dm⁻³
  5. pH = −lg (0.0111) = 1.95

Worked exampleAfter equivalence

Calculate the pH after 30.0 cm³ of the same NaOH is added.

  1. Moles of OH⁻ added = 0.100 × 0.0300 = 3.00 × 10⁻³ mol, so OH⁻ left = 3.00 × 10⁻³ − 2.50 × 10⁻³ = 5.0 × 10⁻⁴ mol.
  2. Total volume = 55.0 cm³ = 0.0550 dm³, so [OH⁻] = 5.0 × 10⁻⁴ ÷ 0.0550 = 9.09 × 10⁻³ mol dm⁻³.
  3. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 9.09 × 10⁻³ = 1.10 × 10⁻¹² mol dm⁻³
  4. pH = 11.96

The pH at a few more volumes shows how sharp the change is.

Volume of NaOH added / cm³ pH
0 1.00
10.0 1.37
20.0 1.95
24.0 2.69
24.9 3.70
25.0 7.00
25.1 10.30
26.0 11.29
30.0 11.96
50.0 12.52

Between 24.9 and 25.1 cm³, 0.2 cm³ of alkali, the pH jumps from 3.70 to 10.30. The curve is steep because pH is a logarithm. Near the equivalence point the H⁺ that remains is tiny, so a tiny addition changes it by a huge factor, and a factor is what the logarithm responds to.

Finding the equivalence volume 14.15

The equivalence volume comes from the moles, not from the picture:

Veq=moles of acid×(mole ratio base : acid)c(base) V_\text{eq} = \frac{\text{moles of acid} \times (\text{mole ratio base : acid})}{c(\text{base})}

Worked exampleA different concentration

What volume of 0.200 mol dm⁻³ NaOH is needed to reach equivalence with 25.0 cm³ of 0.100 mol dm⁻³ HCl?

  1. Moles of HCl = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. The ratio is 1 : 1, so moles of NaOH = 2.50 × 10⁻³ mol.
  2. Volume = 2.50 × 10⁻³ ÷ 0.200 = 0.0125 dm³ = 12.5 cm³

What to put on a sketch 14.15

An exam question that asks you to sketch a curve is awarding marks for features, not artistic skill.

Check yourself

  1. State the difference between the equivalence point and the end point of a titration.
  2. Calculate the volume of 0.100 mol dm⁻³ NaOH needed to reach equivalence with 25.0 cm³ of 0.150 mol dm⁻³ HCl.
  3. 25.0 cm³ of 0.100 mol dm⁻³ HCl is titrated with 0.100 mol dm⁻³ NaOH. Calculate the pH after 15.0 cm³ of NaOH has been added.
  4. For the same titration, calculate the pH after 35.0 cm³ of NaOH has been added.
  5. Explain why the pH changes very little at the start of the titration but very rapidly near the equivalence point.
  6. 25.0 cm³ of 0.100 mol dm⁻³ NaOH is titrated with 0.100 mol dm⁻³ HCl. List the main features of the curve.

Answers cover these until you have tried the questions

  1. The equivalence point is where the reactants have been added in exactly the ratio of the balanced equation. The end point is where the indicator changes colour. They are the same point only if the indicator is well chosen.
  2. Moles of HCl = 0.150 × 0.0250 = 3.75 × 10⁻³ mol, so moles of NaOH = 3.75 × 10⁻³ mol. Volume = 3.75 × 10⁻³ ÷ 0.100 = 0.0375 dm³ = 37.5 cm³.
  3. H⁺ at the start = 2.50 × 10⁻³ mol; OH⁻ added = 0.100 × 0.0150 = 1.50 × 10⁻³ mol; H⁺ left = 1.00 × 10⁻³ mol. Total volume = 40.0 cm³. [H⁺] = 1.00 × 10⁻³ ÷ 0.0400 = 0.0250 mol dm⁻³. pH = −lg 0.0250 = 1.60
  4. OH⁻ added = 0.100 × 0.0350 = 3.50 × 10⁻³ mol; OH⁻ left = 3.50 × 10⁻³ − 2.50 × 10⁻³ = 1.00 × 10⁻³ mol. Total volume = 60.0 cm³, so [OH⁻] = 1.00 × 10⁻³ ÷ 0.0600 = 0.0167 mol dm⁻³. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0167 = 6.0 × 10⁻¹³ mol dm⁻³. pH = 12.22
  5. At the start there is a large amount of H⁺, so adding a little alkali removes only a small fraction of it and the pH moves slowly. Near equivalence almost all the H⁺ has gone, so a tiny volume of alkali removes most of what is left. [H⁺] changes by a factor of millions, and the pH, being a logarithm, changes by several units.
  6. It starts at pH 13.00 (0.100 mol dm⁻³ strong base), falls slowly, then has a vertical section from about pH 11 to pH 3 around 25.0 cm³ (equivalence), where the pH is 7.00. It then levels off near pH 1.5, set by the excess acid. The axes are pH and volume of acid added.