What is titration?
Titration is a way to find how strong (concentrated) a solution is. We react it with another solution whose strength we already know.
- Titrant: the solution of known concentration, kept in the burette.
- Analyte: the solution of unknown concentration, measured with a pipette into a conical flask.
- Indicator: a dye that changes colour when the reaction is complete.
The end point is the moment we see the colour change and stop adding titrant. The equivalence point is the moment when exactly enough titrant has been added to react with all the analyte. A good indicator makes the end point fall almost on the equivalence point.
How to do a titration (step by step)
- Rinse the burette with the titrant and the pipette with the analyte (not with water, which would dilute them).
- Fill the burette, remove air bubbles from the tip, and read the start volume at eye level, at the bottom of the curved surface (meniscus).
- Pipette a fixed volume (for example 25.0 mL) of analyte into a conical flask and add 2–3 drops of indicator.
- Add titrant while swirling. Near the end, add it drop by drop.
- Stop at the first lasting colour change. Read the final volume. The difference is the titre.
- Repeat until two titres agree within 0.10 mL (concordant results) and take their average.
Calculations: moles, ratio and C₁V₁ = C₂V₂
Use three steps:
- Moles of titrant used: n = C × V (V in litres).
- Use the balanced equation to find moles of analyte (the mole ratio).
- Concentration of analyte: C = n ÷ V.
For a 1 : 1 reaction such as HCl + NaOH → NaCl + H₂O, this becomes C₁V₁ = C₂V₂. For H₂SO₄ + 2NaOH, the acid needs twice as many moles of base, so n(acid) = n(base) ÷ 2.
pH curves and the equivalence point
If a pH meter records the pH after each addition, we get a pH curve.
- Strong acid + strong base: starts near pH 1, a very steep jump from about pH 3 to 11, equivalence at pH 7.
- Weak acid + strong base: starts higher (about pH 3), then a gentle flat part called the buffer zone. Equivalence is above 7 (about 8.7) because the salt formed is slightly basic.
- Strong acid + weak base: equivalence below 7.
In the buffer zone the Henderson–Hasselbalch equation works: pH = pKa + log([A⁻]/[HA]). Half-way to equivalence, [A⁻] = [HA], so pH = pKa. This is how chemists measure pKa.
Choosing an indicator: its colour change range must lie inside the steep part. Phenolphthalein (8.2–10) suits weak acid + strong base; methyl orange (3.1–4.4) suits strong acid + weak base; both work for strong + strong.
Other kinds of titration
- Redox titration: an oxidising agent reacts with a reducing agent. Purple potassium permanganate (MnO₄⁻) turns colourless as it reacts with Fe²⁺, so it is its own indicator. The potential of the solution, given by the Nernst equation, jumps at equivalence and can be followed with electrodes.
- Conductometric titration: a conductivity meter is used instead of a colour. Each ion carries current by a fixed amount (Kohlrausch's law: total conductivity is the sum of each ion's share). In HCl + NaOH, fast H⁺ ions are replaced by slower Na⁺, so conductivity falls, then rises again after equivalence when extra OH⁻ is added. The lowest point is the equivalence point. This works even for coloured solutions.
- Precipitation titration: silver nitrate reacts with chloride to form solid AgCl. It works because AgCl is only very slightly soluble (a small solubility product).
Key formulas and definitions
- n = C × V (V in litres)
- C₁V₁ = C₂V₂ (1 : 1 reactions)
- n(analyte) = n(titrant) × (ratio from equation)
- pH = pKa + log([A⁻]/[HA]) (Henderson–Hasselbalch)
- At half-equivalence: pH = pKa
Worked examples
1. 25.0 mL of HCl needs 20.0 mL of 0.100 M NaOH. Find the concentration of HCl.
n(NaOH) = 0.100 × 0.0200 = 0.00200 mol. Ratio 1 : 1, so n(HCl) = 0.00200 mol. C = 0.00200 ÷ 0.0250 = 0.0800 M.
2. Titres of 24.60, 24.20, 24.25 mL are recorded. What titre should be used?
24.60 is a rough first run. 24.20 and 24.25 agree within 0.10 mL, so the mean is (24.20 + 24.25) ÷ 2 = 24.23 mL (to 2 d.p.).
3. 20.0 mL of H₂SO₄ is neutralised by 32.0 mL of 0.150 M NaOH. Find C(H₂SO₄).
n(NaOH) = 0.150 × 0.0320 = 0.00480 mol. H₂SO₄ + 2NaOH, so n(acid) = 0.00480 ÷ 2 = 0.00240 mol. C = 0.00240 ÷ 0.0200 = 0.120 M.
4. A vinegar sample (10.0 mL) needs 16.0 mL of 0.500 M NaOH. Find the concentration of ethanoic acid in g/L (M = 60 g/mol).
n(NaOH) = 0.500 × 0.0160 = 0.00800 mol = n(acid). C = 0.00800 ÷ 0.0100 = 0.800 M. Mass concentration = 0.800 × 60 = 48 g/L.
5. A weak acid HA is titrated with NaOH. At half the equivalence volume the pH is 4.8. What is its Ka?
At half-equivalence pH = pKa, so pKa = 4.8 and Ka = 10⁻⁴·⁸ ≈ 1.6 × 10⁻⁵.
6. 25.0 mL of Fe²⁺ solution needs 18.0 mL of 0.0200 M KMnO₄. MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Find C(Fe²⁺).
n(MnO₄⁻) = 0.0200 × 0.0180 = 3.60 × 10⁻⁴ mol. Ratio 1 : 5, so n(Fe²⁺) = 1.80 × 10⁻³ mol. C = 1.80 × 10⁻³ ÷ 0.0250 = 0.0720 M.
Common mistakes
- Using mL in n = C × V. Change mL to litres (divide by 1000), or use mL on both sides of C₁V₁ = C₂V₂.
- Forgetting the mole ratio. C₁V₁ = C₂V₂ is only for 1 : 1 reactions such as HCl + NaOH.
- Thinking end point and equivalence point are always the same. The end point is what you see; a poor indicator makes it differ from the true equivalence point.
- Thinking the equivalence point is always pH 7. It is 7 only for strong acid + strong base.