Physical Chemistry / Quantitative and practical chemistry
Titrations
Acid-base and redox titrations: standard solutions, indicators, end points and the calculations that follow.
GCSEA-LevelIn a nutshell
A titration is a precise way to find out how much of a substance is in a solution. You slowly add one solution, from a burette, to a measured volume of another until the reaction between them is just complete. That moment, the end point, is usually shown by an indicator changing colour. If you know the concentration of one solution and the volumes used, a short calculation with moles gives the unknown concentration. Titrations check the acidity of vinegar and wine, the amount of vitamin C in fruit juice, the iron content of tablets and the purity of medicines. Done carefully, with repeated runs that agree closely, a titration can be accurate to better than one part in a hundred using simple glassware.
A titration measures exactly what volume of one solution reacts with a known volume of another.
Method (acid–alkali)
1. Use a pipette to measure 25.0 cm^3 of alkali into a conical flask. 2. Add a few drops of indicator, such as phenolphthalein (pink in alkali, colourless in acid) or methyl orange (yellow in alkali, red in acid). 3. Fill a burette with acid and record the starting volume. 4. Add acid while swirling the flask, slowly and drop by drop near the end, until the indicator just changes colour: the end point. 5. Record the final volume. The difference is the titre. 6. Repeat until titres are concordant (within 0.10 cm^3) and average them.
Universal indicator is not used because its gradual colour change gives no sharp end point.
Calculation
25.0 cm^3 of sodium hydroxide is neutralised by 20.0 cm^3 of 0.100 mol dm^{-3} hydrochloric acid.
- Moles of HCl = 0.100 \times 20.0/1000 = 0.00200 mol.
- \mathrm{NaOH + HCl \rightarrow NaCl + H_2O} is a 1:1 ratio, so 0.00200 mol NaOH.
- Concentration = 0.00200 \div (25.0/1000) = 0.0800 mol dm^{-3}.
Remember to divide volumes in cm^3 by 1000 to convert them to dm^3.
Standard solutions
A standard solution has an accurately known concentration. To make one, weigh the solid accurately (by difference), dissolve it in a beaker of distilled water, transfer with washings to a volumetric flask, make up to the mark so the bottom of the meniscus sits on the line, and invert to mix. Primary standards such as sodium carbonate or potassium hydrogen phthalate are pure, stable and not hygroscopic.
Acid–base titrations
Choose an indicator whose colour change lies within the vertical section of the titration curve: phenolphthalein for weak acid–strong base, methyl orange for strong acid–weak base, either for strong–strong.
Redox titrations
- Manganate(VII): \mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}. It is self-indicating: the end point is the first permanent pale pink. Acidify with dilute sulfuric acid, not hydrochloric acid, which would be oxidised to chlorine.
- Iodine–thiosulfate: an oxidant liberates I_2 from excess iodide, then \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}}. Add starch near the end (when the solution is pale yellow); the blue-black colour disappears at the end point.
Worked example
An iron tablet is dissolved in acid and titrated with 0.0200 mol dm^{-3} KMnO_4; the mean titre is 18.50 cm^3. n(\mathrm{MnO_4^-}) = 0.0200 \times 0.01850 = 3.70\times10^{-4} mol, so n(\mathrm{Fe^{2+}}) = 5 \times 3.70\times10^{-4} = 1.85\times10^{-3} mol, which is 1.85\times10^{-3} \times 55.8 = 0.103 g of iron.
Back titrations
When a sample is insoluble or reacts slowly (calcium carbonate in limestone, say), add a known excess of reagent, then titrate the leftover. Reagent used by the sample = amount added − amount left.
Uncertainty
A burette reading is ±0.05 cm^3, and a titre involves two readings, so ±0.10 cm^3. Percentage uncertainty = 0.10/\text{titre} \times 100; larger titres reduce it.
Key points
- A titration finds the volume of one solution that exactly reacts with a known volume of another.
- Repeat until titres are concordant (within 0.10 cm^3) and average those.
- Moles = c \times V/1000 with V in cm^3; use the balanced equation's ratio.
- Choose an indicator whose colour change falls in the steep part of the titration curve.
- Manganate(VII) is self-indicating; iodine–thiosulfate uses starch near the end point.
Check yourself
5 questions
1Why is universal indicator not used in titrations?
A titration needs one sharp colour change at the end point; universal indicator shows a range of colours over the whole pH scale.
2Titres of 24.50, 24.10, 24.15 and 24.05 cm^3 are recorded. What mean titre should be used?
The first (rough) titre is not concordant; the other three agree within 0.10 cm^3, and their mean is 24.10 cm^3.
325.0 cm^3 of H_2SO_4 is neutralised by 30.0 cm^3 of 0.200 mol dm^{-3} NaOH. What is the acid's concentration?
n(\mathrm{NaOH}) = 0.00600 mol; \mathrm{H_2SO_4 + 2NaOH} means 0.00300 mol acid in 0.0250 dm^3, giving 0.120 mol dm^{-3}.
4Why is a manganate(VII) titration acidified with sulfuric acid rather than hydrochloric acid?
MnO_4^- can oxidise Cl^- to Cl_2, so the titre would be too large and the result wrong.
5A titre of 12.50 cm^3 is read from a burette (each reading ±0.05 cm^3). What is the percentage uncertainty in the titre?
Two readings give ±0.10 cm^3, and 0.10/12.50 \times 100 = 0.8\%.
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