Titrations

(Separate Chemistry only) Titrations are assessed on the Chemistry paper only and are not part of Combined Science.

A titration measures the exact volumes of an acid and an alkali that react together. By reacting a solution of known concentration with a solution of unknown concentration and recording the volumes needed, you can work out the unknown concentration or check the reacting volumes precisely.

The specification limits this to strong acids (hydrochloric, nitric or sulfuric) reacting with strong alkalis. The whole point of the technique is accuracy: readings are taken to a fraction of a cubic centimetre, so small errors matter.

A titration finds the precise volumes of acid and alkali that just neutralise each other. That precision is why special apparatus and a sharp indicator are needed.

This lesson also covers Required Practical 2, and the Higher Tier calculations that turn the measured volumes into a concentration.

Two pieces of glassware give the accuracy a titration needs:

Apparatus	Job	Why it is used
Pipette (+ filler)	Measures a fixed volume, e.g. 25.0 cm³, into the flask	Delivers one accurate, repeatable volume
Burette	Adds the second solution a little at a time	Reads volume delivered to 0.05 cm³ so the titre is precise
Conical flask	Holds the measured solution and indicator	Shape lets you swirl without splashing out
White tile	Sits under the flask	Makes the indicator colour change easy to see

The burette is read at the bottom of the meniscus at eye level, to avoid a parallax error. A pipette filler is used rather than mouth suction for safety. Together the pipette and burette let volumes be measured far more precisely than a measuring cylinder could.

The pipette fixes one accurate volume; the burette measures the variable volume needed to reach the end point. That combination is what makes the result reliable.

A titration needs an indicator that gives a single, sharp colour change at the end point. Universal indicator is not suitable because it changes gradually through many colours, so you cannot see one clear point where the reaction is complete.

Suitable single-change indicators for a strong acid and strong alkali are:

Indicator	Colour in acid	Colour in alkali
Methyl orange	Red	Yellow
Phenolphthalein	Colourless	Pink

For example, if the alkali is in the flask and acid is added from the burette using phenolphthalein, the solution starts pink and turns colourless at the end point when the last drop of acid neutralises the alkali.

Universal indicator changes over a range, so it gives no sharp end point. Choose a single-colour-change indicator such as methyl orange or phenolphthalein for a titration.

Required Practical 2 determines the reacting volumes of a strong acid and a strong alkali by titration. A careful, repeatable method is essential.

Method:

1. Use a pipette and filler to measure exactly 25.0 cm³ of the alkali into a clean conical flask. Add a few drops of a suitable indicator (e.g. methyl orange).
2. Fill the burette with the acid, below eye level, and record the starting reading at the bottom of the meniscus.
3. Add the acid from the burette, swirling the flask constantly. Do a quick rough (trial) titration first to find the approximate end point.
4. Repeat carefully, adding the acid drop by drop near the end point, until the indicator just changes colour permanently. Record the final burette reading; the volume added is the titre.
5. Repeat until you have concordant titres (within 0.10 cm³ of each other), then find the mean titre from the concordant results only.

The dependent variable is the volume of acid needed (the titre); the independent variable is which repeat you are doing. Controlled variables include the volume of alkali (fixed by the pipette), the concentration of both solutions, and the indicator used.

The first rough titration is not counted. Only concordant titres, agreeing within 0.10 cm³, are averaged, because they show the result is repeatable and reliable.

(Higher Tier only) Titration calculations are assessed at Higher Tier.

The key relationship is:

$\text{moles}={\text{concentration (mol/dm}}^3)\times {\text{volume (dm}}^3)$

Remember to convert cm³ to dm³ by dividing by 1000.

Worked example. In a titration, 25.0 cm³ of sodium hydroxide solution is exactly neutralised by 20.0 cm³ of hydrochloric acid of concentration 0.100 mol/dm³. Find the concentration of the sodium hydroxide in mol/dm³.

The balanced equation gives a 1:1 mole ratio:

$HCl+NaOH\to NaCl+H_{2}O$

Step 1 — moles of the known solution (HCl): $n(HCl)=0.100\times \frac{20.0}{1000}=0.00200\text{ mol}$

Step 2 — use the mole ratio. It is 1:1, so: $n(NaOH)=0.00200\text{ mol}$

Step 3 — concentration of NaOH: $c(NaOH)=\frac{n}{V}=\frac{0.00200}{\frac{25.0}{1000}}=\frac{0.00200}{0.0250}=0.0800{\text{ mol/dm}}^3$

The sodium hydroxide concentration is 0.0800 mol/dm³.

(Higher Tier only) Converting between mol/dm³ and g/dm³ is assessed at Higher Tier.

To express a concentration in grams per cubic decimetre instead of moles, multiply the concentration in mol/dm³ by the relative formula mass ($M_r$):

${\text{concentration (g/dm}}^3)={\text{concentration (mol/dm}}^3)\times M_r$

Worked example. Convert the sodium hydroxide concentration of 0.0800 mol/dm³ from the previous slide into g/dm³.

Step 1 — relative formula mass of NaOH ($A_r$: Na = 23, O = 16, H = 1): $M_r(NaOH)=23+16+1=40$

Step 2 — multiply: $0.0800\times 40=3.20{\text{ g/dm}}^3$

The concentration is 3.20 g/dm³. This means every 1 dm³ (1000 cm³) of the solution contains 3.20 g of sodium hydroxide.

To go from mol/dm³ to g/dm³, multiply by $M_r$. To go back the other way, divide by $M_r$. Always work out $M_r$ from relative atomic masses first.

1. Using universal indicator

Universal indicator gives a gradual range of colours, so there is no sharp end point. Use a single-colour-change indicator such as methyl orange or phenolphthalein.

2. Including the rough titration in the mean

The first rough titration only locates the approximate end point. Average only the concordant titres (within 0.10 cm³), never the rough result.

3. Forgetting to convert cm³ to dm³ (HT)

The formula moles = concentration × volume needs the volume in dm³. Divide any cm³ value by 1000 first; 25.0 cm³ becomes 0.0250 dm³.

4. Ignoring the mole ratio (HT)

Use the balanced equation. If the ratio is not 1:1, for example sulfuric acid reacting 1:2 with sodium hydroxide, scale the moles accordingly before finding the unknown concentration.

5. Multiplying by the wrong quantity for g/dm³ (HT)

To convert mol/dm³ to g/dm³ you multiply by the relative formula mass, not by the volume. Work out $M_r$ from the relative atomic masses first.

6. Reading the burette wrongly

Read the burette at eye level, at the bottom of the meniscus, to 0.05 cm³. Reading from an angle causes a parallax error and an inaccurate titre.