4.2.5a Titrations
Titration finds the exact volumes that react
Titration
A technique used to determine an unknown concentration by reacting a measured volume of one solution with a solution of known concentration.
Indicator
A substance that changes colour over a particular pH range and can show when the end point of a titration has been reached.
- A titration measures the volume of acid that exactly reacts with a fixed volume of alkali.
- It is used for strong acids and strong alkalis, such as hydrochloric, sulfuric or nitric acid with sodium or potassium hydroxide.
- A suitable single-colour-change indicator, such as phenolphthalein or methyl orange, shows when the reaction is complete.
- Universal indicator is unsuitable because its gradual range of colours makes the exact change hard to judge.
Investigation: finding the volume of acid that neutralises 25 cm3\text{cm}^3cm3 of alkali
- Use a volumetric pipette and filler to run exactly 25.0 cm3\text{cm}^3cm3 of sodium hydroxide solution into a clean conical flask.
- Add a few drops of phenolphthalein, which turns the alkali pink, and stand the flask on a white tile so the colour is easy to see.
- Fill a burette with the dilute acid, run a little through the tap to remove air bubbles, and read the initial volume at eye level from the bottom of the meniscus.
- Run the acid in quickly at first while swirling the flask, then add it drop by drop as the pink colour starts to fade.
- Stop the moment one drop turns the solution permanently colourless, and record the final burette reading.
- Work out the titre by subtracting the initial reading from the final reading, recorded to the nearest 0.05 cm3\text{cm}^3cm3.
- Do a rough titration first, then repeat carefully until two titres agree within 0.10 cm3\text{cm}^3cm3, and take the mean of those concordant results.
- Wear eye protection, because both the acid and the alkali are irritant.
Careful readings and concordant titres give reliable results
End point
The stage in a titration when the indicator shows its first permanent colour change.
Titre
The volume of solution delivered from a burette, calculated by subtracting the initial burette reading from the final reading.
- The end point is the first permanent colour change, from pink to colourless with phenolphthalein.
- Read the burette at eye level from the bottom of the meniscus to avoid a parallax error.
- Make sure there is no air bubble in the burette tip, or the titre will be too large.
- Concordant titres agree within 0.10 cm3\text{cm}^3cm3, and only these are used to calculate the mean.
- The rough titre is never included in the mean.
- Do not use too much indicator, because it can slightly change the volume needed.
- Do not stop at a colour change that fades again when you swirl the flask.
- Do not include the rough titre in your mean.
Why the apparatus is chosen
- The pipette delivers one accurate fixed volume of alkali.
- The burette measures the changing volume of acid to the nearest 0.05 cm3\text{cm}^3cm3.
- A measuring cylinder is not precise enough for either volume in a titration.
- Repeating until titres are concordant reduces the random error in the result.
- What does a titration measure?
- Why is universal indicator unsuitable for a titration?
- What is the colour change of phenolphthalein at the end point?
- How do you work out a titre from two burette readings?
- Which titres are used when calculating the mean?
4.2.5b Calculating quantities in titrations
Titration calculations link volumes through moles
Mole
The amount of a substance, measured in mol, where one mole has a mass in grams numerically equal to its relative formula mass.
Concentration
The amount of solute dissolved in a given volume of solution, often measured in mol/dm3 or g/dm3.
- Start with the reactant whose concentration and volume are both known.
- Find its moles using moles=concentration×volume in cm31000\text{moles} = \dfrac{\text{concentration} \times \text{volume in cm}^3}{1000}moles=1000concentration×volume in cm3
- Use the ratio in the balanced equation to find the moles of the other reactant.
- Find the unknown concentration using concentration=moles×1000volume in cm3\text{concentration} = \dfrac{\text{moles} \times 1000}{\text{volume in cm}^3}concentration=volume in cm3moles×1000
- Always convert volumes from cm3\text{cm}^3cm3 to dm3\text{dm}^3dm3, or keep the ÷1000\div 1000÷1000 in the formula.
- The mole ratio comes from the balanced equation, so balance it first.
Worked example: the concentration of sulfuric acid
Balanced equation
A chemical equation containing the same number of atoms of each element on both sides.
- In a titration, 24.0 cm3\text{cm}^3cm3 of sulfuric acid exactly neutralises 25.0 cm3\text{cm}^3cm3 of 0.100 mol/dm3\text{mol/dm}^3mol/dm3 sodium hydroxide, with equation 2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}2NaOH+H2SO4→Na2SO4+2H2O
- Moles of NaOH=0.100×25.01000=0.00250 mol\text{NaOH} = \dfrac{0.100 \times 25.0}{1000} = 0.00250\ \text{mol}NaOH=10000.100×25.0=0.00250 mol.
- The ratio is 2:12:12:1, so moles of H2SO4=0.002502=0.00125 mol\text{H}_2\text{SO}_4 = \dfrac{0.00250}{2} = 0.00125\ \text{mol}H2SO4=20.00250=0.00125 mol.
- Concentration of H2SO4=0.00125×100024.0=0.0521 mol/dm3\text{H}_2\text{SO}_4 = \dfrac{0.00125 \times 1000}{24.0} = 0.0521\ \text{mol/dm}^3H2SO4=24.00.00125×1000=0.0521 mol/dm3.
- Set the work out in clear steps, because each stage earns method marks even if a later number is wrong.
- Use the mean concordant titre, not the rough titre, as the volume in the calculation.
Converting between mol/dm3\text{mol/dm}^3mol/dm3 and g/dm3\text{g/dm}^3g/dm3
Solute
The substance dissolved in a solvent to form a solution.
- Convert using concentration in g/dm3=concentration in mol/dm3×Mr\text{concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_rconcentration in g/dm3=concentration in mol/dm3×Mr
- For sulfuric acid Mr=98M_r = 98Mr=98, so 0.0521 mol/dm3×98=5.11 g/dm30.0521\ \text{mol/dm}^3 \times 98 = 5.11\ \text{g/dm}^30.0521 mol/dm3×98=5.11 g/dm3.
- Reverse the conversion by dividing the g/dm3\text{g/dm}^3g/dm3 value by the MrM_rMr.
- Write the formula linking moles, concentration and volume in cm3\text{cm}^3cm3.
- In 2NaOH+H2SO42\text{NaOH} + \text{H}_2\text{SO}_42NaOH+H2SO4, how many moles of acid react with 0.00250 mol of alkali?
- How do you convert a concentration in mol/dm3\text{mol/dm}^3mol/dm3 to g/dm3\text{g/dm}^3g/dm3?
- Why should you use the mean concordant titre in the calculation?
- Find the concentration of an acid if 0.00200 mol reacts in 25.0 cm3\text{cm}^3cm3 of solution.