2018 Pure Chemistry Titration Question
Making Biodiesel
Vegetable oils that have been used for cooking can be reacted to make biodiesel for fuel.
They are tri-esters with long hydrocarbon chains. They react with methanol, using a potassium hydroxide catalyst. This reaction is illustrated in the diagram below:
QUESTION 1
Vegetable oils that have been used for cooking contain acids. The oils must be treated to neutralise the acids before they can be used in the reaction to make biodiesel.
Otherwise, using untreated oils affects the reaction speed. Explain how and why. [2 marks]
As the potassium hydroxide catalyst is a base, it would undergo neutralisation with the acids in untreated oils to form salt and water. Without the catalyst, the activation energy would become too high, hence reducing the rate of reaction.
Titration of Acids in Used Vegetable Oils
Before using the oil to make biodiesel, a titration is used to find out how much potassium hydroxide needs to be added to each kg of oil to neutralise the acids.
10.0 g of oil is titrated with aqueous potassium hydroxide with a concentration of 1.00 g/dm3. The results are as follow:
| Burette reading | Rough | Titration 1 | Titration 2 |
|---|---|---|---|
| Final/ cm3 | 22.00 | 43.50 | 21.60 |
| Initial/ cm3 | 0.00 | 22.00 | 0.00 |
| Volume used/ cm3 | 22.00 | 21.50 | 21.60 |
QUESTION 2
Use the titration results to calculate the mass (in grams) of potassium hydroxide that needs to be added to neutralise the acid in 1 kg of oil. [3 marks]
STEP 1: Find average volume of KOH used to titrate 10 g of oil
Average volume of KOH used = 21.50 + 21.60 = 21.55 cm3
STEP 2: Find the mass of KOH used to titrate 10 g of oil
Mass of KOH used = concentration in g/dm3 × volume in dm3 = 1 × (21.55/1000) = 0.02155 g
STEP 3: Find the mass of KOH needed to titrate 1 kg (1000 g) of oil
Mass of KOH needed = 0.02155 ÷ 10 × 1000 = 2.16 g (3 s.f.)
This titration question tests you on proportion, instead of mole concept. You have to be aware that the titration is only done to a fraction of 10 g out of 1 kg. Therefore, after finding the mass of KOH used in the titration, you have to scale it up (step 3).
There is no need to convert the data to number of moles because we don’t have to use the mole ratio to find the amount of oils based on the amount of potassium hydroxide.
2014 Pure Chemistry Titration Question
pH Changes during a Titration
A pH meter and a data logger are used to monitor the pH changes during a series of titrations. In each titration, 0.1 mol/dm3 sodium hydroxide, NaOH, is added from a burette into a solution of a different dilute acid.
During the titration the pH does not change smoothly, as shown by the graph for each titration. Each point is collected and plotted by the data logger.
Titration 1: Strong Acid
Titration 2: Weak Acid
Titration 3: Dibasic, Strong Acid
Sulfuric acid is a strong acid. The ionisation of sulfuric acid can be shown by two equations:
(1) H2SO4 ⟶ H+ + HSO4–
(2) HSO4– ⟶ H+ + SO42-
As both H2SO4 and HSO4– are strong acids, they are both neutralised at the same point in the titration, which means that there is still only a single ‘step’ on the graph.
Titration 4: Dibasic, Weak Acid
The graph for phosphoric acid has two ‘steps’. At ‘step’ one, only one hydrogen ion from H3PO4 has reacted, forming H2PO4–. Subsequently, at ‘step’ 2 another hydrogen ion from H3PO4 has reacted, forming HPO42-.
There appear to be three hydrogen atoms in phosphoric acid that may ionise. However, in practice, only the first two hydrogen atoms can form ions. At the end of the titration, HPO42- ions are left.
QUESTION 3
The information above does not give the concentration of sulfuric acid used in titration 3. What is the concentration? Explain your reasoning in words or by means of a calculation. [2 marks]
In Words
Comparing titration 1 and titration 3, the same amount of sodium hydroxide was used to neutralise hydrochloric acid and sulfuric acid. This means that both acids contained the same amount of hydrogen ions. As their volumes were equal, their concentration of hydrogen ions must be equal.
However, as sulfuric acid is a dibasic acid that releases two hydrogen ions for every acid molecule, the concentration of sulfuric acid must be half of hydrochloric acid, to give the same concentration of hydrogen ions.
Therefore, the concentration of sulfuric acid is 0.1 ÷ 2 = 0.05 mol/dm3
By Calculations
2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O
STEP 1: Convert data to number of moles
No. of moles of NaOH = 0.1 × (20/1000) = 0.002 mol
STEP 2: State the mole ratio
Mole ratio of NaOH to H2SO4 = 2:1
STEP 3: Use the mole ratio to find the number of moles of H2SO4
No. of moles of H2SO4 = 0.002 × (1/2) = 0.001 mol
STEP 4: Use the volume of H2SO4 at the ‘step’ to find its concentration
Concentration of H2SO4 = 0.001 ÷ (20/1000) = 0.05 mol/dm3
QUESTION 4
Hydrochloric acid is a strong acid, but ethanoic acid is a weak acid. What is the difference between a strong acid and a weak acid. [1 mark]
A strong acid ionises completely in water to produce hydrogen ions, while a weak acid ionises partially.
QUESTION 5
Identify two differences between the pH graphs for titration 1 and titration 2. [2 marks]
Starting pH
Titration 1 began at a lower pH of 1, while titration 3 began at a higher pH of 2.6.
Initial pH Change
When the first 20 cm3 of NaOH was added, the pH remained fairly constant at pH 2 for titration 1, but increased almost linearly from pH 2.6 to pH 7 for titration 2.
pH Change during a ‘Step’
Titration 1 has a wider ‘step’ at 20 cm3 of NaOH, from pH 3 to pH 12. However, titration 2 has a narrower ‘step’, from pH 7.8 to 11 at 20 cm3 of NaOH.
QUESTION 6
Write equations to show how phosphoric acid produces the hydrogen ions involved in the two ‘steps’ on the pH graph in titration 4. [2 marks]
‘Step’ 1: H3PO4⟶ H+ + H2PO4–
‘Step’ 2: H2PO4–⟶ H+ + HPO42-
QUESTION 7
Give the formula of the salt formed at the end of titration 4. [1 mark]
Na2HPO4
At the end of titration 4, the anion left behind by the acid is HPO42-. Its negative charge of 2- requires 2 sodium cations, Na+, with a total charge of 2+ to balance.
Endpoints and Indicators
An indicator can also be used to see when a ‘step’ happens in the pH change. The endpoint of each titration happens when the indicator changes colour.
The best indicator for a titration gives a distinct colour change when a ‘step’ occurs. For a titration between a strong acid and a strong alkali, every indicator in the diagram would give an accurate titration volume.
QUESTION 8
Explain why any of the indicators in the diagram can be used to give an accurate titration volume when strong acids are titrated with sodium hydroxide. [2 marks]
Methyl orange, litmus, and, phenolphthalein change colour at around pH 4, 7 and, 9 respectively. All these lie within the wide pH range of the ‘step’, which is from pH 3 to pH 12.
QUESTION 9
Explain why methyl orange would not be suitable to use when titrating ethanoic acid with dilute sodium hydroxide. [1 mark]
Methyl orange changes colour at around pH 4, which is before the pH range of the ‘step’ in titration 2. This means that the colour would have already changed before the endpoint, causing the chemist to add less than the stoichiometric amount of sodium hydroxide to complete the neutralisation.
QUESTION 10
Suggest the best indicator to use when titrating ethanoic acid with dilute sodium hydroxide. [1 mark]
Phenolphthalein. Its colour changes at pH 9, which is within the pH range of the ‘step’ from pH 8 to pH 11.
