Question 1:
In this experiment you are to determine the relative formula mass of an iron(II) salt by titration with potassium manganate(VII).
FA 1 is the iron(II) salt.
FA 2 is 0.0200 mol dm–3 potassium manganate(VII), KMnO4.
FA 3 is dilute sulfuric acid, H2SO4.
(a) Method
Preparing a solution of FA 1
● Weigh the 250 cm3 beaker and record the mass in the space below.
● Add all the FA 1 provided to the beaker. Weigh the beaker with FA 1 and record the mass.
● Calculate the mass of FA 1 used and record this in the space below.
● Use a measuring cylinder to add approximately 100 cm3 of FA 3 to the beaker. Stir until all the solid has dissolved.
● Transfer the solution into the 250 cm3 volumetric (graduated) flask labelled FA 4.
● Wash out the beaker thoroughly using distilled water and add the washings to the volumetric flask. Make the solution up to the mark using distilled water.
● Shake the flask thoroughly to mix the solution before using it for your titrations.
● This solution of the iron(II) salt is FA 4.
| Mass of 250cm3 Beaker / g | 33.42 |
| Mass of 250cm3 Beaker + FA1 / g | 42.86 |
| Mass of FA1 used / g | 9.44 |
Titration
● Pipette 25.0 cm3 of FA 4 into a conical flask.
● Use a measuring cylinder to add 20 cm3 of FA 3 to the flask.
● Fill the burette with FA 2.
● Titrate FA 4 with FA 2 until the solution changes to a permanent pink colour.
● Perform a rough titration and record your burette readings in the space below.
| Initial Burette Reading / cm3 | 0.00 |
| Final Burette Reading / cm3 | 23.50 |
| Titre / cm3 | 23.50 |
● Carry out as many accurate titrations as you think necessary to obtain consistent results.
● Make sure any recorded results show the precision of your practical work.
● Record in a suitable form below all of your burette readings and the volume of FA 2 added in each accurate titration.
| 1 | 2 | |
| Initial Burette Reading / cm3 | 0.10 | 0.10 |
| Final Burette Reading / cm3 | 23.60 | 23.70 |
| Titre / cm3 | 23.50 | 23.60 |
| Best Results (✓) | ✓ | ✓ |
(b) From your accurate titration results, obtain a suitable value to be used in your calculations.
Show clearly how you have obtained this value.
(23.50+23.60)/2 = 23.55cm3
25.0 cm3 of FA 4 required 23.55 cm3 of FA 2 [1]
(c) Calculations
Show your working and appropriate significant figures in the final answer to each step of your calculations.
(i) Calculate the number of moles of potassium manganate(VII) present in the volume of FA 2 calculated in (b).
n = c * v
= 0.0200 * 23.55/1000
= 0.000471 mol
moles of KMnO4 = 0.000471 mol
(ii) The half-equation for the reduction of a manganate(VII) ion is:
Give the half-equation for the oxidation of an iron(II) ion to an iron(III) ion.
Therefore, 1 mole of manganate(VII) ions reacts with 5 moles of iron(II) ions.
(iii) Calculate the number of moles of iron(II) ions present in 25.0 cm3 of solution FA 4.
n(MnO4-) : n(Fe2+) = 1:5
so moles of Fe2+ = 0.000471 * 5 = 0.002355 ~ 0.00236 (correct to 3 sf)
moles of Fe2+ in 25.0 cm3 of FA 4 = 0.00236 mol
(iv) Calculate the number of moles of iron(II) ions present in 250 cm3 of solution FA 4.
no. of moles of Fe2+ in 250cm3 = 0.002355/25 * 250
= 0.02355
~ 0.0236 (correct to 3 sf)
moles of Fe2+ in 250 cm3 of FA 4 = 0.0236 mol
(v) In 1 mole of the iron(II) salt, FA 1, there is 1 mole of iron(II) ions.
Use the mass of FA 1 you weighed out to calculate the relative formula mass of the iron(II) salt.
relative formula mass = …………………..
[5]