Titration – Exam Practice

Aim

The purpose of this experiment is to determine x in the formula Fe(NH4)2(SO4)2•xH2O by titration against a standard solution of potassium manganate (VII) (permanganate).

Introduction

This experiment provides useful experience in preparation for a practical examination. The directions are similar to those that would be given by an examination board.

A is a solution of ammonium iron(II)sulfate, Fe(NH4)2(SO4)2•xH2O, the precise concentration of which (in g dm-3) is given by the teacher.

B is a solution of potassium manganate(VII) (permanganate), KMnO4, the precise concentration of which (in mol dm-3) is given by the teacher.

Requirements

  • safety glasses
  • pipette, 25 cm3
  • pipette filler
  • 3 conical flasks, 250 cm3
  • ammonium iron(II) Sulfate solution, A (IRRITANT)
  • sulfuric acid, dilute, 1 M H2SO4
  • burette, 50 cm3
  • funnel, small
  • potassium manganate (VII) solution, B (0.01 mol dm-3) (LOW HAZARD)
  • white tile
  • wash-bottle of distilled water

Procedure

Pipette 25 cm3 of the ammonium iron(II) Sulfate solution, A, into a conical flask and add an equal volume of dilute sulfuric acid. Titrate with potassium manganate(VII) solution, B, until a permanent faint pink colour appears. Repeat the titration twice and enter your results in a copy of your results table.

The overall equation for the reaction is:

MnO4– (aq) + 5Fe2+ (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l)

2016-10-19-1

Table of Results (PDF)

Calculation

Use your results to determine x in the formula Fe(NH4)2(SO4)2•xH2O. You should set out your calculations so that every step in your working is clearly shown. If you cannot work out a method of calculation, use the suggestions below. (These would probably not be given in an examination.)

Calculation steps

  1. From the titre and the equation for the reaction calculate the concentration of Fe2+ ions (y mol dm-3).
  2. From the concentration calculate the mass of anhydrous Fe(NH4)2(SO4)2 in one litre of solution.
  3. Subtract the mass obtained in step 2 from the mass of the salt in one litre. This difference, z g, divided by 18 g mol-1 gives the amount of water of crystallization in y mol of the salt.
  4. x mol is the amount of water in 1 mol of the salt.

Water of crystallization exam practice