Redox titration

Result

Concentration of potassium manganate (VII) solution, B:  0.0189 ( mol dm3)

Concentration of ammonium iron (II) sulphate solution, A: 39.214 (g dm3)

Calculation:

                MnO4- (aq) + 5Fe 2+ (aq) +8H+(aq) → Mn2+(aq) +5Fe3+(aq)+4H2O(l)

We can find the no. of moles of MnO4- by:

                                                                        0.0189 × 26.05 =  4.293 × 10-4

                                                                                        1000

   According to the equation above, 1 mol of MnO4- react with 5 moles of Fe 2+

So, the no of moles of Fe 2+ is :

                                                    4.293 × 10-3× 5 = 2.46× 10-3

                                        By the finding the no of moles of Fe 2+, we can find the Molarity by using the formula,

                                                    Molarity = No of moles

                                                                        Volume

                              Molarity of Fe 2+ = 2.46× 10-3

                                                               25∕1000  

                                                          = 0.0985 mol dm-3

By finding the Molarity we can find the mass of anhydrous Fe (NH4)2(SO4)2 in one dm3.

                    Molar mass of Fe (NH4)2(SO4)2 = 55.8+ (14.0+4) + (32.1 + 16×4) ×2

                                                                       = 55.8+ 36+192.2

                                                                       = 284 g mol

                       Molarity of Fe 2+ = molarity of Fe (NH4)2(SO4)2

      Therefore, Molarity of Fe (NH4)2(SO4)2 = 0.0985 mol dm-3

The No of moles of Fe (NH4)2(SO4)2  can be found by the formula:

                                           Molarity × Volume = no of moles

                                        0.0985 × 1 dm-3 = 0.0985 moles

               Therefore the no of moles of Fe (NH4)2(SO4)2 is 0.0985 moles.

    The mass of anhydrous Fe (NH4)2(SO4)2  can be found by:

                              0.0985 mol × 284 g mol = 27.97g

        Therefore the mass of anhydrous Fe (NH4)2(SO4)2 in 1 dm-3 is 28 g.

            Using the concentration of ammonium iron (II) sulphate solution and the mass of anhydrous Fe (NH4)2(SO4)2 we can obtain volume of the Fe (NH4)2(SO4)2•xH2O

Let the volume of the solution of ammonium iron (II) sulphate be V:

                        39.214 g dm-3 = 27.97g

                                                     V

                      39.214 g dm-3 V= 27.97

                                            V= 0.7132 dm-3      

                    The no of moles of Fe (NH4)2(SO4)2•xH2O can be found by the formula:

                                               Molarity × Volume = no of moles

                        The Molarity of  Fe (NH4)2(SO4)2•xH2O= Molarity of Fe 2+

                       0.0985mol dm3× 0.7132 dm3 = 0.703mol

Therefore the no of moles of Fe (NH4)2(SO4)2•xH2O is 0.0703mol

The molar mass of Fe (NH4)2(SO4)2•xH2O Can be found by:

Let the molar mass of Fe (NH4)2(SO4)2•xH2O be MM

                                                   27.97 = 0.0703

                                                    MM

                                                    27.97=0.0703MM

                                                 MM= 398.1

Therefore the total molar mass Fe (NH4)2(SO4)2•xH2O is 398.

                               The Molar mass of Fe (NH4)2(SO4)2•xH2O is :

284+x(2+16)

                                                          = 284 + 18x

                       By using the total molar mass:

                                                  284 + 18x = 398

                                                       18x= 398-284

                                                       18x= 114

                                                         x=6.33

                                                            x=6

                               Therefore the value of x is 6.

Therefore the formula is Fe (NH4)2(SO4)2•6H2O.

Discussion:

A redox titration involves a reducing agent and an oxidizing agent. In this experiment, ammonium iron (II) sulphate was the reducing agent and potassium manganate was the oxidizing agent.

Observations:

The solution in the conical flask was titrated with potassium permanganate, the solution in the flask first turn into a yellowish colour; this was because of the formation of Iron (III) ions formed during the redox reaction.

According to the following equation:

                         Fe 2+ (aq)+ e-→ Fe3+(aq)

The formation of Fe3+ caused the solution to turn yellow when it was titrated against the potassium manganate (VII).

Solution gradually turned into pale pink because of the formation of Mn2+

The excess Mn2+ caused the solution to turn to pale pink. According to the eqation.

                    MnO4- (aq) + 8H+ (aq) + 5e- → 4H2O (l) + Mn2+ (aq)

Questions:

1. Explain why it is unnecessary to use any indicator in this experiment.

  →This is because potassium permanganate can be its own self indicator because it can form a pale pink colour due to the formation of Mn2+ when it is being reduced

       2. Explain why dilute sulphuric acid should be added to ammonium iron (II) sulphate solution before titration.

→ It prevents premature oxidation of iron (II) ions as hydrated ammonium iron (II) ions may react with air to form Iron (III) ions.

   It also ensures the complete reaction for manganate from manganate (VII) to Mn2+

2. Suggest one possible error in the experiment

→ The possible error of this experiment is that maybe the pipette was not washed properly as the same pipette was used to pipette 25cm3 of ammonium iron (II) sulphate and 25cm3 of sulphuric acid. The solution could have already reacted when pipette.

Conclusion:

 The value of x in the formula of Fe (NH4)2(SO4)2•xH2O is 6.