Experiment 2

Topic : Volumetric analysis – Redox

Purpose : To determine the ratio of the number of moles of hydroxyammonium

ions to the number of moles of iron (III) ions participating in the

reaction.

Materials : KA 1 is a solution containing 1.58 g of potassium manganate (VIII)

per 500 cm3

KA 2 is a solution prepared by boiling 4.00 g of hydroxyammonium

sulphate, (NH3OH)2 SO4 per dm3 with excess iron (III) ammonium

sulphate and dilute sulphuric acid

KA 3 is 1.0 mol dm-3

sulphuric acid

Apparatus : One 25 cm3 pipette and pipette filler

Three titration flasks

One 50 cm3 burette

One retort stand and clamp

One 50 cm3

measuring cylinder

One white tile

One wash bottle filled with distilled water

Introduction : In the presence of hydrogen ions, the hydroxyammonium ion,

NH3OH+ will reduce iron (III) ion to iron (II) ion while the

NH3OH+ itself will be oxidised to dinitrogen oxide.

Procedure : (a) 25.0 cm3 of KA 2 solution is pipetted into a titration flask.

(b) Approximately 25 cm3 of KA 3 is added into KA 2 and this mixture

is titrated with solution KA 1.

(c) The titration is repeated as many times to achieve accurate results.

(d) The titration readings are recorded in the table below.

Sample report

Results :

(i) 25.0 cm3 of KA 2 required 25.05 cm

3 of KA 1 for a complete reaction.

(ii) Calculate your average titre value showing the suitable titre values that

you use.

Average titre value = 25.00 25.10

2

+

= 25.50 cm3

Questions : (a) Calculate the concentration, in mol dm-3

, of manganate (VII) ions in

solution KA 1.

The concentration of potassium manganate (VII) in mol dm-3

= n

v

= 3

1.58 mol

158

0.5 dm

= 0.02 mol dm-3

Accurate Titration number Rough

1 2

Final reading/cm3 25.00 25.00 25.10

Initial reading/cm3 0.00 0.00 0.00

Volume of KA 1/cm3 25.00 25.00 25.10

n = The number of moles of KMnO4

v = The volume of KMnO4

(b) Calculate the concentration, in mol dm-3

, of iron (II) ions in solution

KA 2.

The redox equation of the reaction between iron (II) ions and the

manganate (VII) ions are as follow.

5Fe2+

+ MnO4- + 8H

+ � 5Fe

3+ + Mn

2+ + 4H2O

Using the formula a a

b b

m v a

m v b=

Where m1 = concentration of KMnO4 in mol dm-3

m2 = concentration of iron(II) ions in mol dm-3

v1 = The volume of KA 1 in cm3

v2 = The volume of KA 2 in cm3

ma = 0.020 mol dm-3

mb = ?

v1 = 25.05

v2 = 25.0

a = 1

b = 5

Hence, 2

0.020 25.05 1

25.0 5m

×=

×

∴ m2 = 0.100 mol dm-3

(c) Calculate the mass of iron(II) ions in 1 dm3

of KA 2.

The mass of iron(II) ions in 1 dm3

of KA 2

= the number of moles of iron(II) ions x relative atomic mass

= 0.100 mol x 55.8

= 5.58g

(d) Determine the number of moles of iron(III) ions required to oxidise

1 mol of hydroxyammonium ions

Fe2+

� Fe3+

+ e-

1 mol of iron (II) ions gives 1 moles of iron(III) ions

∴ The concentration of iron(II) ions

= The concentration of iron(II) ions

= 0.100 mol dm-3

The concentration of hydroxyammonium sulphate, (NH3OH)2 SO4

= 4.00 g dm-3

= 4

2(14.0 3.0 16.0 1.0) 32.1 4(16.0)+ + + + +

= 4.00

164.1

= 0.02438 mol dm-3

The concentration of NH3OH+ ions

= 2 x 0.02438

= 0.04876 mol dm-3

The number of moles of iron (III) ions required to oxidise 1 mole of

hydroxyammonium ions

= 0.100

0.04876

= 2.05 mol

≈ 2 mol

(e) Write a balanced redox equation between NH3OH+

ions and Fe3+

ions.

Reduction : 4Fe2+

+ 4e- � 4Fe

2+

Oxidation : 2NH3OH+ � N2O + H2O + 6H

+ + 4e

-

Redox Equation : 2NH3OH+ + 4Fe

3+ � N2O +H2O + 6H

+ + 4Fe

2+

(f) Why would the titration not require an external indicator?

The titration does not require an external indicator because potassium

manganate (VII) can act as indicator. Potassium managanate (VII)

turns colourless when manganate (VII) ions are reduced. The end

point can be determined when the solution turns to pink colour.

Conclusion : The ratio of number of moles of hydroxyammonium ions to the

number of moles of iron (III) ions participating in the reaction is

1:2