Volumetric Analysis 11

EXPERIMENT - 11

Estimation of Potassium Permanganate Aim :- To estimate the amount of KMnO 4 present in 250ml of given solution by

using 0.1M Ferrous Ammonium Sulphate Solution (Mohr’s Salt Solution). Apparatus and Chemicals required :-

Burette, Burette stand, Pipette, Conical flask, Glazed tile, given KMnO 4 solution, 0.1M Mohr’s salt solution, Dil.H 2SO4 solution. Principle :- KMnO4 reacts with FeSO 4 in the presence of H 2SO4 by the following

equation. 2 KMnO4 + 8 H 2 SO4 2 moles

+

10 FeSO4

→  K 2 SO4 +

2 MnSO4

10 moles

+

5 Fe2 ( SO4 ) 3

+

8 H 2 O

Here, 2 moles KMnO 4 reacts with 10 moles FeSO 4 Procedure :- The solution taken in burette is given KMnO 4 solution. 20ml of 

given 0.1M Mohr’s salt solution is transferred into a conical flask with the help of a pipette. Then test tube full of Dil.H 2SO4 is added to it. Here no external indicator is required. Here KMnO 4 acts as self indicator in this redox titration. This solution is titrated against given KMnO 4 solution taken in burette. Initially the titration should be very slow. Because it is an auto catalysed reaction. At the end point the colour of the solution changes from colourless to pale pink colour. The volume of KMnO 4 run down from the burette is noted. The titrations are repeated till consecutive readings are obtained Tabular Form :-

S.No

Volume of  Mohr’s salt solution in (ml)

Burette readings Initial

‘a’ ml

Final ‘b’ ml

Volume of  KMnO4 run down (b-a) ml

1

20 ml

2

20 ml

3

20 ml

∴

Formula :-

Volume of  KMnO4 solution = _______________________

V 1 M 1

=

n1

V 2 M 2 n2

V1 = Volume of  KMnO4 solution = ____________________ ml M  1

n1

= Molarity of  KMnO4 solution = ?

= No.of moles of  KMnO4 in the equation = 2 mole

V2 = Volume of  Mohr’s salt solution = 20 ml M 2 n2

= Molarity of  Mohr’s salt solution = 0.1M

= No.of moles of  Mohr’s salt in the equation = 10 mole ∴

Molarity of  KMnO4,

=

20 × 0.1 10

×

M 1

=

V 2 M 2

×

n2

n1 V 1

2 V 1

= ________________ M ∴

Molarity of  KMnO4=

M 1

= _________________ M

Amount of  KMnO4 present in 250 ml of given solution = Molarity of  KMnO4 × Gram mo1.wt. Of  KMnO4 × =

M 1 ×

158 4

= ________________ grams

250 1000

Report :-

Amount of  KMnO4 present in 250ml of given solution = _______ gms