Volumetric Analysis 12

EXPERIMENT - 12

Estimation of Oxalic Acid Aim :- To estimate the amount of Oxalic acid present in 250ml of given solution

by using 0.02M KMnO 4 solution. Apparatus and Chemicals required :-

Burette, Burette stand, Pipette, Conical flask, Glazed tile, given Oxalic acid solution, 0.02M KMnO 4 solution, Dil.H 2SO4 solution Principle Principle :- KMnO4 reacts Oxalic ic acid acid in the the pres presen ence ce of H 2SO4 by the reacts with Oxal

following equation. 2 KMnO4 + 3 H 2 SO4 2 moles

+

5 H 2 C 2 O4

→  K 2 SO4 +

2MnSO4

+

10CO2

+

8 H 2 O

5moles

Here, 2 moles KMnO 4 reacts with 5 moles H 2C2O4 Procedure :- The solution taken in burette is given 0.02M KMnO 4 solution. 20ml of given Oxalic acid solution is transferred into a conical flask with the help of a pipette. Then test tube full of Dil.H2SO4 is added to it, and then heat the solution to reach the sharp boiling point (Until bubbles are observed). Here no external indicator is required. Here KMnO 4 acts as self indicator in this redox titration. This solution is titrated against 0.02M KMnO 4 solution taken in burette. Initially the titration should be very slow. Because it is an auto catalysed reaction. reaction. At the end point the colour colour of the solution changes changes from colourless 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  Oxalic acid 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

=

V 2 M 2

n1

n2

V1 = Volume of  KMnO4 solution = ____________________ ml M  1

n1

= Molarity of  KMnO4 solution = 0.02M

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

V2 = Volume of  Oxalic acid solution = 20 ml M 2 n2

= Molarity of  Oxalic acid solution =?

= No.of moles of  Oxalic aicd in the equation = 5 moles ∴

Molarity of Oxalic acid, =

V 1

×

0.02

2

×

M  2

=

V 1 M 1 n1

×

n2 V 2

5 20

= ________________ M ∴

Molarity of  Oxalic acid =

M  2

= _________________ M

Amount of  Oxalic acid present in 250 ml of given solution = Molarity of  Oxalic acid =

M 2

×

×

Gram Mo1.wt. of  Oxalic acid

126 4

= ________________ grams

×

250 1000

Report :-

Amount of  Oxalic acid present in 250ml of given solution = _______ gms