Estimation of oxalic acid experimental data

Estimation of oxalic acid by titrating with KMnO4.

1. AIM:Estimationofoxalicacidbytitratingwith
KMnO
4.
THEORY:
Thetitrationofpotassiumpermanganate(KMnO
4)againstoxalicacid(C
2H
2O
4)isan
exampleofaredoxreaction.Inthistitration,MnisreducedandCisoxidizedfrom
KMnO
4and(COOH)
2,respectively.Initially,KMnO
4reactswithoxalateionstoform
Mn
2+
,K
+
,CO
2,andwater,whichresultinacolourlesssolution.Aftercomplete
consumptionofoxalateionsattheendpoint,anextradropofKMnO
4turnsthe
solutionpinkwhichindicatescompleteoxidationofoxalateions.Asthereactionis
sluggishatroomtemperatureoxalicacidalongwithsulphuricacidisheatedtoabout
60°Cbeforethetitration.TheaqueoussolutionofKMnO
4needstobestandardized
usingsodiumoxalate.
1

2. Observations: StandardizationofKMnO
4
•Burette solution: ~0.05N KMnO
4 solution
•Conical flask solution: 10 mL of 0.05N Na
2C
2O
4
+ 20 mL 2N H
2SO
4+ heat (~60 °C)
•Indicator: KMnO
4acts as self indicator
•Colourchange: Colourlessto light pink
•Reaction:
2(MnO
4)
-
+ 5(C
2O
4)
2-
+ 16H
+
→2Mn
2+
+ 10CO
2+ 8H
2O
Burette
reading
Piolet
reading
I (mL) II (mL) III (mL) Constant
(mL)
Final
9.0-10.0 mL
20.0 30.2 40.3
10.1Initial 10.0 20.1 30.2
Difference 10.0 10.1 10.1
Observations table

3. Observations: Estimation of oxalicacid
•Burette solution: ~0.05N KMnO
4 solution
•Conical flask solution: 10 mL H
2C
2O
4+ 20 mL 2N H
2SO
4+ heat (~60 °C)
•Indicator: KMnO
4acts as self indicator
•Colourchange: Colourlessto light pink
•Reaction:
2KMnO
4+5H
2C
2O
4+3H
2SO
4→2MnSO
4+K
2SO
4+10CO
2+8H
2O
Burette
reading
Piolet
reading
I (mL) II (mL) III (mL) Constant
(mL)
Final
9.0-10.0 mL
20.0 30.0 40.0
10.0Initial 10.0 20.0 30.0
Difference 10.0 10.0 10.0
Observations table

4. Calculations:
1.Equivalentweight(E.W.)ofoxalicacid
(H
2C
2O
4.2H
2O)
Molecularweight(M.W.)=(2×1)+(2×12)+(4×16)+(4×1)+(2×16)
=126
Equivalentweight=(126/2)=126g/2=63g

5. 2.Preparationof100mLof0.05NNsodiumoxalate
solution.
Molecularweight(M.W.(=134g
Equivalentweight=(M.W./2)=134g/2=67g
1000mL =1NNa
2C
2O
4 =67g
100mL =0.05NNa
2C
2O
4 =‘x’g
Therefore,x=(0.05NX100mLX67g)/(1NX1000mL)=0.335g
0.335gofsodiumoxalatedissolvedanddiluteduptothemarktoprepare
0.05Nstandardsolution.

6. 3. Exact normality of ~0.05N KMnO
4.
10 mL of 0.05N Na
2C
2O
4.2H
2O solution requires 10.1mL
of 0.05N KMnO
4solution
(sodium oxalate) N
1V
1 = N
2V
2(B.R of standardization)(KMnO
4)
0.05 N X 10.0 = N
2X 10.1
N
2= 0.0495 N
Exact normativity (strength) of KMnO
4is 0.0495 N

7. 4. Strength of unknown concentration of oxalic acid.
10 mL of unknown concentration of oxalic acid solution requires 10
mL of 0.0495N KMnO
4solution
(oxalic acid) N
1V
1 = N
2V
2(B.R of estimation)(KMnO
4solution)
N
1×10 = 0.0495×10
N
1= 0.0495 N
Strength of unknown concentration of oxalic acid is 0.0495 N

8. 4. Amount of oxalic acid in 100 mL.
Amount of oxalic acid in 1000 mL = N X E. W.
= 0.0495 X 63 g
= 3.119 g/1000 mL
So,
3.119g→1000mL
0.3119g→100mL
0.312g→100mL
100mLofgivensolutioncontains0.312gofoxalicacid.

9. Results:
•The exact normality of given KMnO
4solution
is 0.0495N.
•Amount of oxalic acid in 100 mL is 0.312 g.