Please use this example to do your lab report.
Title: Titration of H2C2O4.H2O with NaOH
Aim: Predict from calculation and measure the volume of NaOH
needed to neutralize 0.020dm3 of H2C2O4.2H2O.
Introduction:
Materials List:
H2C2O4.2H2O powder
NaOH powder
Phenolphthalein indicator solution
Apparatus List:
0.020dm3 pipette (20ml)
pipette pump
burette, 0.050dm3(50ml)
2beakers, 250ml
Procedures:
1.
2.
Observations:
Data:
NaOH solution: Mass= 0.41grams in 0.100dm3 water solution
C2O4H2.2H2O: Mass=1.26grams in 0.100dm3 water solution
Titration
Run1
Run 2
Titration Data
Start Volume, dm3
Finish Volume, dm3 Volume NaOH
Used
0.00
0.042
0.042
0.05
0.048
0.044
Average Volume
0.043
Data Processing (Use your values below)
Given that we use 0.020dm3 of oxalic acid in the beaker under the burette, what volume
do we predict that we will titrate of NaOH? We know the Mr for NaOH and oxalic acid
or H2C2O4.2H2O.
Now the concentration of acid and base are:
molesA
1.26 /126
CA 

 .1 mole/dm3
0.100dm3
0.100
CB 
molesB
0.41/ 40.0

 .102 mole/dm3
0.100dm3
0.100
The neutralization equation for the titration of oxalic acid with NaOH is as
follows:
H2C2O4.2H2O+ 2NaOHNa2C2O4.2H2O+2H2O
CAVA 1

CBVB 2
Here CA and CB are the concentration of acid and base. NaOH is the base.
H2C2O4.H2O is the acid.
2
CAVA
 VB
CB
We used 0.020dm3 or 20ml of oxalic acid into which we added phenolphthalein indicator
when we did the titration. Therefore
CAVA
CB
0.1
VB  2*
*0.020
0.1
VB  0.040dm3 or 40ml
VB  2
Conclusions (How does the volume calculated and measured
compare? Are they close or far apart?)