lab oxalic acid

Oxalic Acid Lab

Aim: Use acid base titration to determine the number of water molecules in hydrated hydrochloric acid.

Apparatus required:
Oxalic acid solution 250 cm3
Weighing bottle
Digital balance
Beaker (250 cm3)
Distilled Water
Volumetric Flask 250cm3
Filter funnel
Pipette
Burette 50cm3
Retort Stand
Beakers 100cm3
Standardized sodium hydroxide solution 0.1M
Pipette filter
Conical flasks 250cm3
Phenolphthalein Indicator

Procedure

1) Rinse the burette with distilled water and then with oxalic acid solution and fill the burette with oxalic acid solution.

2) Rinse the 10.0 cm3 pipette with sodium hydroxide solution, and transfer it to a conical flask.

3) Add two drops of phenolphthalein indicator. Titrate the solution to the end- point, as shown by the disappearance of faint pink colour. Use the first flask for a trial run.

4) Repeat steps 2-3 three times and record the results.

Table 1.1

Trial
Initial Reading of Burette
Final Reading of Burette
Amount of Oxalic Acid Used
Color Change
1.
50 ± 0.5 cm3
40.6
9.4 ± 1 cm3
Colorless to light pink
2.
50 ± 0.5 cm3
40.6
9.4 ± 1 cm3
Colorless to light pink
3.
50 ± 0.5 cm3
40.5
9.5 ± 1 cm3
Colorless to light pink

It is important to note that the endpoint of the titration or the point of neutralization occurs when the solution in the conical flask turn a light pink color. This is when you stop and immediately record the volume of oxalic acid used.

The formula:
H2C2O4 (aq) + 2 NaOH (aq) --> Na2C2O4 (aq) + 2 H2O (l)

Table 1.2

NaOH
(COOH)2
Concentratiom
0.1M
-
Volume
10 mL
9.43 ± 3cm3
Moles
2
1

Solving the problem:
(M1 x V1 )/n1 = (M2 x V2 )/n2

M2 = (M1V1n2)/n1V2

Molarity of oxalic acid = 0.05319 M

Molar mass (M) of oxalic acid is = (90 + 18x)
0.05319 = (1.6)/M x (1000/250)
M = (1.5 x4) / 0.05319
(90+18x) = 6.4/0.05319
(90+18x) = 120.323
18x = 30.323
X = 30.323/19
X = 1.6800
This will