Chem Lab 2 Formal Report

Purpose

Use a standardized potassium permanganate solution to analyze an unknown iron solution by using redox titration.

Theory

Reduction-oxidation titration is an analytical method based on electrons transferring between an oxidizing agent and a reducing agent in solutions. Chemical analysis can be built on Redox titration if four criteria are met:
a) The reaction is completed in a thermodynamically spontaneous condition.
b) The reaction is fast enough to give an operational result instantly.
c) The reaction does not have any side reaction.
d) The reaction has a satisfactory indicator.

Standardization of KMnO4 solution with Na2C2O4
2KMnO4+5Na2C2O2+8H2SO4→2MnSO+K2SO4+5Na2SO4+10CO2+8H2O
KMnO4 solution is unstable in air because it is oxidized to MnO2 spontaneously by a reducing substance in distilled water.
Standardization will be interfered by this brown colour MnO2 especially in the presence of light. So MnO2 has to be removed by filtering the solution since it is insoluble in water.
In that hot Na2C2O4 decomposes into CO2 and H2O in the air by the presence of acid:
C2O42-+O2+2H3O+→H2O2+2C2O+2H2O
The titration is started at room temperature.
There is a titration error since the end point occurs before or after the equivalence point. A blank solution with 1.0M H2SO4 is used to calculate the amount of drops still needed to add or subtract from the end point of titration.

Redox titration of Fe2+ with KMnO4
MnO4-+8H3O++5Fe2+→5Fe3++Mn2++12H2O
KMnO4 is oxidized to MnO2 in neutral solution and MnO2 has brown colour which is difficult to observe colour changing during titration. Instead, KMnO4 is oxidized to colorless Mn2+ in acidic solution which is easy to observe a change in colour while titrating.
The solution gradually decolorized after reaching the permanganate end point due to the reaction between MnO4- and Mn2+, so the end point is read as quickly as possible.
2 mL of H3PO4 solution reacts with Fe3+ to produce Fe (PHO4)2- , decolorizing the orange colour of Fe3+. So the colour change will be visible at the end point.
There are two side reactions formed during titration:
2MnO4-+3Mn2++6H2O→5MnO2+4H3O+
10Cl- +2MnO4-+16H+→2Mn2++8H2O+5Cl2

Procedure

0.02M KMnO4 solution was filtered into a clean and dry beaker (250mL) and mix thoroughly. 55 mL of H2SO4 (concentrated) was added by a TA to 900 mL of water while stirring. Then the solution was filled with distilled water to obtain a 1M H2SO4 solution with total volume of 1000 mL in an Erlenmeyer flask. Three samples of dried Na2C2O4 (0.25 g to 0.30 g) were weighed by difference using an analytical balance. Samples were added into three 500 mL Erlenmeyer flasks and filled with 250 mL of the 1M H2SO4 solution separately. Na2C2O4 was completely dissolved into colorless solution before introducing 90% of the KMnO4 solution (calculated by using weight and molar weight of dried Na2C2O4 and concentration of KMnO4 solution) from burette. The mixed solutions were stirred occasionally until purple colour was disappeared. After heating each solution to 55-60℃, it was titrated by drops until the transparent solution turned pale pink which persisting for 30 seconds. Each reading was recorded as the end point of each sample. A blank solution (the remaining 1M H2SO4 solution) was added one drop of KMnO4 solution to test the end point. The number of drops that turned solution into pale pink was recorded as well.
A numbered sample of an iron ore was obtained from a TA and the specific number was recorded. Two samples of ore (around 1.5g) were weighed by difference using an analytical balance. Each sample was transferred to a 250 mL Erlenmeyer flask and 10 mL of 3M H2SO4, 2 mL H3PO4 and distilled water was added into each individual to dissolve it (distilled water was added at first then acids were added). Next, each solution was titrated with standardized KMnO4 solution. Titration stopped when the last drop that turned solution to pale pink after swirling for 15 seconds.

Observations

Na2C2O4 sample is in white colour and powdered shape.
Fe sample is also white powder with sample number 215.
Sample content
Color
Shape
Number
Na2C2O4 white powder
/
Fe white powder
215
Table 1. Sample number and physical description of each sample

In standardization of KMnO4 solution, after adding 90% of KMnO4 solution, the colour changes from colorless to brown and finally back to colorless while swirling. During titration, solution changes from colorless to pale pink in the end.
In titration of the Fe2+ sample, the initial colour for titration solution is colorless. In the end point, solution changes to pale pink at last.
Stages
Initial colour
While adding 90% of KMnO4 solution
After adding 90% of KMnO4 solution
Final colour at end point
Titration of standardization of KMnO4 solution
Colorless
Brown
Colorless
Pale pink
Titration of the Fe2+ sample
Colorless
/
/
Pale pink
Table 2. Color change of the titration solution during two different stages

Data
Sample content
Sample number
Accepted % value
Fe2+
# 215
2.96%
Table 3. Sample number and accepted % value for Fe2 Sample number
Initial mass (±0.0001g)
Final mass (±0.0001g)
Sample mass (±0.0002g)
Sample 1
10.8851
10.6076
0.2775
Sample 2
10.6076
10.3503
0.2572
Sample 3
10.3503
10.1070
0.2433
Table 4. Masses of Na2C2O4 samples.

90% volume
10% volume

Sample number
Initial volume (±0.05mL)
Final volume (±0.05mL)
Total volume (±0.10mL)
Initial volume (±0.05mL)
Final volume (±0.05mL)
Total volume (±0.10mL)
Total
added
(±0.20mL)

Sample 1
0.00
37.28
37.28
35.20
38.30
3.10
40.38

Sample 2
0.00
34.55
34.55
32.70
35.20
2.50
37.05

Sample 3
0.00
32.68
32.68
38.30
42.20
3.90
36.58

Table 5. Volumes of standardization of KMnO4 trials.
Sample number
Initial mass (±0.0001g)
Final mass (±0.0001g)
Sample mass (±0.0002g)
Sample 1
14.0380
13.8964
0.1416
Sample 2
13.8964
13.7497
0.1467
Table 6. Masses of Fe2+ initial samples.

Sample number
Initial volume (±0.05g)
Final volume (±0.05g)
Total volume (±0.10g)
Sample 1
0.00
0.90
0.90
Sample 2
0.90
1.80
0.90
Table 7. Volume of standardized KMnO4 solution for Fe2+ titrations.

Calculations

1. Volume (90%consumption) of KMnO4
= [[(mass Na2C2O4/ MW Na2C2O4) (2/5)]/ (0.02)] (0.9) (1000)

Sample 1:
Volume
= [[(0.2775g)/ (133.998gmol-1) (2/5)]/ (0.02molL-1)] (0.9) (1000mL/L)
=37.28mL

Sample 2:
Volume
= [[(0.2572g)/ (133.998gmol-1) (2/5)]/ (0.02molL-1)] (0.9) (1000mL/L)
=34.55mL

Sample 3:
Volume
= [[(0.2433g)/ (133.998gmol-1) (2/5)]/ (0.02molL-1)] (0.9) (1000mL/L)
=32.68mL

2. Concentration of KMnO4
= [[(mass Na2C2O4/ MW Na2C2O4) (2/5)]/ [(total volume of KMnO4)/ (1000)]

Sample 1:
Concentration
= [[(0.2775g)/ (133.998gmol-1) (2/5)]/ [(40.38mL)/ (1000mL/L)]
= 0.0205molL-1

% uncertainty= (uncertainty/actual value) (100%) = [(0.0002/0.2775) + (0.20/40.38)] (100%) = 0.57%

Sample 2:
Concentration
= [[(0.2572g)/ (133.998gmol-1) (2/5)]/ [(37.05mL)/ (1000mL/L)]
= 0.0207molL-1

% uncertainty= (uncertainty/actual value) (100%) = [(0.0002/0.2572) + (0.20/37.05)] (100%) = 0.62%

Sample 3:
Concentration
= [[(0.2433g)/ (133.998gmol-1) (2/5)]/ [(36.58mL)/ (1000mL/L)]
= 0.0199molL-1

% uncertainty= (uncertainty/actual value) (100%) = [(0.0002/0.2433) + (0.20/36.58)] (100%) = 0.63%

Average of concentrations = (0.0205molL-1+0.0207molL-1+0.0199molL-1)/3 = 0.0204molL-1

Relative spread = [(highest value – lowest value)/ average value] (1000ppt) = [(0.0207 molL-1)-(0.0199 molL-1)/ 0.0204molL-1] (1000ppt) = 39.2ppt

3. % Fe = [[(volume of KMnO4/1000) (average concentration of KMnO4) (Molar weight of Fe)]/ (mass of Fe initial sample)] (100%) Sample 1:
% Fe
= [[(0.9mL/1000mLL-1) (0.0204molL-1) (55.845gmol-1)]/ (0.1416g)] (100%)
= 0.72%

% uncertainty= (uncertainty/actual value) (100%) = [(0.1/0.9) + (0.0002/0.1416)] (100%) = 11.3%

Sample 2:
% Fe
= [[(0.9mL/1000mLL-1) (0.0204molL-1) (55.845gmol-1)]/ (0.1467g)] (100%)
= 0.70%

% uncertainty= (uncertainty/actual value) (100%) = [(0.1/0.9) + (0.0002/0.1467)] (100%) = 11.2%

Average of % Fe = (0.72%+0.70%)/2 = 0.71%

% error = [(approximate value-exact value)/exact value] (100%) = [(0.71%-2.96%)/2.96%] (100%) = -76.0%

Relative spread = [(highest value – lowest value)/ average value] (1000ppt) = [(0.72%-0.70%)/0.71%] (1000ppt) = 28.2ppt

Discussion

The accepted value of % Fe sample is 2.96% but the result our group calculated was 0.71% in average. The reasons that the experimental value is lower than the real value might be the following:
Firstly, since Fe2+ has strong reducing power it is easy to be oxidized by oxygen in the air. Secondly, colour changing of solution from transparent to pale pink is too quick to be observed, so there is a titration error. Thirdly, the reading on burette is hard to read horizontally. Because the top of the solution is much higher so we have to raise our eyes to read the numbers.

Conclusion

To summarize, the real value of Fe sample # 215 is 2.96% while the average experimental value is 0.71%. Uncertainties of total mass and total volume are ± 0.0002g and ±0.10mL respectively. % error is -76.0% and relative spread for concentration of KMnO4 and % Fe are 39.2ppt and 28.2ppt.