Experiment 19:
Aim:
To prepare potassium permanganate solution
Equipment:
Refer to STAWA book page 148
Procedure:
Refer to STAWA book page 148
Processing of Results and Questions:
The solution was first boiled for 10 minutes to speed up the
rate of reaction. This is because at room temperature, the solution has a very
slow rate of reaction.
Experiment 20:
Aim:
To prepare standard oxalic acid solution
Equipment:
Refer to STAWA book page 149
Procedure:
Refer to STAWA book page 149
Processing of Results and Questions:
m(Oxalic Acid) = 1.58 g
M(Oxalic Acid) = 126.068 gmol -1
n(Oxalic Acid) = 0.0125 mol
V = 0.25 L
c(Oxalic Acid) = n ÷ V
c = 0.0125 ÷ 0.25
c(Oxalic Acid) = 0.05013 mol L -1
M(Oxalic Acid) = 126.068 gmol -1
n(Oxalic Acid) = 0.0125 mol
V = 0.25 L
c(Oxalic Acid) = n ÷ V
c = 0.0125 ÷ 0.25
c(Oxalic Acid) = 0.05013 mol L -1
Experiment 21:
Aim:
Standardising potassium permanganate solution
Equipment:
Refer to STAWA book page 150
Procedure:
Refer to STAWA book page 151
Results:
|
Rough Estimate
|
Accurate 1
|
Accurate 2
|
Accurate 3
|
Final Reading (mL)
|
40.5
|
40.5
|
40.7
|
40.6
|
Initial Reading (mL)
|
20
|
20
|
20
|
20
|
Titration Volume
(mL)
|
20.5
|
20.5
|
20.7
|
20.6
|
Processing of Results and Questions:
1.
c(Oxalic Acid) = 0.05013
V(Oxalic Acid) = 0.02
n(Oxalic Acid) = 0.001 mol
n(Permanganate) = [n(Oxalic Acid) ÷ 5] x 2
n(Permanganate) = 0.000401 mol
V(Permanganate) = 0.0206 L
c(Permanganate) = n(Permanganate) ÷ V(Permanganate)
c(Permanganate) = 0.000401 ÷ 0.0206
c(Permanganate) = 0.0195 mol L -1
V(Oxalic Acid) = 0.02
n(Oxalic Acid) = 0.001 mol
n(Permanganate) = [n(Oxalic Acid) ÷ 5] x 2
n(Permanganate) = 0.000401 mol
V(Permanganate) = 0.0206 L
c(Permanganate) = n(Permanganate) ÷ V(Permanganate)
c(Permanganate) = 0.000401 ÷ 0.0206
c(Permanganate) = 0.0195 mol L -1
2a)
Condy’s
Crystals, a name given to potassium permanganate has a pinkish, purple colour.
2b) Manganese
oxide has a pale yellowish brown colour
2c)
MnO4- + 8H + +
5e - --> Mn2+ +
4H2O
H2S --> S + 2H+ + 2e -
2MnO4- + 6H + + 5H2S --> 2Mn2+ + 8H2O + 5S
H2S --> S + 2H+ + 2e -
2MnO4- + 6H + + 5H2S --> 2Mn2+ + 8H2O + 5S
2d) A
very horrendous smell of skunk scent and other smells such as garlic.
3.
Yes oxalic acid can be used as a primary
standard to standardise a basic solution such as sodium hydroxide. This is
because oxalic acid is pure and can be accurately weighed out on a scale.
Experiment 22:
Aim:
Analysis of iron content
Equipment:
Refer to STAWA book page 152
Procedure:
Refer to STAWA book page 152-153
Results:
|
Rough Estimate
|
Accurate 1
|
Accurate 2
|
Accurate 3
|
Final Reading (mL)
|
43.0
|
43.7
|
43.5
|
43.6
|
Initial Reading (mL)
|
30
|
30
|
30
|
30
|
Titration Volume
(mL)
|
13.0
|
13.7
|
13.5
|
13.6
|
Processing of Results and Questions:
1.
MnO4- + 8H + +
5e - --> Mn2+ +
4H2O
Fe2+ --> Fe3+ + e -
MnO4- + 8H + + 5Fe2+ --> Mn2+ + 4H2O + 5Fe3+
Fe2+ --> Fe3+ + e -
MnO4- + 8H + + 5Fe2+ --> Mn2+ + 4H2O + 5Fe3+
2.
c(Permanganate) = 0.0195 mol L -1
V(Permanganate) = 0.0136 L
n(Permanganate) = c(Permanganate) x V(Permanganate)
n(Permanganate) = 0.000265 mol
V(Permanganate) = 0.0136 L
n(Permanganate) = c(Permanganate) x V(Permanganate)
n(Permanganate) = 0.000265 mol
3.
n(Iron) = n(Permanganate) x 5
n(Iron) = 0.001324 mol in 20 ml
n(Iron) = 0.001324 mol in 20 ml
4.
n(Iron) = 0.016548 mol in 250 ml == ([0.001324
÷ 20] x 250)
M(Iron) = 55.85 gmol -1
m(Iron) = n(Iron) x M(Iron)
m(Iron) = 0.924224 g
M(Iron) = 55.85 gmol -1
m(Iron) = n(Iron) x M(Iron)
m(Iron) = 0.924224 g
5.
0.924224 ÷ 1 = 0.924224
0.924224 x 100 = 92.4%
0.924224 x 100 = 92.4%
6.
Iron can be oxidised from its surroundings.
Oxygen is a main source of this impurity.
7.
Fe + H2SO4 --> FeSO + H2
Fe + 2H+ --> Fe2+ + H2
Fe + 2H+ --> Fe2+ + H2
Investigation 12:
Aim:
To determine the concentration of hydrogen peroxide
Equipment:
·
Beakers
·
Conical Flask
·
Funnel
·
Volumetric Flask (500ml)
·
Pipette
·
Pipette Filler
·
Burette and Stand
·
Sulfuric Acid
·
Potassium Permanganate
·
Hydrogen Peroxide
·
Distilled Water
·
Safety Equipment
Procedure:
1.
Put on safety equipment
2.
Take out 2 ml of concentrated hydrogen peroxide
3.
Mix 20 ml of sulphuric acid with 30 ml of
potassium permanganate solution
4.
Add 2 ml intervals of the solution to the
hydrogen peroxide
5.
Do this until you see the first undissolved
purple colour in the hydrogen peroxide solution
6.
Place 5 ml of hydrogen peroxide in volumetric
flask and dilute it (should be by a factor of 100), we diluted ours by a factor
of 102
7.
Titrate with sulphuric acid and potassium
permanganate solution
Results:
|
Rough Estimate
|
Accurate 1
|
Accurate 2
|
Accurate 3
|
Final Reading (mL)
|
44.0
|
43.7
|
43.7
|
43.8
|
Initial Reading (mL)
|
20
|
20
|
20
|
20
|
Titration Volume
(mL)
|
24
|
23.7
|
23.7
|
23.8
|
Processing the Data:
1.
MnO4- + 8H + +
5e - --> Mn2+ +
4H2O
H2O2 --> O2 + 2H+ + 2e -
2MnO4- + 6H + + 5H2O2 --> 2Mn2+ + 8H2O + 5O2
H2O2 --> O2 + 2H+ + 2e -
2MnO4- + 6H + + 5H2O2 --> 2Mn2+ + 8H2O + 5O2
2.
c(Permanganate) = 0.0195 mol L -1
V(Permanganate) = 0.023733 L
n(Permanganate) = c(Permanganate) x V(Permanganate)
n(Permanganate) = 0.000462 mol
n(Hydrogen Peroxide) = [n(Permanganate) ÷ 2] x 5
n(Hydrogen Peroxide) = 0.001155 mol in 20 ml
n(Hydrogen Peroxide) = 0.029456 mol in 510 ml ([0.001155 ÷ 20] x 510)
c(Hydrogen Peroxide) = n(Hydrogen Peroxide) ÷ V(Hydrogen Peroxide)
V(Hydrogen Peroxide) = 0.51 L
c(Hydrogen Peroxide) = 0.058912 mol L -1 (dilute)
c(Hydrogen Peroxide) = 6.01 mol L -1 (concentrated) ([0.057757 ÷ 5] x 510)
V(Permanganate) = 0.023733 L
n(Permanganate) = c(Permanganate) x V(Permanganate)
n(Permanganate) = 0.000462 mol
n(Hydrogen Peroxide) = [n(Permanganate) ÷ 2] x 5
n(Hydrogen Peroxide) = 0.001155 mol in 20 ml
n(Hydrogen Peroxide) = 0.029456 mol in 510 ml ([0.001155 ÷ 20] x 510)
c(Hydrogen Peroxide) = n(Hydrogen Peroxide) ÷ V(Hydrogen Peroxide)
V(Hydrogen Peroxide) = 0.51 L
c(Hydrogen Peroxide) = 0.058912 mol L -1 (dilute)
c(Hydrogen Peroxide) = 6.01 mol L -1 (concentrated) ([0.057757 ÷ 5] x 510)
3.
c(Hydrogen Peroxide) x M(Hydrogen Peroxide) = g
mol -1
M(Hydrogen Peroxide) = 34.016 g mol -1
c(Hydrogen Peroxide) = 204.4 gL -1
M(Hydrogen Peroxide) = 34.016 g mol -1
c(Hydrogen Peroxide) = 204.4 gL -1
Evaluating the Investigation:
1.
Based on our results and the time it took us to
finish the investigation, I would say that we worked fairly efficiently.
Therefore, I do not see any places to make improvements. The only minor
‘problem’ we hit was the fact that we diluted the hydrogen peroxide too much.
After filling the volumetric flask with distilled water, we found out that we
had 510 ml of standard solution instead of 500 ml. This only changed our
calculation number.
2.
[c(Hydrogen Peroxide) ÷ 1000] x 100 = % by mass
% by mass = 20.44%
The bottle stated a 30%, however, we got 20%. This can easily be explained through chemistry. As soon as the seal cap is opened, gases start to escape the bottle. Therefore the percentage drops over time since the gases escape. This is the reason why our results are lower than the one stated on the bottle.
% by mass = 20.44%
The bottle stated a 30%, however, we got 20%. This can easily be explained through chemistry. As soon as the seal cap is opened, gases start to escape the bottle. Therefore the percentage drops over time since the gases escape. This is the reason why our results are lower than the one stated on the bottle.
3.
We were the first group to finish this
experiment. Compared to the other
groups, our results were more accurate since we used the hydrogen peroxide
sooner; therefore, fewer gases had escaped. Overall, the only error that may
have occurred in our experiment would be from the hydrogen peroxide solution;
since it wasn’t seal closed when we used the solution.
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