|May21-12, 05:58 PM||#1|
My group messed this experiment up and now i have to get a good grade(i receive 50% on all of my labs!). I would highly appreciate your input.
Our teacher gave us a procedure sheet, which we "accurate followed"
This is what we did:(If this is not enough i have a few more steps that were done before but i do not think they are important..)
Redox titration;Analysis of Hydrogen Peroxide
Take 125Ml Erlenmeyer flask, transfer 1.0ml of the commercial hydrogen peroxide. commercial??
Add 25Ml of distilled water to the flask
Measure 5Ml of 6M sulfuric acid into a cylinder and than add it to the Erlenmeyer flask.
Open the buret stopcock and allow 5-8ml of potassium permanganate to flow into a flask.
Add as much as of KMn04 as needed to see a color change(turned pink) The burret was labeled in ml units.
We did those steps 3 times to the best results.
We also recorded a volume of KMNo4 used and averaged it. the result was 16.5
I am kinda confused what should i do in order to calculate the % error, number of moles of H2O2;KMn04 by using (C=N/V) C=n/v; C of MnO4 was given at 0.025M.
Here are the numbers that were given to us:
Concentration of KMnO4 =0.025M
Theoretical Moles of H2O2(8.88*10 to the power of negative)
%uncertainty of a burret is plus/minus 0.05ml. -My teacher gave us a formula:(0.05+0.05):avg volume of KMnO4*100 not sure what do i determine with this formula(% uncertainty of a solution??)
Here are the 1/2oxidation and 1/2 reduction equations(written on my paper by a teacher):
½ reduction: H2O2---> O2 + 2H+ + 2e
½ reduction: MnO4- + 8H+ + 5e = Mn2+ + 4H2O -Purple
Full balanced equation(done by me)
2MnO4 + 16H + 5H202--->5O2 + 10H + 2Mn2+ + 8H2O
I have calculated moles of H2O2
H2O2:0.025*0.0165ml(average volume of KMNo4used):5and the result *2=0.000165 moles
Now i have to find moles of MNo4; % uncertainty and % error.
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