Sunday, 3 February 2013

Third Lab Sesion - Redox titration of hydrogen peroxide and potassium permanganate

Eugenia Fernández de los Ronderos Jiménez and Fco Javier Mondaza Hernández 


Objective: To determine the volume of KMnO4 needed for 4mL of H2SO4 and the concentration of hydrogen peroxide.

Background:
Titration is a method of analysis which allow us to determine the precise endpoint of a reaction, and therefore, the precise quantity of reactant in the titration flask.
A burette is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction. (DARMOUTH COLLEGE, 2012)

Materials

-Stand
-Erlenmeyer flask
-Burette
-Clamp stand
-Beaker
-Pippete 
-Clamp
-KMnO (0,05 M)
H2SO(4 M)
-100 mL volumetric flask
-Comercial oxygenated water (H2O2)



                                           KMnO4                                   Beaker, Pippete and Erlenmeyer flask.
 Procedure:
1. Set up the material. (picture).
2. Fill the burette with KMnO4.




3. Adjust the level of the burette to 0.
4. Take 4 mL of  H2Owith the pippete and add it to the Erlenmeyer flask.
5. Add 4 mL (we do not need to be accurate) of the H2SO to the Erlenmeyer flask.
6. Titrate the redox solution, until no reaction is taking place. For this, you have to open the tap carefully and see how the "KMnO" behaves when it is in contact with the solution.

Image 10
Final look of the solution.

Results







Our results were the following one:

4 mL H2SO4 + 4 mL H2O2 + 29,2 mL KMnO4 

And the reaction observed is this one:

KMnO4H2SO4 + H2O2 --> MnSO4 + O2   H2O + K2SO4

Carrying out a simple balancing redox reaction, we can balance it in order to continue with our calculations:


2KMnO4 +3 H2SO4 + 5H2O2 --> 2MnSO4 + 5O2 + 8H2O + K2SO4


-Therefore, we can calculate the moles of KMnO4:

M = moles/ volume in L. // 0.05 = x/ 0.0292 //0.00146 moles KMnO4

-Taking into account the ratio between 
KMnOand H2O2, we calculate the moles of the last one:

 5 H2O2 / KMnO4   --> (0.00146 *5) :2 = moles H2O2 // 0.0073 : 2 = moles H2O//

0.00365 = moles H2O2

-Therefore, we can now calculate the molarity of  H2O2:

M = moles /volume in L // M = 0.00365 / 0,004 L // M  H2O2  = 0.9125

-According to the theoretical relation between H2Oand the O2 produced (1 volume  H2Oproduces 10 volumes of  O2) we carry out the following reasoning:

 H2O2 = n O2

V O2 (Standard conditions for gases) = 3,65 m moles * 22.4 ml/mol =  81,76 mL O2

-Finally, we calculate if the mL of O2 produced were the expected ones:

Expected mL O2: 4mL H2O2 * 10 = 40 mL

% error: (Experimental value - Theoretical value) / Theoretical value *100 = (81,76 - 40) / 40  *100= 41,76 / 40 *100 = 104, 4%


Conclusions:
In this experiment we have observed that there is always a relation between the reactants of a chemical reaction. Due to this, relations, we can calculate  different characteristics of the compounds using easy formulas.
Also, we have learned how to carry out a chemical procedure; tritation. As we have seen, this process consists of determining the precise quantity of a reactant needed to carry out a determined chemical reaction. We have learned this process could be really useful to calculate extra-characteristics of the reactants we are using.
Besides these points, we have calculated the molarity of H2O2 (0.9125 M) and we have been able to check if the volume of O2 was the correct one according to the commercial acid relation. We found out it was not (the percentage error was more than 100%), so we conclude that the hydrogen peroxide we were using was not a commercial one (so its concentration was different), or that there was not standard conditions for the O2 produced (so the results were altered).


Bibliography:

DARMOUTH COLLEGE (2012) : Titration Technique. Taken from:


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