Blog #3: Question Response

Question asked by Maggie Calvert:

Sodium oxalate, Na2C2O4, in solution is oxidized to CO2(g) by MnO4- which is reduced to Mn2+. A 50.1ml volume of a solution of MnO4- is required to titrate a 0.339g sample of sodium oxalate. This solution of MnO4- is used to analyze uranium-containing samples. A 4.62g sample of a uranium-containing material requires 32.5ml of the solution for titration. The oxidation of the uranium can be represented by the change UO2+-->UO2 2+. Calculate the percentage of uranium in the sample.

Solution:

First the entire reaction was wrote out as
2 KMnO4 + 5 Na2C2O4 + 8 H2SO4 ---> 2 MnSO4 + K2SO4 + 5 Na2SO4 + 10 CO2 + 8 H2O
The molar mass of sodium oxalate and potassium permanganate were calculated:

Molar mass of Na2C2O4 is 134 g/mol

Molar mass of KMnO4 is 158 g/mol 

 

Using the reaction above, it can be seen that

2 moles of KMnO4 reacts with  5 *134 of Na2C2O4  

X moles of KMnO4 reacts with  5*0.339 g of Na2C2O4  

X = ( 2*5*0.339g) / (5*134g)

   = .00056 moles of KMnO4 

 

So Molarity of MnO4 - , M = no.of moles / Volume of solution in L

                                           = .00056 moles / 0.0501 L

                                           = 0.01108 M 

 

The reaction of uranium sulfate with potassium permanganate with water is wrote as:

5 U(SO4)2 + 2 KMnO4 + 2 H2O ----> 2 H2SO4 + K2SO4 + 2 MnSO4 + 5 UO2SO4

Volume of MnO4- required for this titration is , V = 32.5 mL = 0.0325 L

So no.of moles of MnO4- , n = Molarity * Volume in L

                                             = 0.01108 M * 0.0325 L

                                             = .00036  moles of MnO4- 

 

Using the equation directly above, 

5 moles of U(SO4)2  reacts with 2 moles of MnO4 -

So Y moles of U(SO4)2  reacts with .00036 moles of MnO4-

Y = ( 5*.00036 moles ) / 2

   = .0009 moles

Molar mass of U(SO4)2  = 430 g

So mass of U(SO4)2  is , m = No.of moles * Molar mass = .3876 g 

Therefore the percentage of Uranium in the sample is calculated by

( .3876g/ 4.62g) *100 = 8.39 % 

 

Insightful Response:

This was a very good and hard (haha) question as it utilized being able to write the proper equations and balance them for the reactions, using the info learned from this chapter, as well as incorporating things we learned previously in chem.

Sources:

 

Chemistry Book, Chapter 18 Electrochemistryfor being able to write out the reactions

Using previous knowledge of titrations and basic chemistry for mathematical calculations.