Before beginning this experiment in the laboratory, you should be able to answer

Question

Before beginning this experiment in the laboratory, you should be able to answer the following questions. 1. If 0.4586 g of sodium oxalate, Na_2C_2O_4, requires 33.67 mL. of a KMnO_4 solution to reach the end point, what is the molarity of the KMnO_4 solution? 2. Titration of an oxalate sample gave the following percentages: 15.53%, 15.55%, and 15.56%. Calculate the average and the standard deviation. 3. Why does the solution decolorize upon standing after the equivalence point has been reached? 4. Why is the KMnO_4 solution filtered, and why should it not be stored in a rubber-stoppered bottle? 5. What volume of 0, 100 M KMnO_4 would be required to titrate 0.23 g of K_2[Cu(C_2 O_4)2] middot 2H_2 O? 6. Calculate the percent C_2O_4^2 in each of the following: H_2 C_2 O_4, Na_2 C_2 O_4, K_2 C_2 O_4, and K_3[AI(C_2 O_4)_3] middot 3H_2 O.

Explanation / Answer

1)

Oxidation Half-      

C2O4 2- = 2 CO2 + 2e-

Reduction half -

MnO4 - + 8 H+ + 5e- = Mn2+ + 4 H2O

Oxidation and reduction must occur together, so it is redox processes. If we combine the two half-reactions above, we would find with a balanced net ionic equation if we have a total of 10 electrons exchanged.

5 C2O42- + 2 MnO4 - + 8 H+ = 10 CO2 + 2 Mn2+ + 8 H2 O

Now, calculate moles of sodium oxalate-

Molar mass of sodium oxalate = 134 g/mol

Moles of sodium oxalate = 0.4586 g/134 g/ mol

                                           = 0.00342 moles.

Now calculate moles of MnO4

Moles MnO4- = 0.00342 x 2 /5=0.001368

M = 0.001368 / 0.03367 L=0.04066M

Molarity of KMnO4 solution is 0.041M

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