Balance the following reaction for the redox titrations performed in this experi

Question

Balance the following reaction for the redox titrations performed in this experiment (Hint: Use half reactions to obtain the correct coefficients): If 25.14 mL of the KMn04 solution is needed to titrate a 0.8273 g sample of Na2C204, what is the molarity of the permanganate solution? Hint: Calculate the moles of sodium oxalate, a2C204. The moles of KMn04 are some ratio of this number (you will need to use your balanced equation from Question #1-Na2C204 and C2042- are equivalent). Then, molarity = moles KMn04/Liters KMn04. Suppose that in Step #2 of this experiment, you measure out a 0.1084 g sample of your complex and you determine that it takes 11.79 mL of a 0.02396 M KMn04 solution to reach the titration endpoint. Calculate the millimoles of oxalate in this sample. (Hint: First calculate the # moles of KMn04 from the concentration and volume values given, Then multiply by the proper molar ratio to obtain the moles of oxalate. See your balanced equation from Question #1.) Calculate the weight% and number of millimoles of oxalate that would be present in a 100.0 g sample of the complex in Question #3 above. (Hint: Equate the ratios of the sample masses to millimoles and solve.)

Explanation / Answer

1)

{H2C2O4(aq) = 2CO2(g) + 2H+ + 2e-} x 5

{8H+ + 5e-+ MnO4- = Mn2+ +4H2O} x 2

5H2C2O4(aq) + 2MnO4-(aq) + 6H+(aq) = 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

2)

from Q#1 we can observe that 5 moles of C2O4- is needed to titrate 2 moles of MnO4-

moles of C2O4- = 0.8273/134 = 6.17*10^-3 moles

moles of MnO4- = (2/5)*moles of C2O4-

= 2.47*10-3 moles

Molarity = 2.47*10-3/25.14*10^-3

= 0.098 M

3)

Moles of KMnO4 used is = 11.79*10^-3*0.02396

= 2.82*10^-4 moles

moles of C2O4- = 2.5*Moles of KMnO4 used

= 7.06*10^-4 moles

4)

wt of oxalate in above sample = moles of C2O4-*88 gm

= 0.062 gm

wt of oxalate in 100 gm of sample = 100*0.062/0.1084

= 57.33 gm

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