A student weighs by difference 0.1956g of sodium oxalate into a 100ml Volumetric

Question

A student weighs by difference 0.1956g of sodium oxalate into a 100ml Volumetric flask and dilutes to the mark 10mL of HCI and 90mL of distilled water. The student then pipetted a 25mL aliquot of the ocalate solution to a 125mL Erlenmeyer flask. Approximately how many mL's of 0.02 M KMnO4 would be requied to reach the equivalence point of the trtration?

The balanced alf reactions are:

Reduction: MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

Oxidation: C2O4 2- ---> 2CO2+ 2e-

Combining (and balancing) the two half reactions yields:

2MnO4 2- + 5 C2O4 2- + 16H+ ---> 2Mn2+ + 10CO2 + 8 H2O

Explanation / Answer

Reduction: MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

Oxidation: C2O4 2- ---> 2CO2+ 2e-

Combining (and balancing) the two half reactions yields:

2MnO4 2- + 5 C2O4 2- + 16H+ ---> 2Mn2+ + 10CO2 + 8 H2O

no of moles of Na2C2O4 = W/G.M.Wt

                                         = 0.1956/134   = 0.00146 moles

from balanced equation

                 5 moles of Na2C2O4 react with 2 moles of KMnO4

                 0.00146 moles of Na2C2O4 react with = 2*0.00146/5 = 0.000584 moles of KMnO4

             molarity = no of moles/volume in L

             0.02 = 0.000584/volume in L

             volume in L = 0.000584/0.02 = 0.0292 L = 29.2ml >>>> answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.