Thank you very much Calculating the molarity of KMnO_4 Suppose you weighed out 0

Question

Thank you very much Calculating the molarity of KMnO_4 Suppose you weighed out 0.2546 g of primary standard Na_2C_2O_4 that was 99.85% pure and it took 34.22 mL of your approximately 0.02 M KMnO_4 to reach the endpoint. 1) Write the chemical equation for the reaction between Na_2C_2O_4 and 4 under acidic conditions. 2) How many grams of Na_2C_2O are in your 0.2546 g sample of Standard? 3) How many moles of Na_2C_2O_4 are in your 0.2546 g sample of Standard. The molar mass of sodium oxalate is 134.00 grams/mole? 4) What is the exact molarity of your KMnO_4? Calculating the percent Na_2C_2O_4 in Unknown C Suppose you did several more standardization titrations and you found that the average molarity of your 4 was 0.02229 M. Then you used this standardized KMnO_4 to titrate 0.8665 grams of your Unknown C and it took 39.36 mL off your KMnO_4 to reach the endpoint. 1) How many moles of KMnO_4 are in the 39.36mL of your standardized KMnO_4? 2) How many moles of Na_2C_2O4 were in your 0.8665 gr

Explanation / Answer

1. The titration of Na2C2O4 with KMnO4 is a redox reaction and it requires an acidic medium. Let us take H2SO4 for maintaining the acidic condition. The net balanced reaction would be:

8 H2SO4 + 2 KMnO4 + 5 Na2C2O4 2 MnSO4 + 10 CO2 + K2SO4 + 5 Na2SO4 + 8 H2O

Individual oxidation and reduction reactions are:

Oxidation:   C2O42- 2 CO2 + 2 e-

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

2.   As the Na2C2O4 sample is 99.85% pure , hence the 0.2546g sample will actually have (99.85/100) * 0.2546= 0.2543g of Na2C2O4 .

3. Moles of Na2C2O4 present in 0.2546 g of sample= weight of sample/ molecular weight of Na2C2O4 i.e.    Number of moles of Na2C2O4 = 0.2546/134 = 0.0019 moles

4. For calculating the exact molarity of KMnO4, we need the volume of flask in which standard Na2C2O4 solution is made. Suppose we dissolve the 0.2546 g of Na2C2O4 in 100 mL flask. Then the molarity of resulting Na2C2O4 solution would be 0.0019/ 0.1000 L = 0.019 M. The formulae for calculating molarity in a redox reaction is:           M1V1/n1 = M2V2/n2   where M= molarity of solution, V= volume of solution used in titration, n= number of electrons involved in the reaction.

Suppose if wee take 20 mL of the 0.019 M Na2C2O4 solution and it requires 34.22 mL of KMnO4 of unknown molarity to titrate the Na2C2O4 (20 mL and 0.0019M) completely , then the molarity of KMnO4 is calculated as: (0.019 * 20) / 2 = (M * 34.22)/5    or M=   0.028M

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