Question

of Oxalate Before beginning this experiment in the laboratory, you should be able to answer the following questions 1. If0.4586 g of sodium oxalate, Na C204, requires 33.67 mL of a KMno4 solution to reach the end point, what is the molarity of the KMnO4 solution? Titration of an oxalate sample gave the following percentages: 15.53%, 15.55%, and 15.56%. Calculate the average and the standard deviation. 2. 3. Why does the solution decolorize upon standing after the equivalence point has been reached? 4. Why is the KMnO, solution filtered, and why should it not be stored in a rubber-stoppered botle? Report Sheet Oxidation-Reduction Titrations I: Determination of Oxalate What volume of o.100 M KMnO4 would be required to titrate 0 5. Calculate the percer t C2042-in each ofthe follo in g H2C204,NaC20,KC204- dkjAI(C204)3] 3H20 6. 7. If 28.70 mL of 0.0200 M KMnO, is required to titrate a 0.250 g sample of K,[Fe(C204)%1-3H2O, what is the percent C20,2 in the complex? 8. What is the percent purity of the complex in question 7?

Answer 1

1^st Answer

Weight of sodium oxalate(W) – 0.4856gms

Volume of KMNO4(R ) - 33.67 ml             

Molarity of KMNO4 = W / R x 0.0671

Where

W = Weight of sodium oxalate

R = Volume of KMNO4

So Molarity of KMNO4 = 0.4856/33.67 x 0.06701

= 0.4856/2.26

=0.21 M

2^nd Answer

Purity of oxalate in

Trail 1- 15.53 %

Trail 2- 15.55 %

Trail 3- 15.56 %

Average Purity of oxalate Xi = (15.53+15.55+15.56)/ 3

                                              =15.55%

X	15.53	15.55	15.56
(X-Xi)²	0.000278	0.000011	0.000178

∑(X-Xi)² = 0.000467

Standard Deviation = Square root of ( (X-Xi)²/n)

Where X= purity of oxalate in each trail

Xi= Average Purity of oxalate

n = No of trails = 3

Standard Deviation = Square root of(0.000467/3)

= 0.0125

3^rd Answer

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

_After reaching end point all KMnO4 converted to MnSO4

MnSO4 = colourless