Question

A.) what is the normality of Na2C2O4 solution if you use 1.14g of Na2C2O4 to prepare 100.00 mL of solution?

B.) what is the normality of the stock KMnO4 solution if it took 31.20 mL of the KMnO4 solution to titrate a 20.00 mL sample of the oxalate solution prepared in part a?

Answer 1

Answer for (A):

2 MnO4^- + 5 Na2C2O4 + 16 H⁺ → 2 Mn⁺ + 10 CO2 + 8 H2O + 10 Na⁺

Normality = Number of equivalent wts/ Volume in L

Equivalent wt = molar mass/no. of Hydrogens in an acid

Number of equivalent wts = Mass used/ Equivalent wt

Na2C2O4 = Mol. Wt = 134 g/mol

Mass of Na2C2O used = 1.14 g

Equivalent wt of Na2C2O4 = 134 / 2 = 67

Number of equivalent wts of Na2C2O4 = 1.14/67 = 0.0170

Volume of solution made = 100 ml = 0.1 L

Normality of Na2C2O4 = 0.017/0.1 = 0.17 N

Answer for (B):

Normality of Na2C2O4 = 0.017/0.1 = 0.17 N……………………….N1

Volume of Na2C2O4 solution used = 20 ml………………………….V1

Volume of KMnO4 solution used = 31.2 ml………………………….V2

Normality of KMnO4 solution used = …………………………………….N2

Using N1V1 = N2V2 equation one may find normality of KMnO4 solution used.

Therefore Normality of KMnO4 solution used = 0.17 x 20/31.2 = 0.109 N

However, the normality obtained above is valid only when equivalent weights are equal.   

According to balanced equation 2.5 equivalents of Na2C2O4 are oxidized by 1 equivalent of

KMnO4.

Therefore Normality of KMnO4 solution used = 2.5 (0.17 x 20/31.2) = 0.2725 N

Xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Second solution for (B):

To verify above calculation alternatively one may calculate molarity of solutions first and then

convert them into normality solutions.

Moles of Na2C2O4 = 1.14 g / 134 = 0.0085 Moles

Volume = 100 ml = 0.1 L

Molarity of Na2C2O4 solution = 0.0085 Mols/ 0.1 = 0.085 M……………M1

Volume used = 20 ml………………………………………V1

KMnO4 volume used = 31.2 ml…………………………V2

At this stage KMnO4 solution molarity can be calculated using equation: M1V1 = M2V2

KMnO4 solution molarity, M2 = 0.085 x 20/31.2 = 0.0545 M

Molarity of KMnO4 = 0.0545 M

Normality of KMnO4 solution used = M x No. of transferred in redox reaction

According to above balanced equation the number of electrons transferred from MnO4^- in the   

redox reaction are 5. Hence

Normality of KMnO4 = 0.0545 M x 5 = 0.2725 N