Question

In a volumetric analysis experiment, a solution of sodium oxalate (Na2C2O4) in acidic solution is titrated with a solution of potassium permanganate (KMnO4) according to the following balanced chemical equation:
2KMnO4(aq) + 8H2SO4(aq) + 5Na2C2O4(aq) → 2MnSO4(aq) + 8H2O(l) + 10CO2(g) + 5Na2SO4(aq) + K2SO4(aq)
What volume of 0.0393 M KMnO4 is required to titrate 0.142 g of Na2C2O4 dissolved in 30.0 mL of solution? PICK ONE.

A

27.0 mL

B

10.8 mL

C

30.0 mL

D

1.45 mL

E

3.61 mL

Answer 1

The reaction is

2KMnO4(aq) + 8H2SO4(aq) + 5Na2C2O4(aq) → 2MnSO4(aq) + 8H2O(l) + 10CO2(g) + 5Na2SO4(aq) + K2SO4(aq)

Given molarity of KMnO4=0.0393 M,

Mass of Na2C2O4=0.142 g and molar mass=134 g/mol.

Therefore moles of Na2C2O4=mass/molar mass=0.142 g/134 g/mol=1.059x10^-3 mol.

And volume=30 mL=0.030 L.

Molarity of Na2C2O4=moles/volume=(1.059x10^-3 mol/0.030 L)=0.0353 M.

Here we can use the formula

M1V1/n1=M2V2/n2

For KMnO4, M1=0.0393 M, n1=2 and V1=?.

For Na2C2O4, M1=0.0353 M, n2=5 and V2=30 mL.

Then V1=(M2V2 n1)/(n2 M1)

V1=(0.0353 M x 30 mL x 2)/(5 x 0.0393 M)

V1=10.78 mL ~ 10.8 mL.

(B) is correct.

Please let me know if you have any doubt. Thanks.