Question

The concentration of a solution of potassium permanganate, KMnO4, can be determined by titration against a known amount of oxalic acid, H2C2O4, according to the following equation:
5H2C2O4(aq)+2KMnO4(aq)+3H2SO4(aq)→10CO2(g)+2MnSO4(aq)+K2SO4(aq)+8H2O(l)

What is the concentration of a KMnO4 solution if 27.20 mL reacts with 0.5610 g of oxalic acid?

Potassium permanganate (KMnO4) reacts with oxalic acid (H2C2O4) in aqueous sulfuric acid according to the following equation:
2KMnO4+5H2C2O4+3H2SO4→2MnSO4+10CO2+8H2O+K2SO4

How many milliliters of a 0.240 M KMnO4 solution are needed to react completely with 3.220 g of oxalic acid?

Answer 1

A. Number of moles of 0.5610 g of oxalic acid = 0.5610g/90g/mol = 0.0062mol

From the steochometric of the ractirea

5 mole oxalic acid react with 2 mole potassium permanganate

So 1 mole oxalic acid react with 2/5 mole potassium permanganate

Hence 0.0062 mole oxalic acid react with 2×0.0062/5 mole potassium permagperman

Number of mole of potassium permanganate = 2×0.0062/5= 0.002493 mole

Concentration of potassium permanganate = (number of moles of potassium permanganate)×1000/volume of soutisol in ml

Concentration of potassium permanganate = 0.002493mol×(1000/27.20 liter) = 0.091mol/ liter

B.. number of mole of oxalic acid is = 3.220g/90 g/mol =0.03577

From the steochometric of the reaction

5 mole of oxalic acid react with 2 mole of potassium permanganate

So 1 mole of oxalic acid react with 2/5 mole of potassium permanganate

Hence 0.03577 mole of oxalic acid react with 2×0.03577/5 mole of potassium permanganate

Number of mole of potassium permanganate = 0.01431 mole

We know that 1M = 1 mole / litet

So 0.240M = 0.240 mole/ liter

Concentration = number of mole / volume

0.240mole/ liter = 0.01431mole/volume

Volume = 0.01431mole/0.240mole/liter

Volume = 0.05962 liter

1liter = 1000ml

0.05962liter= 1000×0.05962 ml= 59.62ml