Question

A sample of pure KHC2O4.H2C2O4.2H2O (3 replaceable hydrogens) require 46.2ml of 0.1M NaOH solution for titration.How many ml of 0.1M KMnO4 will have same size of smaple react with?

Answer 1

First of all you need Molarity of the standardized KMnO4 solution to solve the second problem.

So the solution of first problem:

 

i)To Calculate number of moles of FeSO4

Fe+H2SO4→FeSO4+H2

No of moles of Fe=Weight/Atomic weight=0.158/26=0.006076 moles

From the above equation 1 mole of Fe gives 1 mole of FeSO4

therefore 0.006076 moles of Fe will give 0.006076 moles of FeSO4

 

ii)To calculate no of moles of KMnO4

2KMnO4+10FeSO4+8H2SO4→2MnSO4+5Fe2(SO4)3+K2SO4+8H2O

From the above equation 10 moles of FeSO4 reacts with 2 moles of KMnO4

Therefore 0.006076 moles will react with=2/10 X 0.006076=0.0012152 moles of KMnO4

Molarity=No of Moles X 1000/Vol(ml)=0.0012152 X 1000/31.23= 0.03891

 

 

Solution of Second problem:

Now we have molarity of KMnO4 solution

we can calculate no of moles from it.

No of moles=molarity X Volume(ml)/1000

Volume of KMnO4 now=36.56 ml

So Moles=0.03891 X 36.56/1000=0.0014225

 

2KMnO4+5H2C2O4+3H2SO4→2MnSO4+K2SO4+10CO2+8H2O

From this equation:

2 moles of KMnO4 react with 5 moles of Oxalic acid.

Therefore 0.0014225 moles will react with=5/2 X 0.0014225=0.003556 moles of Oxalic acid

1 mol of Oxalic acid=H2C2O4=90gm

therefore 0.003556 moles= 90 X 0.003556=0.32gm

 

when rhubarb leaves were reacted 0.32 gm of oxalic was there to react.

So % of oxalic acid=0.32 X 100/12.38 =2.5%

 

There is significance of mentioning that "this is no hydrate" because oxalic acid is generally available as hydrate.

So if you take hydrate the stoichiometry of the reaction will change and everything changes then.