solution:

volume v₁= 50ml

concentration n₁= 0.5N

volume v₂=?

concentration n₂=o.1N

Now we have

molar equivalent of seminormal solution= molar equivalent of deci normal solution

or, 0.5* 50= 0.1*v

or, 250= v

Thus, volume of water should be added=250-50 =200 ml

Solution:

Molecular weight of oxalic acid= 126

Equivalent weight of oxalic acid =63

Volume of solution to be prepared ,V =250ml

Conc.of solution to be prepared, N=0.1N

Now ,

We have

Mass of oxalic acid (g)=Equivalent weight volume normality 1000

= 632500.11000

1.575g

An example of redox titration is titration of oxalic acid solution against KMno₄solution in acidic medium.

It is called so because ,during titration ,oxidation occur on one of the species and reduction on the other species, making it redox reaction and thus redox titration.

The requisites for a substance to be a primary solution are:

• It should be in available in pure state ,
• Its composition should be constant during storage and weighing,
• It should be readily soluble in water and not be decomposed by water,
• It should have high equivalent weight and should not be hygroscopic or deliquescent.

The solution containing 1/10 (one tenth) of one gram equivalent weight of substance dissolved in 1 litre of volume is called decinormal solution. It is represented as N/10 or 0.1N .