calcuate the normality of 0.2M solution of h3p04 ?

Answer 1

The Normality of H3PO4 (f.wt. = 98 g/mol) depends upon what type of reaction it is associated with. If it is in a metathesis reaction, the equivalent weight is the mass of reagent delivering 1 mole of + charge.
=> H3PO4 => 3H^+ + PO4^- => Eqv Wt = (98g/mol)/(3mole +) = 32.7g/eqv wt.
Given 0.20M H3PO4 => (0.20mol/L)(98g/mol)=1.96g/L => N(0.20M H3PO4)=(1.96g/L)/(32.7g/eqv wt)=0.60N H3PO4

If the reagent is in an oxidation-reduction reaction, however, the equivalent weight is the weight of reagent that can deliver or gain 1 mole of electrons in the redox process.

Here's an example ...
3P4 + 5HNO3 +38H2O => 12H3PO4 + 5NO + 45H^+
P4 => 4H3PO4 + 4(5e^-)
(1/4)P4 => H3PO4 + 5e^-
1 eqv wt in this redox process = [98g/5]=19.6g. => Normality of a 0.20M H3PO4 solution = (0.20mol/L)(98g/mol)/(19.6g/eqv wt) = 0.10eqv wts/Liter = 0.10N H3PO4