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Question: What is the normality of 0.2M \[{H_3}P{O_3}\] ​ solution? A.0.2 N B.0.1 N C.0.4 N D.0.6 N...

What is the normality of 0.2M H3PO3{H_3}P{O_3} ​ solution?
A.0.2 N
B.0.1 N
C.0.4 N
D.0.6 N

Explanation

Solution

We need to know the concept of Basicity of an acid to solve this question. It is the number of hydrogen ions which can be produced from one molecule of the acid. For example Acetic acid is acid which means that as it can lose one proton or hydrogen atom, it forms an acetate ion. H3PO3{H_3}P{O_3} is a dibasic acid which is evident from its structure.

Complete answer:
From the figure given above we can say that H3PO3{H_3}P{O_3} is a dibasic acid because both of the OH - OH group can easily lose its hydrogen as H+{H^ + } ions making it acidic
We also know that normality can be expressed by the formula
Normality=Molarity×n factor\Rightarrow Normality = Molarity \times n{\text{ }}factor
Where n factor is the number of replaceable H+{H^ + } ions in that molecule.
Since we are given H3PO3{H_3}P{O_3} we can say that the value of n factor will be 2
It is also given to us that the molarity of the given solution is 0.2 M.
Thus we can write that
Normality=0.2M×2\Rightarrow Normality = 0.2M \times 2
Normality=0.4 N\Rightarrow Normality = 0.4{\text{ }}N
Thus we can say that the normality of the given solution of 0.2M H3PO3{H_3}P{O_3} is 0.4 N
Thus the correct option is C .

Note:
For acids we can say that n factor is the number of replaceable H+{H^ + } ions in that molecule.
For bases we can say that n factor is the number OHO{H^ - } of ions that the molecule of base would give when dissolved in a suitable solvent.
For Redox reactions n factor will be the change in the oxidation number of the reducing or oxidising agent