Law of Equivalence

In continuation to the previous blog on Redox reactions ( Haven't read it yet? Click here! ) , we will be reading about the :-

The Law of Equivalence

When do we use it? When balancing the equation ( and using mole concept) becomes too tedious of a task

In any reaction :-

A + B -----> C + D

Equivalence of A = Equivalence of B = Equivalence of C = Equivalence of D

But, what is Equivalence?

Number of Equivalence = Given mass

Equivalent mass

Equivalent mass = Molecular mass

n-factor

By combining both the above equations , we get:-

Number of equivalence = No. of moles * n-factor

How to find n-factor?

(I) ACIDS

n factor = No. of replaceable 'H' or number of Hydrogen replaced in the reaction

= Basicity of acid

Example -- in H2SO4 , there are 2 replaceable hydrogens , hence n factor =2

(II) BASES

n factor = In a reaction, number of OH replaced

= Acidicity of base

Example -- NaOH, there is 1 replaceable hydroxide, hence n factor =1

(III) IONS

n factor = Total charge on cation or anion

(IV) SALTS

n factor = Total positive charge or negative charge displaced per molecule

(V) OXIDISING OR REDUCING AGENT

n factor = It is the number of electrons involved per mole of oxidising agent or reducing agent

SHORT FORMS TO REMEMBER

In a reaction

1) Equivalence of A = Equivalence of B

2) Moles * n factor of A = Moles * n factor of B

3) Given mass of A/ Equivalent mass of A = Given mass of B / Equivalent mass of B

4) Normality * Volume of A = Normality * Volume of B

5) Molarity * Volumes (in lit) * n factor of A = Molarity * Volumes * n factor of B

A very short blog, I know, but this topic, by itself, carries humungouse weightage in both competitive and board exams so wanted to dedicate a seperate page to it

Again another chem meme to end this blog in a good note