Question

What is the normality of 2.5 M sulphuric acid

Answer 1

Given:

The molarity of the sulphuric acid solution = 2.5 M

To Find:

The normality of the given sulphuric acid solution.

Calculation:

- Sulphuric Acid (H₂SO₄) dissociates as follows:

H₂SO₄ ⇄ 2H⁺ + SO4²⁻

- So, for sulphuric acid , the value of n = 2

- We know the relation between normality and molarity as:

N = n × M

⇒ N = 2 × 2.5

⇒ N = 5 N

- So, the normality of the given sulphuric acid solution is 5 N.

Answer 2

The value of normality for sulfuric acid is 5 N.

Explanation:

The extent of dissociation of sulfuric acid into ions is given.

H2SO4 ⇄ 2H⁺ + SO4^2-

Here the value of n = 2 for the H2SO4.

N is the number of equivalents of sulfuric acid per mole.  

The relation between normality and molarity is given below.

• N = n × M
• N = 2 × 2.5
• N = 5 N

Thus the value of normality for sulfuric acid is 5 N.