A sulphuric acid solution has pH = 3. Its normality is:
(a) ${{10}^{-3}}$
(b) $5\times {{10}^{-3}}$
(c) $5\times {{10}^{-4}}$
(d) ${{10}^{-2}}$

Hint: Sulphuric acid having pH=3 is acidic in nature. First, find the molarity using this pH and then calculate the normality using the molarity.

Complete step by step answer:
First, we should know what pH is. pH gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic and above 7 it represents basic.
Now, from the statement, we know that sulphuric acid having pH = 3 is acidic in nature.
 Normality may be defined as the no moles of gram equivalents of the solute dissolved in 1litre(1000ml) of the solvent and in order to find this, we will first have to calculate the molarity(no of moles the solute to the total no of moles of the solvent dissolved in 1litre of the solvent). we can easily find the molarity, as we know the pH of the sulphuric acid and from that, we can easily calculate the concentration of hydrogen and then can easily calculate its molarity.
 As we know, pH = 3
Taking log both sides, we get
log pH= log 3
$[{{H}^{+}}]=\dfrac{1}{3}$
$[{{H}^{+}}]=1\times {{10}^{-{{3}^{{}}}}}$mol/l
Calculation of the molarity:
Consider the following reaction:
${{H}_{2}}S{{O}_{4}}\to 2{{H}^{+}}+SO_{4}^{2-}$
Molarity=$\dfrac{\text{no of moles of the solute }}{\begin{align}
 & \text{total no of moles dissolved in 1 litre} \\
 & \text{ of the solvent } \\
\end{align}}$

Molarity= $\dfrac{1\times {{10}^{-3}}}{2}$
= $\dfrac{1}{2000}M$
Normality and molarity are related to each other as:
 Normality = 2 × Molarity (for ${{H}_{2}}S{{O}_{4}}$)
=$\dfrac{2}{2000}$
 =$\dfrac{1}{1000}N$
 =$1\times {{10}^{-3}}$N
i.e. =${{10}^{-3}}N$
So, the correct answer is “Option A”.

Note: Molarity and normality are two different words. Molarity may be defined as the number of moles of the solute dissolved in 1 liter of the solution whereas normality may be defined as the number of moles of gram equivalent of the solute dissolved in 1 liter of the solution.