What is the normality of 2.5 M sulphuric acid?

Given data

• The molarity$\left(M\right)$ of Sulphuric acid$\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\right)}$ solution is $\text{2.5 M}$

Formula used

• Normality in terms of molarity is as follows:
• $$\begin{gathered} \mathrm{Normality}\left(\mathrm{N}\right)=\mathrm{No}.\mathrm{of}{\mathrm{e}}^{-}\mathrm{transfer}\left(\mathrm{n}\right)\times \mathrm{Molarity}\left(\mathrm{M}\right) \\ \mathrm{N}=\mathrm{n}\times \mathrm{M} \end{gathered}$$

Definition and reaction used

• Normality: It is defined as the number of one mole of gram equivalent of solute present in one litre of the solution.
• The number of gram equivalent is equal to the number of moles of a reactive unit of the given compound.
• The chemical reaction will dissociate as follows:
• $$\begin{gathered} {\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{HSO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right) \\ {\mathrm{HSO}}_4^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right) \end{gathered}$$
• Thus, the number of moles of the reactive unit$\left(\mathrm{n}\right)=2$ and the number of electrons transferred is also 2.

Calculating the normality

• Substituting all the values we get,
• $$\begin{gathered} \mathrm{N}=\mathrm{n}\times \mathrm{M} \\ \mathrm{N}=2\times 2.5 \\ \mathrm{N}=5 \end{gathered}$$

Therefore, the normality of $\text{2.5 M}$ Sulphuric acid is $5\mathrm{N}$.