Question

What is the normality of $2.5M$ sulphuric acid.

Answer 1

Normality is a method of measuring the concentration of any solution. Since sulphuric acid is an acid, its dissociation into its ions can be used to determine the mole equivalents or the $n - factor$ of sulphuric acid.

Complete answer:
The concentration of a solution tells us to what extent a solvent contains the solute. A concentrated solution has a high amount of solute dissolved in it and a dilute solution is less concentrated and has a lesser amount of solute and an excess of solvent.
There are various methods of measuring the concentration of a solution. Two such methods are molarity and normality that are interrelated to one another.
Molarity is the number of moles of the solute that are dissolved in one liter of the solution. Normality is the number of gram equivalents or mole equivalents of the solute dissolved in one liter of the solution.
Sulphuric acid is a strong diprotic acid that dissociates to give two moles of protons (hydrogen ions) and one mole of sulphate ions per mole of the acid.
${H_2}S{O_4}(aq) \to 2{H^ + } + S{O_4}^{2 - }$
The normality can be related the molarity of a solution by the following formula:
$Normality = n - factor \times molarity$
The $n - factor$ of a mineral acid is equal to its proticity i.e. number of protons released per mole of an acid. Since sulphuric acid is a diprotic acid its proticity is equal to two.
Thus the $n - factor$ of sulphuric acid is $2$ .
By inserting the value of $n - factor$ in the formula for normality we get,
$Normality = 2 \times 2.5N$
$Normality = 5N$
Hence the normality of $2.5M$ is $5N$ .

Note:
The $n - factor$ is the measure of the number of electrons involved in a reaction per molecule. There are different methods of calculating it depending upon the type of reaction. In redox reaction we actually calculate the number of electrons exchanged between the oxidizing and reducing species, but in an acid it can be calculated by simply calculating the proticity.