Question

If 20 mL, 2M ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ is diluted to 1 litre. Normality of the resulting solution.

Answer 1

Hint: Molarity is defined as the ratio of the number of moles of the solute to the volume of the solution in litres. Whereas normality is defined as the ratio of the gram equivalent of the solute to the n-factor or acidity or basicity of the molecule.

Complete answer:
-It is given in the question that molarity of the sulphuric acid is 2M, the volume of the sulphuric acid is 20 mL, the volume of the solution is 1`L.
-So, here we have to find the molarity of the solution, let the molarity is equal to the \[{{\text{M}}_{1}}\text{ }\times \text{ }{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}\text{ }\times \text{ }{{\text{V}}_{2}}\].
-So, it is given in the question that the value of the \[{{\text{M}}_{1}}\] is 2M, \[{{\text{V}}_{1}}\]is equal to the 0.02 L and the value of \[{{\text{V}}_{2}}\] is 0.02 L so by using the formula of the relation between molarity and normality we will get
\[{{\text{M}}_{1}}\text{ }\times \text{ }{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}\text{ }\times \text{ }{{\text{V}}_{2}}\]
\[\begin{align}
  & \text{2 }\times \text{ 0}\text{.02 = }{{\text{M}}_{2}}\text{ }\times \text{ 1} \\
 & {{\text{M}}_{2}}\text{ = 0}\text{.04M} \\
\end{align}\]
-Now, we know that the normality of the solution can be calculated by the formula:
$\text{N = M }\times \text{ n}$

-Here, N is the normality of the solution, M is the molarity of the solution and n is the n-factor or no. of hydrogen ion or hydroxyl ion released by the compound.
-So, when sulphuric dissociates then it gives 2 electrons, we can see it through the reaction:
${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{ = 2}{{\text{H}}^{+}}\text{ + SO}_{4}^{-2}$
-So, we can see that a total of 2 electrons are participating in the reaction so the value of n will be 2.

Now, the normality will be:
$\text{N = 0}\text{.04 }\times \text{ 2 = 0}\text{.08N}$
Therefore, the normality of the solution in which the 20 mL of 2M ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ is diluted in 1-litre water is 0.08M.

Note: While determining the value of n we should remember that the acidic species will yield only hydrogen ion so we have to calculate the no. of hydrogen ions whereas the basic species will yield hydroxyl ion in the reaction so we have to calculate no. of hydroxyl ions.