Question

Find the normality of oxalic acid solution containing 63 g of crystalline oxalic acid in 500 ml of solution.

Answer 1

Hint- In order to find the normality, first we will determine oxalic acid equivalent mass, then we proceed further by using the normality formula which is mentioned in solution as well. We will use the data given and use the formula for volume in ml.

Complete answer:
Given that oxalic acid solution contains 63 g of crystalline oxalic acid in 500 ml of solution.
We know that
Normality of any solution is given as:
$\Rightarrow {\text{Normality}} = \dfrac{{{\text{Given weight}} \times 1000}}{{{\text{Equivalent weight}} \times {\text{Volume in }}ml}}$
Now we will calculate the equivalent weight of oxalic acid.
As we know that the equivalent weight for an acid is given as:
$\Rightarrow {\text{Equivalent mass}} = \dfrac{{{\text{Molecular mass}}}}{{{\text{Basicity}}}}$
Oxalic acid equivalent mass

$$\Rightarrow {\text{Equivalent mass of oxalic acid}} = \dfrac{{{\text{Molecular mass of oxalic acid}}}}{{{\text{Basicity of oxalic acid}}}} \\\ \Rightarrow {\text{Equivalent mass of oxalic acid}} = \dfrac{{126}}{2} = 63gm/eqiv \\\$$

Substitute the value of equivalent mass and given volume in above formula, we get the normality as:
Normality:
$\because {\text{Normality}} = \dfrac{{{\text{Given weight}} \times 1000}}{{{\text{Equivalent weight}} \times {\text{Volume in }}ml}} \\\ = \dfrac{{63 \times 1000}}{{63 \times 500}} \\\ = 2N \\\$

Hence the value of normality of oxalic acid solution is 2N.

Note- Normality is described as the number of gram or mole equivalents of solute present in one liter of a solution. When we say equivalent, it is the number of moles of reactive units in a compound. Normality is used to measure the concentration of a solution. It is mainly used as a measure of reactive species in a solution and during titration reactions or particularly in situations involving acid-base chemistry.