Question

Define normality.
\[6.3gms\] of oxalic acid is present in \[500ml\] of solution its normality is:

Answer 1

Hint: We must know about the word normality. Generally, normality is a quantitative measurement for chemical solutions. We can use the below formula to determine the normality of the solution.
\[Normality{\text{ }}of{\text{ }}solution{\text{ }}N = \dfrac{{Weight}}{{equivalent{\text{ }}wt.}} \times \dfrac{{1000}}{{volume(ml)}}\]

Complete step by step answer:
The concentration of any solution that is acknowledged in terms of gram equivalents is the normality of the solution.
We can use normality as the number of mole equivalents per liter of solution. Usually, the normality of the solution is represented by N or equivalent per liter (\[eq/L\])units.
We know that the chemical formula of oxalic acid is \[{C_2}{H_2}{O_4}.2{H_2}O\]
Ans also the molecular mass of\[{C_2}{H_2}{O_4}.2{H_2}O\] =\[126g\]
Now, to calculate normality, we will use below formula
\[Normality{\text{ }}of{\text{ }}oxalic{\text{ }}acid,N = \dfrac{{Wt.{\text{ }}of{\text{ }}oxalic{\text{ }}acid}}{{equivalent{\text{ }}wt.}} \times \dfrac{{1000}}{{volume(ml)}}\]
To solve this, we must have a value of the equivalent weight of oxalic acid.
So, equivalent weight of Oxalic acid = \[\dfrac{{molecular{\text{ }}mass}}{{Basicity}}\]
∴ equivalent weight of Oxalic acid = \[\dfrac{{126}}{2}\] =\[63{\text{ }}gm/eq\].
Now, substituting the values, in the formula of normality equation, we get

Normality= \[\dfrac{{6.3}}{{63}} \times \dfrac{{1000}}{{500}} = 0.1 \times 2{\text{ }} = 0.2{\text{ }}gm.mole{\text{ }}per{\text{ }}liter\]
Hence, the Normality of oxalic acid is\[0.2{\text{ }}gram{\text{ }}moles{\text{ }}per{\text{ }}litre\].
Additional information:
We can use normality instead of molarity because often \[1{\text{ }}mole\]of acid does not neutralize \[1{\text{ }}mole\]of base.
Below given are different formulas that are used to calculate Normality of a solution depending upon given data.
i.\[Normality{\text{ }} = {\text{ }}Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents{\text{ }} \times {\text{ }}{\left[ {volume{\text{ }}of{\text{ }}solution{\text{ }}in{\text{ }}litres} \right]^{ - 1}}\]
ii.\[Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents{\text{ }} = {\text{ }}weight{\text{ }}of{\text{ }}solute\; \times {\text{ }}{\left[ {Equivalent{\text{ }}weight{\text{ }}of{\text{ }}solute} \right]^{ - 1}}\]
iii.\[N{\text{ }} = {\text{ }}Weight{\text{ }}of{\text{ }}Solute{\text{ }}\left( {gram} \right){\text{ }} \times {\text{ }}\left[ {Equivalent{\text{ }}weight{\text{ }} \times {\text{ }}Volume{\text{ }}\left( L \right)} \right]\]
iv.\[N{\text{ }} = {\text{ }}Molarity{\text{ }} \times {\text{ }}Molar{\text{ }}mass{\text{ }} \times {\text{ }}{\left[ {Equivalent{\text{ }}mass} \right]^{ - 1}}\]
v.\[N{\text{ }} = {\text{ }}Molarity\; \times {\text{ }}Basicity\; = {\text{ }}Molarity{\text{ }} \times {\text{ }}Acidity\]

Note:
We can use normality to determine the concentrations of the solution in acid-base titration chemistry. For example, we can use normality to determine the number of ions that will get precipitated in precipitation reactions. Also in redox reactions to determine the number of electrons that a reducing or an oxidizing agent can donate or accept.