Normality Formula

Normality Formula - Normality Definition and Formula

Normality like molarity is a way of expressing the concentration of solute in a solution. If you can recollect, molarity was defined as the number of moles (n) of solute present in 1lt. of the solution. Now to define ‘normality’ or ‘equivalent concentration’, instead of using number of moles (n), we will use number of equivalents (eq) of solutes.
So, Normality (N) = \[\frac{{{\text{Number}}\,{\text{of}}\,{\text{equivalents}}}}{{{\text{Volume}}\,{\text{in}}\,{\text{litres}}}}\]
Now, calculating the number of equivalents is similar to calculating number of moles
\[{\text{Number}}\,{\text{of}}\,{\text{moles}}\,\left( n \right) = \frac{{{\text{given}}\,{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}};\]
\[{\text{Number}}\,{\text{of}}\,{\text{equivalents}}\,\left( {Eq} \right) = \frac{{{\text{given}}\,{\text{mass}}}}{{{\text{equivalent}}\,{\text{mass}}}}\]
The next problem is how do we find the Equivalent Mass or Equivalent Weight (E.W) or gram equivalent.
Equivalent Weight is defined as Molar mass divided by equivalence factor (f).
\[{\text{E}}.\,{\text{W}} = \frac{{{\text{Molar}}\,{\text{mass}}}}{{{\text{Valency}}\,{\text{factor}}\,\left( f \right)}}\]
This equivalence factor will differ for different species and also the type of reaction under consideration.
But let us understand this with a very easy example:

Case I:
Equivalent weight of NaCl (ionic salt)
The valency factor (f) of an ionic salt is always equal to the total charge on anion ( in this case) or cation ( in this case)
So if you can observe, the valency factor (f) for NaCl =1.
Hence its Equivalent Weight = \[{\text{E}}.\,{\text{W}} = \frac{{{\text{Molar}}\,{\text{mass}}}}{{{\text{Valency}}\,{\text{factor}}\,\left( f \right)}}\]
\[ = \frac{{56.5}}{1} = 56.5\]
Similarly, for MgCl2, valency factor (f) = 2
and E.W \[ = \frac{{95}}{2} = 47.5\]

Case II:
For acids or bases the valency factor (f) is equal to number of moles of \[{H^ + }\,\,{\text{or}}\,\,O{H^ - }\] ions 1 mole of acid or base can furnish in an aqueous medium.
e.g. 1 mole of H2SO4 will furnish 2 moles of ions, so its (f) = 2, E.W = 98/2 = 49 gm/eq
1 mole of NaOH will furnish 1 mole of ions, so its (f) = 1, E.W = 40/1 = 40 gm/eq
1 mole of H3PO3 will produce only 2 moles of ions (check the structure of acid), so its (f) = 2,
E.W = 82/2 = 41

So, what we can gather from here is that we need to know the acidity of the base and the basicity of an acid in order to determine the valency factor (f)

Case III:
For Redox Reaction, the (f) factor for any species is the total change in the oxidation state it has undergone in that reaction.
e.g. KMnO4 as an oxidising agent in acidic media
\[Mn{O^{4 - }} + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O\]
KMnO4 gained 5 electrons from reductant.
Therefore (f) = 5
Hence E.W = 158.04/5 = 31.61 grams/equivalent.

I know there is a lot to remember in this and it looks more tricky than calculating molarity, but with few examples and little practice you will get a hang of it!

Example: What is the normality of the following of 1lt. aqueous solution containing 80 gm of NaOH dissolved in it.
Solution:
Firstly, In order to determine the Normality (N), we need to find the number of equivalents.
Now we know that the (f) for NaOH = 1
E.W = 40/1 = 40gm
\[{\text{Number}}\,{\text{of}}\,{\text{equivalents}}\,\left( {Eq} \right) = \frac{{{\text{given}}\,{\text{mass}}}}{{{\text{equivalent}}\,{\text{mass}}}}\]
 = 80/40 = 2 Eq
Hence, N = \[\frac{{{\text{Number}}\,{\text{of}}\,{\text{equivalents}}}}{{{\text{Volume}}\,{\text{in}}\,{\text{litres}}}}\] = 2/1 = 2 Eq/lt.

Question: What is the normality of 0.1381 M NaOH?
Options:
(a) 0.5231 N
(b) 0.6123 N
(c) 0.7775 N
(d) 0.1381
Answer: (d)
Solution:
0.1381 mol/L x (1 eq/1mol) = 0.1381 eq/L = 0.1381 N
Since the Equivalent Weight of NaOH is equal to Molar Mass hence the molarity is equal to Normality in this case.