# Normality Definition , Formula , Formality Formula, Solved Examples

## NORMALITY

It is the number of gram equivalent present in per litre solution. It is used for acid or base solution

where n =  number of H+ in Acid  , OH- in base and for salt,charge present in ionic forms
Its unit is N or eq /L .It depends on the chemical reaction being studied and depends on temperature too.
This is used in acid-base reactions and redox reactions

## Difference Between Molarity and Normality

 Molarity Normality number of moles of solute per liter of solution. number of gram equivalent of solute per liter of solution. Molarity is a measurement of the moles in the total volume of the solution, Normality  is a measurement of the gram equivalent in relationship to the total volume of the solution. Unit is Moles L-1 Unit is eq L-1

## Relation between Normality and Molarity

Here is Normality formula in terms of molarity
$Normality = n \times Molarity$
where n =  number of H+ in Acid  , OH- in base and for salt,charge present in ionic forms

## Solved  Examples

Question 1
Calculate the equivalent weight of following:-
(i)H2 SO4
(ii)NaCl
(iii) NaOH
(iv) Na2 CO3
Solution
i) Molar Mass of H2 SO4=98
n = 2 as it has 2 H+ ion
$Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {98}{2} = 49$

ii)   Molar Mass of NaCl =23+35.5= 58.5
n = 1
$Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {58.5}{1} = 58.5$

iii)
Molar Mass of NaOH =40
n = 1 as it has 1 OH- ion
$Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {40}{1} = 40$

iv)
Molar Mass of Na2 CO3=106
The salt Na2 CO3 ionizes to form 2Na+ and CO3 (–2). So, the charge present on both is 2
n = 2
$Equivalent \; Weight= \frac {Molar \; Mass}{n} = \frac {106}{2} = 53$
Question 2
Calculate the normality of NaOH solution Formed by dissolving 0.2 gm Na OH to make 250 ml solution
Solution
Normality(N) = (no.Gram Equivalent of solute)/(Volume of Solution in litre)
No. of Gram Eq. of Solute = weight/Equivalent weight
Now, Equivalent weight=$\frac {Molar \; Mass}{n} = \frac {23+16+1}{1}=40$
So, N = (No.gram eq.mass)/(Vol (l)
=Weight/(Equivalent weight)  ×  1000/(V ml)
=(2/40) X (1000/250)
= .2 N

Question 3
Calculate the number of moles & molarity of N/2   500 ml solution of H2 SO4
Solution
N = n ×M
1/2=2 ×M
1/4=M
Now Molarity of solution
= No of Moles/Volume in litre /> 1/4=n/.5
or n = 1/8

## Normality equation

To calculate the volume of a definite solution required to prepare solution of other Normality, the following equation is used:
N1 V1 = N2 V2
where N1= initial Normality
N2= Normality of the new solution
V1= initial volume
V2= volume of the new solution

Suppose we have mix three solution of same solute of Normality N1 ,N2 ,N3 and Volume V1,V2 ,V3
Normality of the resulting solution is given by
$N_R= \frac {N_1V_1 +N_2V_2 + N_3V_3}{V_1 + V_2 + V_3}$
Suppose we have mix three solution of different solute(acid) of Molarity M1 ,M2 ,M3 , Volume V1,V2 ,V3
and no of H+ as n1 ,n2 ,n3 Normaility of the resulting solution is given by
$N_R= \frac {n_1M_1V_1 +n_2M_2V_2 + n_3M_3V_3}{V_1 + V_2 + V_3}$

Question 1
Find out the volumes of two HCl solutions P (0.5 N) and Q (0.1 N) to be mixed for preparing 2 L of 0.2 N HCl solution
Solution
Let y L of P and (2 -y) L of Q are mixed.
N1 V1 + N2 V2= NV
$0.5 \times y + 0.1 \times (2 -y) = 0.2 \times 2$
$(0.5 -0.1) y = 0.4 -0.2$
$y = 0.5 L$
Therefore,
0.5 L of A and 1.5 L of B should be mixed.

## Acid-Base Titration equation using Normality

In titration, acid and base react with each other and neutralize the solution.The below equation can be used to find the Normality of the acid and base in titration
Na Va= NbVb
Where
Na =Normality of Acid solution
Va= Volume of Acid solution
Nb=Normality of base solution
Vb= Volume of base solution

## FORMALITY (F)

Formality is used for ionic compounds solution like NaCl It is the number of moles of solute (ionic compound) present in 1 litre solution
Molar mass of ionic compound is known as formula Mass i. e. (FM)
So it is the number of Formula mass in gram of solute in 1 litre solution unit = ‘F’ or ‘Moles L-1

### Quiz Time

Question 1 1 M of fe(OH)3 Solution is A) 2 N
B) 3 N
C) 1 N
D) .333 N
Question 2 .5 gram equivalent of H2S is equal to ?
A) .25 Moles of H2S
B) 1 moles of H2S
C) .05 moles of H2S
D) None of the above
Question 3 300 ml 0.2 M HCl and 200 ml of 0.03M H2SO4 are mixed. The normality of the resulting mixture will be
A) .044 N
B) .72 N
C) .84 N
D) .144 N
Question 4 Find the Normality of the solution containing .5 gm of NaOH in 1L solution
A) .0125 N
B) .125 N
C) .5 N
D) .0250 N
Question 5 how much water is to be added to prepare a .25N HCL solution from .5 N HCL 1 L solution
A) 500 ml
B) 100 ml
C) 1000 ml
D) 250 ml
Question 620 ml of a 0.125 N HCl solution were neutralised by 25 ml of a KOH solution. What is the normaility of KOH solution
A) .01N
B) .1 N
C) .2 N
D) None of the above