Normality
Defination
The strength of solution measured in terms of gram equivalent per liter is called Normality.
The number of gram or mole equivalents of solute present in one Liter of a solution.
A solution having 1 gram equivalent of the dissolved solute in 1 liter of its solution is called 1 Normal solution.
Normality Depend On Two Factor
1. Dilution
2. Temperature
Formula
\[N=\frac{Number\;Of\;Grams\;Equivalent}{Volume\;of\;solution\;in\;liter}\]
\[N=\frac{Weight\;Of\;Subs/Equivalent\;Weight}{Volume\;in\;ml/1000}\;\;\]
\[N=\frac{Weight\;Of\;Subs\tan ce\times1000}{Volume\;in\;ml\times\;Equivalent\;weight}\;\;\]
OR
\[N=\frac{Molarity\times\;Molar\;Mass}{Equivalent\;Weight}\;\;\]
OR
N= Molarity × Acidity
N = Molarity × basicity
Uses
Normality is often used in the Acid-Base titration.
Ex. when titrating a solution of hydrochloric acid (HCL) with sodium hydroxide(NaOH) the normality of the acid and base are considered.
Calculation
1.) 2 Normal Solution How Much Gram NaOH Dissolve in 1 Liter water will become 2 Normality solution ?
N = 2
Volume:1 Liter = 1000 ml
Weight = ? Gram
First define equivalent weight of NaOH,
Molar mass of NaOH = 23 g/mol (for Na) + 16 g/mol (for O) + 1 g/mol (for H)
= 40/1
Eq = 40 g/mol
\[N=\frac{Weight\;Of\;Subs\tan ce\times1000}{Volume\;in\;ml\times\;Equivalent\;weight}\;\;\]
\[2=\frac{Weight\;Of\;Subs\tan ce\times1000}{1000\times\;40}\;\;\]
2× 40 = Weight
Weight = 80 Gram
02.) Calculate Normality 98 Gram of H2SO4 Dissolve in 5 Liter ?
Weight= 98 Gram
Equivalent Weight = 98/2 =49 g/mol
Volume = 5 Liter =5000 ml
\[N=\frac{Weight\;Of\;Subs\tan ce\times1000}{Volume\;in\;ml\times\;Equivalent\;weight}\;\;\]
\[N=\frac{98\times1000}{5000\times\;49}\;\;\]
Normality= 0.4 N
03.) How much Water will add to 1000 ml of 0.5 N Na2CO3 to get 0.1N Solution ?
N1 V1 = N2 V2
N1 = 0.5 N
V1 = 1000 ml
N2 = 0.1 N
V2 =?
N1V1=N2V2
0.5 × 1000 = 0.1 × V2
V2 = (1000×0.5)/0.1
V2 = 5000 ml
