NORMALITY
The normality ( N) of a solution expresses the number of
equivalents of solute contained in one liter of solution.
Or the number of milliequivalents of solute contained in
one milliliter of solution.
For example 0.2 N hydrochloric acid solution contains
0.2 equivalent ( eq ) of this solute per liter.
Equivalents of solute
Normality =
N=
‫ــــــــــــــــــــــــــــــــــــــــــــــ‬
Liters of solution
Equivalent = Wt. g
M.W
Example:
What is the equivalent weight of these acids and buses:
a) HNO3
b) H2SO4
c) Mg(OH)2
Answer:
In each case we divide the formula weight (
consider M.wt) aby the number of H+ or OH- ions
produced by the acid or base:
63
a)
HNO3 : Eq.wt. =
63
=
1
b) H2SO4 : Wq.wt.
=
98
=
49
2
c) Mg ( OH)2 : Eq.wt. =
58
2
=
29
Example 2 :
What is the normality of a solution made by
dissolving 8.5g of H2SO4 in inough water to
malce 500ml of solution ?
Answer:
First, we need the number of equivalents of H2SO4
per liter of solution, and we know (from the above
example) that the equivalent weight of H2SO4 is 49.
Thus: Equivalent of
H2 SO4 = weight ( g)
= 8.5 = 0.17
Eg . wt
49
To find normality N we divide equivalents pf H2SO4
by liters of solution: (Note: 500ml = 0.5L)
N=
0.17
0.5
= 0.34 N.
** The relationship between Normality and Molarity:
Where n = the number of replaceable H+ or OHper molecule ( for acids and bases )
Or n the number of electrons lost or gained per
molecule ( for oxidizing and reducing agents ).
The molarity and normality are related by :
N = nM
Example:
What is the normality of the 0.01M solution of
H2SO4?
Answer:
N = nM in
H2SO4 n = 2
: N = 2 x 0.01 = 0.02 N
Answer 1:a) M = mol
liter
mole =
0.04 = mol
0.5
0.02 =
g .
mwt
moles = 0.5 x 0.04 = 0.02
g
40
g = 40 x 0.02 = 0.8 g
We need 0.8 g to dissolve it in water and dilute to 500
ml to get the desired conc. 0.04M NaOH.
b) * NaOH contains one OH per molecule, thus :
M = N cind the solution is 0.04N
* The solution contains 0.8g in the 500 ml
1.6 g /liter
* %(w/v) = g per 100 ml
1.6 g/
100 ml
liter = 1.6/ 1000 ml = 0.16 g /
0.16 %
* mg % = mg per 100ml
0.16 g /100
160 mg / 100 ml =
160 mg %
Example 2 :
- How many ml of 5M H2 SO4 are required to
made 1500 ml of a 0.002 M H2 SO4 solution ?
Answer 2:- C x V (for conc. ) = C x V (for dilute)
5 M x V = 0.002M x 1500 ml
V = 0.002 × 1500 = 0.6 ml
5
We need 0.6 ml of the conc. Solution and dilute it
to 1.5 Kuters.