NORMAL SOLUTION
Section I
A normal solution is a solution that contains an equivalent weight of solute in one liter of solution. The
mathematical expression is given in Eq. 1.
Eq. 1
Normality (N) =
equivalents
liters
or Normality (N) =
Number of equivalents
number of liters
Example 1
49 grams of sulfuric acid (H2SO4) were dissolved in 1 liter of solution. Calculate the normality of the
solution.
Solution
From Eq. 1
N=
equivalents
liters
The equivalent weight of H2SO4 is equal to its molecular weight divided by the number of replaceable
98
hydrogens; that is:
 49.
2
In this solution we have one equivalent weight of the acid dissolved in one liter of solution; therefore,
the normality of the solution is 1.
The number of equivalents of a solute is equal to grams of the solute divided by its equivalent weight.
The mathematical expression is given in Eq. 2.
Eq. 2
Equivalents =
grams of solute
equivalent weight of solute
Example 2
How many equivalents are there in 33.26 grams of phosphoric acid (H 3PO4)?
Solution
The equivalent weight of phosphoric acid is equal to its molecular weight divided by the number of
98
replaceable hydrogens.
 32.6
3
Substitute the values of Eq. 2
Equivalents =
grams of solute
equivalent weight of solute
thus, equivalents =
33.26 (grams of solute)
1
32.6 (equivalen t wt. of solute)
The number of liters is equal to the number of milliliters of solution divided by 1000. The
mathematical expression is given in Eq. 3.
Eq. 3
liters =
milliliters
1000
Substituting Eq. 2 and 3 into Eq. 1 gives Eq. 4.
Eq. 4
grams of solute
equivalent weight of solute
Normality =
milliliters of solution
1000
Rearranging Eq. 4 gives us Eq. 5.
Eq. 5
N=
(1000)(grams of solute)
(equivalent wt. of solute)(ml of solution)
Solving for grams of solute given in Eq. 5, Eq. 6 is obtained.
Eq. 6
Grams of solute =
( N)(eq. wt. of solute)(ml of solution)
1000
Example 3
5 grams of sodium hydroxide (NaOH) were dissolved in enough water to give 200 ml of solution.
Calculate the normality of the solution.
Solution
The equivalent weight of NaOH is equal to its molecular weight divided by the number of replaceable
hydroxide ions. That is the equivalent weight of NaOH is the same as its molecular weight. Substitute the
values in Eq. 5.
N=
(1000)(5)
 .625
(40)(200)
Example 4
How many grams of calcium hydroxide, Ca(OH)2 are required to prepare 150 ml of .15 N Ca(OH) 2
solution?
Solution
The equivalent weight of calcium hydroxide, Ca(OH)2 is equal to its molecular weight divided by the
number of replaceable hydroxyi ions. Hence, the equivalent weight of calcium hydroxide, Ca(OH) 2 is equal
74
to
 37.
2
Substituting the values in Eq. 6.
Grams of solute =
(.15)(37)(150)
 .83
1000
NORMALITY
Section II
1.
98 g of H2SO4 were dissolved in enough water to form 1 liter of solution. Calculate the normality
of the solution.
2.
6.3 g of HNO3 were dissolved in enough water to form 1 liter of solution. Calculate the normality
of the solution.
3.
.98 g of H3PO4 were dissolved in enough water to give 100 ml of solution. Calculate the normality
of the solution.
4.
How many grams of NaOH are required to prepare 200 ml of .05 normal NaOH solution?
5.
5 g of NaOH are dissolved in enough water to form 112 g of solution whose density is 1.08 g/ml.
Calculate the normality of the solution.
6.
How would you prepare 120 ml of .15 N Ca(OH) 2 solution?
7.
19.6 g of H2SO4 were dissolved in enough water to form 200 ml of solution. Calculate the
normality and the molarity of the solution.
Section 1 Normality
1.
2N
2.
.1 N
3.
.3N
4.
.4 grams
5.
1.2 N
6.
Dissolve .67 g of Ca(OH)2 in 120 ml of solution
7.
(a) 2N
(b) 1M
Equivalent Weights
Section VII
1.
Calculate the equivalent weight of
a.
H2SO4
B. HCl
c.
H3PO4
e.
NaOH
d.
Ca(OH)2
f.
Al(OH)3
2.
3.245 g of metal replaced .058 g of hydrogen. Calculate the Eq. Wt. of the metal.
3.
11.33 g of Fe are contained in 16.2 g of the iron oxide. Calculate the Eq. Wt. of Fe.
4.
Determine the Eq. Wt. of tin in:
a.
b.
an oxide of tin that contains 88.12% tin
an oxide of tin that contains 78.77% tin
Section VII Equivalent weights
1.
(a) 49
2.
55.95
3.
18.61
4.
(a) 59.35
(b) 36.5
(b) 29.68
(c) 32.67
(d) 37
(e) 40
(f) 26
TITRATION
Section III
Titration is the volumetric determination of the concentration of unknown substances by the addition of a
standard solution until the end point is reached. The end point is the point at which the color of the
indicator changes. At the end point of the normality of acid times the milliliters of the acid is equal to the
normality of the base times the milliliter of the base. The mathematical expression is given in Eq. 1.
Eq. 1
NAmlA = NBmlB
Example 1
How many milliliters of .1 N NaOH are required to neutralize 50 ml of .05 N HCl?
Solution
Substitute the values given in Eq. 1
NAmlA = NBmlB
(.05)(50) = (.1)(mlB)
divide both sides of the equation by .1
.05  50
 ml B
.1
25 = mlB
Example 2
How many ml of .2M H2SO4 are required to neutralize 50 ml of .1M sodium hydroxide (NaOH)?
Solution
Note that the strength of the acid and that of the base are expressed in molarity. The first step is to
express the strength of the acid and the strength of the base in normality. Sulfuric acid (H 2SO4) has two
replaceable hydrogen ions; hence, the normality is twice the molarity, the normality and molarity of the
sodium hydroxide are the same, because there is one replaceable hydroxide ion.
Substitute the values in Eq. 1.
(.4)(mlA) = (50)(.1)
Divide both sides of the equation by .4.
50.1
.4
mlA = 12.5
mlA =
It is important to remember the difference between the normality and molarity of a solution. Normality
(N) is defined as equivalents per liter. We can also define normality in terms of milli-equivalents per
milliliter. If we divide the equivalent weight by 1000, we end up with milliequivalents. If we divide the
liters by 1000, we end up with milliliters just as the molarity can be defined in terms of millimoles divided
by milliliters. Milliequivalents is the product of (ml)(N); hence, at the end point of titration the
milliequivalents of an acid is equal to the milliequivalents of base.
TITRATION
Section IV
1.
How many ml of .05 N H2SO4 are required to neutralize 100 ml of .1 N Ca(OH)2?
2.
How many ml of 1.5 N NaOH are required to neutralize 25 ml of .25 N of HCl?
3.
How many ml of .05 M H2SO4 are required to neutralize 40 ml of .05 M NaOH?
4.
How many milliequivalents of NaOH are required to neutralize 45 ml of .05 N HCl?
5.
.5 grams of NaOH were dissolved in enough water to form 25 ml of solution. How many ml of
this solution are required to neutralize 100 ml of .2 N HCl?
Section IV Titration
1.
200 ml
2.
4.17 ml
3.
20 ml
4.
2.25 meq.
5.
40 ml
DILUTION I
Section V
Dilution is the process of decreasing the concentration of a solution by adding water. In dilution, the
milliequivalents of the solution remains the same while the volume of the solution changes. The
mathematical expression for the dilution of an acid solution is given in Equation 1.
Eq. 1
mlA1  NA1 = mlA2  NA2
The mathematical expression for the dilution of a base solution is given in Eq. 2.
Eq. 2
mlB1  NB1 = mlB2  NB2
Example 1
50 ml of .1N H2SO4 were diluted to 150 ml with water. Calculate the normality of the diluted solution.
Solution
50  .1 = 150 (NA2)
divide both sides of the equation by 150
.1  50
= (NA2)
150
.03 = NA2
Example 2
20 ml of .1M Ca(OH)2 were diluted to 40 ml Calculate the normality of the diluted solution.
Solution
The strength of the concentrated solution is expressed in molarity and the strength of the diluted
solution is expressed in normality. It is important in dilution problems to express the concentration of both
solutions in the same units. It is reasonable then to change .1M Ca(OH) 2 to .2N and substitute the value in
Eq. 2.
mlB1  NB1 = mlB2  NB2
(20) (.2) = (40) (NB2)
Divide both sides of the equation by 40
Therefore NB2 =
NB2 = .1
20.2
40
DILUTION II
Section VI
1.
10 ml of 5 N HCl were diluted with water to give 500 ml of solution. Calculate the normality of
the diluted acid.
2.
Concentrated HCl is 12 N. How many ml of the concentrated HCl are needed to prepare 100 ml of
.2 N?
3.
Concentrated H2SO 4 is 18 M. How many ml of the concentrated H2SO4 are needed to prepare 500
ml of .36 N?
4.
Concentrated H2SO4 has a density of 1.8 g/ml and is 98% H2SO4 by weight. How many ml of the
concentrated solution are needed to prepare 1 liter of .72 N solution.
Section VI Dilution
1.
.1N
2.
1.67 ml
3.
5 ml
4.
20 ml