Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1. 2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube? 1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2. Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3. Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen 2SO 2 + O 2 → 2SO 3 1cm 3 of O 2 = 2cm 3 20cm 3 of O 2 = 2 x 20 = 40cm 3
A standard solution
is a solution which has a known concentration, which means the amount in mole
or mass concentration of the substance which is dissolved in a known volume of
a solvent is known.
If you want to
prepare a standard solution of an acid, the knowledge of the relationship
between amount, molar concentration and volume is very necessary
C = n/V
Where: n = amount (mole)
V =
volume (dm3)
C = molar
concentration (mole per dm3)
Experiment has shown that
C1V1
= C2V2
Since the amount of solute in a solution is not affected by
increase in the volume of the solvent. This formula becomes the basic principle
of dilution. It is also known as dilution formula.
Information required
for preparation of standard solution of acid is listed below:
·
Specific gravity of the acid
·
Percentage purity of the acid
·
Molar mass of the acid
·
Concentration of the required dilute acid
·
Volume of the dilute solution required
Specific gravity of a substance is the ratio of density of
the substance to the density of water at a specified temperature. Specific gravity
shows what mass of a substance is contained in 1cm3 of a solution. For
example, if the specific gravity of a substance is 1.67, it means that 1.67g of
that substance is contained in 1cm3 of water.
Molar mass of a substance is mass of one mole of the substance
expressed in grams
The Percentage Purity
and Specific gravity of common Acids
|
Acid |
Molar mass (g/mol) |
Percentage purity |
Specific gravity |
|
HCl |
36.5 |
32% -36% |
1.18 |
|
H2SO4 |
98 |
96% -98% |
1.84 |
|
HNO3 |
63 |
65% -72% |
1.047- 1.052 |
Example question 1
What volume of concentrated hydrochloric acid would be
required to prepare 2 dm3 of 0.6mol/dm3 solution? (Specific
gravity =1.18, percentage purity =36%, molar mass =36.5 g/mol)
Answer
Molar mass of HCl = 36.5g/mol
Specific gravity = 1.18
1cm3 of the stock of solution contains = 1.18g of
HCl
1000cm3 solution contains = 1.18 x 1000 =1180g
Mass of pure hydrochloric acid in 1dm3 of solution
= 36/100 x1180 = 424.8g
Amount of pure hydrochloric acid = 424.8/36.5 = 11.64mol
Using dilution formula
C1
V1 =C2 V2
C1 = 11.64 mol
V1 = unknown
C2 = 0.6mol/dm3
V2 = 2000 cm3 (2dm3)
11.64 x V1 = 0.6 x 2000
V1 = 1200/11.64
= 103.1cm3
103.1cm3 of hydrochloric acid would be diluted to
2dm3 in order to produce 0.6mol/dm3
Example question 2
Calculate the volume of concentrated trioxonitrate (V) acid
required to prepare 2dm3 of 0.5mol/dm3 of solution. (Specific
gravity =1.052, Percentage purity = 72%, molar mass = 63g/mol)
Answer
Molar mass of HNO3 = 63g/mol
Specific gravity of HNO3 = 1.052
1cm3 of the stock solution contains = 1.052g
1000cm3 solution contains =1.052 x 1000 = 1052g
Mass of pure HNO3 in 1dm3 solution = 72/100 x 1052 =
757.4g
Amount of HNO3 in 1dm3 of solution =
757.4/63 = 12.02 mole
Using dilution formula
C1
V1 = C2 V2
C1 = 12.02 mol
V1= unknown
C2 = 0.5 mol/dm3
V2 = 2000cm3 (2dm3)
12.02 x V1 = 0.5 x 2000
12.02V1 = 1000
V1 = 1000/12.02 = 83.2cm3
83.2 cm3 of the concentrated acid would be diluted
to 2dm3 in order to produce 0.5 mol/dm3 of solution.
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