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
Quantitative analysis deals with the
determination of the amount of each component in a sample of matter. It is
divided into two which include:
a.
Gravimetric analysis
b.
Volumetric analysis
Gravimetric
analysis: is interested in the measurement of mass of
the substances.
Volumetric
analysis: is interested in the measurement of the volume
of solutions.
Titration
is the method used in volumetric analysis. This is done by careful addition of
standard solution from a graduated container until the reaction is complete. A
complete reaction is indicated by a colour change in the resulting solution.
Some apparatus used for volumetric analysis
These are burette, pipette, conical flask,
beaker, funnel, wash bottle and volumetric flask.
Standard
solution is a solution of a known concentration.
Basic
Principles of Volumetric Analysis
Test substance or analyte, is normally a base in
acid-base titration, is pipetted into the conical flask. The burette contains a
standard solution of the acid in acid-base titration, which is usually called titrant.
The requirements of a titration are as follows:
1.
There must be a well-defined
reaction between the analyte and titrant. For example reaction of hydrochloric acid
and sodium hydroxide . NaOH + HCl …….. NaCl + H2O
2.
The reaction must be fast
3.
The reaction must not have side
reaction
4.
The reaction must be specific
5.
There must be specific change in
some properties of the solution at the end of the reaction
6.
The end-point is the point at which
the reaction is considered to be complete and it is detected by colour change
or change in the property of the solution.
Example
Question
A is a solution of hydrochloric acid. B is a
solution containing 6.0g of sodium hydroxide in 1dm3 of solution.
a.
Put A in the burette and titrate
with 20cm3 or 25cm3 portions of B using methyl orange as indicator.
Record the volume of your pipette. Tabulate your readings and calculate the
average volume of a used.
b.
From your results and information provided,
calculate the
i)
Concentration of B in mol/dm3
ii)
Concentration of A in mol/dm3
iii)
Amount of A which reacted with 25cm3
of B
Equation for the reaction
NaOH(aq) + HCl(aq) ……….
NaCl(aq)+ H2O(l)
Answer
Volume of pipette used = 25cm3
Indicator used = methyl orange
|
Burette
redings |
Rough |
First |
Second |
Third |
|
Final
reading (cm3) |
24.50 |
22.10 |
22.10 |
22.20 |
|
Initial
reading (cm3) |
0.00 |
0.00 |
0.00 |
0.00 |
|
Volume
of A used |
24.50 |
22.10 |
22.10 |
22.20 |
Average volume of A used = 22.10 + 22.10+
22.20/3
= 22.13 cm3
i)
Concentration of B in mol/dm3
40g of NaOH
= 1mol/dm3
6.0g of NaOH = 1/40 x 6 = 0.15 mol/dm3
ii) Concentration of A in mol/dm3
CAVA/CBVB =
a/b
CA =?
VA = 22.13 cm3
a= 1
CB = 0.15 cm3
VB = 25 cm3
b=1
CA x 22.13\ 0.15 x 25 = 1\1
CA = 0.15 x 25 x 1\22.13 x 1 = 0.17 mol/dm3
iii) Amount of A which reacted with 25 cm3
of B
1000 cm3 of A contains = 0.17 moles
22.13 cm3 will be = 0.17\1000 x 22.13
= 3.7621\1000
= 0.00376 moles
Comments
Post a Comment