Solutions and Concentration - Solution Composition
Chemical reactions often take
place in aqueous solutions. To perform stoichiometric
calculations in such cases the amounts of chemicals present in solution
– the concentration of solution -
must be known.
Concentration of a solution is a
measurement stating the amount of a solute present in a known amount of
solution:
Concentration = amount of solute / amount of
solution
The terms solute and solution are
usually used for liquid samples but they can be extended to gaseous and solid
samples.
The most common units of
concentration are given in Table I.1:
Common Units of
Concentration
|
||
Name
|
Symbol
|
Units
|
molarity
|
moles solute / liters of
solution
|
M
|
molality
|
moles solute / kg
solvent
|
m
|
normality
|
number of equivalent
weights of solute / liters of solution
|
N
|
formality
|
number of formal weights
of solute / liters of solution
|
F
|
weight %
|
g solute / 100 g of
solution
|
%
w/w
|
volume %
|
ml solute / 100 ml
solution
|
%
v/v
|
weight-to-volume %
|
g solute / 100 ml
solution
|
%
w/v
|
parts per million
|
g solute / 106
g solution
|
ppm
|
parts per billion
|
g solute / 109
g solution
|
ppb
|
Note: Another way of describing
solution concentration is the mole
fraction (xi)
Molarity (M) is defined as the number of moles of solute per liter of
solution.
i.e by dissolving 0.1 mol NaOH in
1 l of H2O gives a solution that contains 0.1 mol Na+ and
0.1 mol of OH- in 1 l. The concentration of the solution is
[Na+] = 0.1 M and [OH-] = 0.1 M.
Since molarity depends on the
volume of the solution it changes slightly with temperature.
Another way of describing solution
concentration is molality
(m) which is the number of moles of solute per kilogram of
solvent.
Molality is independent of temperature since it depends on
mass.
In very dilute aqueous solutions
the molarity (M) and molality (m) are nearly the same.
Example #1
A solution of 1M H2SO4 has density 1.04 g/cm3. Calculate the (%w/w) concentration of the solution.
Given
|
[H2SO4] = 1M
d = 1.04
g/cm3
PH2O =
41 mmHg
MW
H2SO4 = 98 g/mole
|
Asked
for
|
(%w/w) = ?
|
From the definition of (%w/w) =
g solute / 100 g of solution
(1)
The mass of solute (g solute) is
unknown but it can be calculated.
Since [H2SO4] = 1M ⇒ 1000 cm3 of H2SO4
solution contain 1 mole “pure”
H2SO4 (2)
The mass of 1 mole
“pure” H2SO4 can be calculated as shown
below:
mass (g) = mole
* MW = 1 mole * 98 g/mole = 98
g (3)
From (2) and d =
m/V = 1.04 g/cm3 the mass of 1000 cm3 of
H2SO4can be calculated:
m = d * V = 1.04 g/cm3 * 1000 cm3 = 1040 g of H2SO4
solution (4)
From (2), (3) and (4):
Mass of 1040 g of
H2SO4 solution contain 98 g of “pure” H2SO4
Mass of 100 g of H2SO4 solution
contain x = ? g of “pure” H2SO4
x = 98 g “pure” H2SO4 * (100g of H2SO4 / 1040g of H2SO4) = 9.42 g of “pure” H2SO4
Therefore,
(%w/w) =
9.42
Provided that the theory and the
definitions of solution concentration units is understood a % solution calculator
can be used.
Meant to be used in both the
teaching and research laboratory, a %
solution calculator can be utilized to perform
a number of different calculations for preparing percent (%) solutions when
starting with the solid or liquid material.
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