College of Dentistry
Inorganic Chemistry
Assistant Lecture
Aayad Amaar
Concentration
Most chemical reactions occur in a liquid solvent/solute (i.e.,
solution) environment. Typically the solute(s) in a solution will be
the reactants. Since stoichiometric calculations require amounts of
reactants, we need a way to express amounts of reactants when
they are in solution. The way chemists do this is through the
concept of concentration. Concentration is a way to express the
amount of solute per unit amount of solution/solvent. There are
several ways that concentration can be expressed. A few of these
are:
ppm (parts per million)
ppb (parts per billion)
Weight/weight percent (w/w %)
Weight/volume percent (w/v %)
Volume/volume percent (v/v %)
molarity
ppm and ppb
Parts per million (ppm) and parts per billion (ppb) are examples
of expressing concentrations by mass. These units turn out to be
convenient when the solute concentrations are very small (almost
trace amounts). For example, if a solution has 1 ppm solute this
would mean that 1 g of solution would have one "millionth" gram
of solute. Equivalently, 1 kg of this solution will have 1 mg of
solute etc... By definition we have:
For example, suppose a 155.3 g sample of pond water is found to
have 1.7x10-4 g of phosphates. What is the concentration of
phosphates in ppm?
College of Dentistry
Inorganic Chemistry
A similar procedure would be followed to calculate ppb. In the
above example the pond water would be 1100 ppb.
Now suppose we have 400 g sample of pond water and it has a
concentration of 3.5 ppm dissolved nitrates. What is the mass of
dissolved nitrates in this sample?
Weight/Weight %
This concentration unit is similar to ppm or ppb except it
focuses on the solute as a percent (by mass) of the total solution.
It is appropriate for relatively large solute concentrations, by
definition we have
As an example considers 5 g sugar dissolved in 20 g of water.
What is the w/w% concentration of sugar in this solution?
Now suppose we have 450 g of NaCl solution that is 35 NaCl
w/w %. What is the mass of NaCl?
Weight/Volume Percent
The concentration of a solution is defined as the amount of solute
dissolved in a specified amount of solution,
If we define the amount of solute as the mass of solute (in grams)
and the amount of solution in volume units (milliliters),
concentration is expressed as the ratio
College of Dentistry
Inorganic Chemistry
This concentration can then be expressed as a percentage by
multiplying the ratio by the factor 100%. This result in
The percent concentration expressed in this way is called
weight/volume percent, or % (W/V). Thus
Example
Volume/Volume %
When the solute is a liquid sometimes it is convenient to
express its concentration in volume/volume percent (v/v %). The
definition of v/v % is
Wine has about 12 mL of alcohol (ethanol) per 100 mL of
solution. Wine would have the following v/v % alcohol content:
Molarity
Molarity is the most useful concentration for chemical reaction
in solution because it directly relates moles of solute to volume of
solution. The definition of molarity is
College of Dentistry
Inorganic Chemistry
As an example, suppose we dissolve 23 g of ammonium chloride
(NH4Cl) in enough water to make 145 mL of solution. What is the
molarity of ammonium chloride in this solution?
Now, suppose we have a beaker with 175 mL of a 0.55 M HCl
solution. How many moles of HCl is in this beaker?
As a final example suppose we have a solution of 0.135 M
NaCl and we need 1.2 moles of NaCl. What volume of the NaCl
solution is required?
Dilution
Often it is necessary to take a concentrated solution and dilute
it. However, we want to dilute it in a controlled way so that we
know the concentration after dilution. The way this is done can be
extracted from the following figure of dilution:
The solute is concentrated in the beaker on the left. Adding
water dilutes the solution as shown with the beaker on the right.
However, note that although the concentration changes upon
dilution, the number of solute molecules does not. In other words
College of Dentistry
Inorganic Chemistry
the number of moles of solute is the same before and after
dilution. Since Moles = Molarity x Volume (i.e., moles= M x V)
we end up with the following equation relating molarity and
volume before and after dilution:
Mi x Vi = Mf x Vf
Where i and f stand for initial and final. Suppose we need 150 mL
of 0.25 M NaCl. On the shelf we find a bottle of 2M NaCl. What
do we do? the concentrated molarity is Mi and the volume needed
is Vi. We need to determine Vi and can do so by rearranging the
above equation and doing the resulting calculation:
Thus, we need 18.8 mL of the 2M NaCl solution, put it in a
beaker and add enough water to make 150 mL of solution. The
resulting solution will have a molarity of 0.25 M NaCl.
This document was created with Win2PDF available at http://www.daneprairie.com.
The unregistered version of Win2PDF is for evaluation or non-commercial use only.