16.2. Solution Concentration http://www.ck12.org
complex mixture. Common units for this type of concentration areparts per million(ppm) and parts per billion
(ppb). Let’s say we have a sample of water in which Pb^2 +is present at a concentration of 20 ppm. This would mean
that if we randomly took a million particles from our solution, 20 of them (on average) would be Pb^2 +ions. The rest
would be mostly water molecules, with a possibility of the presence of various other ions. The maximum allowable
concentrations of various toxic gases in the air or heavy metal ions in drinking water are often reported in units of
ppm or ppb.
If given information on aqueous solution concentration by ppm, these units can be converted to molarity or molality
using the following conversion factor:
1 ppm= 1 mg/ LThis conversion is derived from the fact that ppm is really just a ratio between amount of solute and amount of
solution. If you had a 1 ppm solution, this would indicate 1 g of solute per 1× 106 g of solution. Assuming an
aqueous solution density of 1 g/mL as is standard for water, this indicates 1 g of solute per 1× 106 mL of solution, or
1 mg of solute per 1 L of solution. You can use this conversion unit and a similar conversion method for converting
concentration from ppm or ppb.
Dilutions
Many chemicals that we use on a daily basis are transported in a concentrated form but used in a more diluted form.
For example, concentrated cleaners are often diluted before they are used. To perform adilution, pure solvent is
added to a concentrated solution in order to make a less concentrated (more dilute) solution. The resultant solution
will contain the same amount of solute but a greater amount of solvent. It will therefore have a lower concentration
than the original solution. When performing a simple dilution, the concentration and volume of the initial solution
are related to the new concentration and volume as follows:
M 1 V 1 = M 2 V 2
M 1 = initial molarity, V 1 = initial volume
M 2 = final molarity, V 2 = final volume
This relationship holds true due to the fact that the moles of solute stays constant through a dilution process.
Example 16.12
50.0 mL of a 0.40 M NaCl solution is diluted to 1000.0 mL. What is the concentration of NaCl in the new solution?
Answer:
Determine the values of each variable in the dilution equation above, and then solve for the unknown variable.
V 1 = 50.0 mL, M 1 = 0.40 M, V 2 = 1000.0 mL, and M 2 =?
Solving the above equation for M 2 gives us the following:
M 2 =
M 1 V 1
V 2
=
( 50 .0 mL)( 0 .40 M)
(1000 mL)