Chem 35.5: Expression Solution Concentration
In the study of chemistry, the concentration of a solution is a macroscopic property that can be expressed in many
different ways. The exercise given below (from Petrucci’s text) should help you understand the various types of units
and what each means. Please take a good look as you will be expected to handle problems of this sort.
Problem
An ethanol-water solution is prepared by dissolving 10.00 mL of ethanol, C2 H5 OH (d=0.789 g/mL) in a sufficient
volume of water to produce 100.0 mL of a solution of a with a density of 0.982 g/mL. What is the concentration of
EtOH in the solution expressed as a) volume % b) mass % c) mass/volume % d) mole fraction e) mole percent f)
molarity g) molality ?
Here is my approach (not the only way, of course). Remember the Law of Conservation of Mass (Mass of the Solution
= Mass of the Solute + Mass of the Solvent ) and the fact that you can relate the mass of any substance (be it the
solute, solvent or solution) to its volume via density and vice versa. I then write out what I need to substitute into the
definitions.
0.0.1
Solute
Volume C2 H5 OH = 10.00 mL (given)
Mass C2 H5 OH = 10.00 mL × density = 10.00 mL × 0.789 g/mL = 7.89 g C2 H5 OH
Moles C2 H5 OH = 7.89 g × Molar Mass C2 H5 OH = 7.89 g C2 H5 OH × 46.07 g/mol = 0.171 mol C2 H5 OH
0.0.2
Solution
Volume of Solution = 100.0 mL (given)
Mass of Solution = Volume of Solution × Density of Solution = 100.0 g solution × 0.982 g/mL = 98.2 g solution
0.0.3
Solvent
Mass H2 O (Solvent) = Mass Solution -Mass Solute = 98.2 g - 7.89 g = 90.3 g H2 O
Moles H2 O (Solvent) = 90.3 g H2 O × 1 mol H2 O/18.02 g H2 O = 5.01 mol H2 O
It’s really downhill from here. Focus on the definitions and what constitutes the solute, solvent and solution both
in the text and notes and then substitute and keep the units consistent by writing the units out!
(a) % Volume = v/v % =
Vol of solute
10.00 ml C2 H5 OH
× 100 =
× 100 = 10.00% C2 H5 OH
Vol of solution
100.00 mL solution
(b) % Mass = w/w % =
7.89 g C2 H5 OH
mass of solute
× 100 =
× 100 = 8.03% C2 H5 OH
Mass of Solution
98.2 g solution
(c) % Mass/volume = m/v % =
(d) Mole fraction =
mass of solute
7.89 g C2 H5 OH
× 100 =
× 100 = 7.89% C2 H5 OH
mL of Solution
100.00 mL solution
moles of solute
0.171 mol C2 H5 OH
=
= 0.0330
total # of moles of (solute + solvent)
5.18 mol H2 O
(e) % mole = Mole fraction × 100 =
0.171 mol C2 H5 OH
× 100 = 3.30%
5.01 mol H2 O
1
(f) molarity = M =
moles of solute
0.171 mol C2 H5 OH
=
= 1.71 M C2 H5 OH
Liters of Solution
0.1000 L solution
(g) molality = m =
moles of solute
0.171 mol C2 H5 OH
× 100 =
× 100 = 1.89 m C2 H5 OH
Kg Solvent
0.0903 kg H2 O
Units of Minute Concentrations–pph, ppt, ppm, ppb, pptr
When the concentration of a solute in a solution is very small, we often use different terminology and units that are
easier to communicate and report. Parts per thousand, parts per million, parts per billion, and parts per trillion are all
units used in water and air analysis for contaminants in the environment, or in blood chemistry analysis.
Generally speaking, you can think of all of the latter as a specific type of percent by weight (w/w). For example,
you know (or should know) that 1% (w/w) of a solute means 1 part of solute per 100 parts of solution (or sample). If
we use grams to measure the parts we can for example express 2.5 % by weight (w/w) as:
2.5% (w/w) =
2.5 g solute
× 100 = 2.5 parts per hundred = 2.5 pph
100 g solution
Both pph and % weight names can be used but pph is normally not uses. It is just a matter of tradition and practice
even though pph is correct. The important thing is that you should recognize what pph means and how it can be stated
in different ways as shown below:
1 pph =
1 g solute
10 g solute
10 g solute
=
=
100 g solution
1000 g solution
1 kg solution
We can now extend our definitions and units to cover a wider range of concentrations that are routinely encountered
in the real world—parts per thousand (ppt), parts per million (ppm), parts per billion (ppb) and parts per trillion (pptr).
By analogy we can write:
ppt =
1 g solute
10−3 g
1g
g solute
=
=
=
× 103
3
10 g solution
g
kg
g solvent
ppm =
1 g solute
10−6 g
10−3 g
1 mg
g solute
=
=
=
=
× 106
6
10 g solution
g
1000 g
kg
g solvent
ppb =
1 g solute
10−9 g
10−6 g
1 µg
g solute
=
=
=
=
× 109
9
10 g solution
g
1000 g
kg
g solvent
pptr =
1 g solute
10−12 g
10−9 g
1 ng
g solute
=
=
=
=
× 1012
12
10 g solution
g
1000 g
kg
g solvent
Because most real world analytical measurements use water as the solvent, we assume the density of water to be
1 g/mL and convert the grams normally used in the denominator (i.e. the mass of the solution) to mL. If we do this,
1 ppm = 1 gram solute/106 mL solution. In any particular problem that you encounter, you choose between units of
volume or mass by understanding the context of the problem and whether we are dealing with aqueous solutions or
solid solutions (more often than not, it is aqueous).
1 ppm =
1 mg
1g
=
6
10 mL
L
2
Sample Problem Using ppm, ppb, pptr
There are basically two types of problems encountered with units of this sort. Suppose we have a solution that contains
0.0054 g Zn2+ dissolved in a total solution volume of 0.0025 kg H2 O. Let’s calculate the concentration of the resulting
solution in pph, ppt, ppm and ppb.
My approach is as follows:
1. if you are given masses, convert both to the same units (grams, kilograms, etc) or if the problem involves aqueous
solutions, remember to assume a density of water of 1 g/mL and replace grams or kilograms of solvent with mL
or L, respectively.
2. Take the ratio of solute to solvent and remember to use the proper metric prefix, 102 for pph, 103 for ppt, 106
for ppm, 109 for ppb and 1012 for pptr
For the problem stated above, we have:
Solute = 0.0054 g Zn2+ and Solution = 0.0025 kg
• Convert masses to the same units in this case convert 0.0025 kg H2 O to 2.5 g, and because it is aqueous use 1.0
g/mL and convert the solution mass units to = 2.5 mL H2 O
• Remember the definitions and prefixes for pph, ppt, ppm, ppb
pph =
0.0054 g Zn2+
× 100 = 0.216 pph
2.5 mL H2 O
ppt =
0.0054 g Zn2+
× 103 = 216 ppt
2.5 mL H2 O
ppm =
0.0054 g Zn2+
× 106 = 216 × 103 ppm
2.5 mL H2 O
ppb =
0.0054 g Zn2+
× 109 = 216 × 106 ppb
2.5 mL H2 O
The other type of problem you might encounter is as follows. Suppose you are given an aqueous solution with a
stated concentration of 100 ppm of dissolved O2 and you are asked to calculate the number of moles of O2 in 1.5 liter
of solution. Here you can proceed in several ways. From the definition of ppm you know:
100 g of O2 in 106 mL of solution, or if you understand the definition well enough, you might also realized that 100
ppm also means 100 mg O2 per L of solution. Using either quantity with the factor label method, you can calculate
the number of moles of O2 .
100. g O2
103 mL solution
1 mole O2
×
× 1.5 L solution ×
= 4.7 × 10−3 mol O2
mL solution
1 L solution
32 g O2
106
3