11.2 Quantitative Units of Concentration

Rather than qualitative terms (Section 11.1 "Some Definitions"), we need quantitative ways to express the amount of solute in a solution; that is, we need specific units of concentration. In this section, we will introduce several common and useful units of concentration.

MolarityThe number of moles of solute divided by the number of liters of solution. (M) is defined as the number of moles of solute divided by the number of liters of solution:

$$\text{molarity = }\frac{\text{moles of solute}}{\text{liters of solution}}$$

which can be simplified as

$$\text{M}=\frac{\text{mol}}{\text{L}}\text{, or mol/L}$$

As with any mathematical equation, if you know any two quantities, you can calculate the third, unknown, quantity.

For example, suppose you have 0.500 L of solution that has 0.24 mol of NaOH dissolved in it. The concentration of the solution can be calculated as follows:

$$\text{molarity}=\frac{\text{0}\text{.24 mol NaOH}}{\text{0}\text{.500 L}}=0.48\text{ M NaOH}$$

The concentration of the solution is 0.48 M, which is spoken as “zero point forty-eight molarity” or “zero point forty-eight molar.” If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol):

$$\text{22}\text{.4}\cancel{\text{ g HCl}}\times \frac{\text{1 mol HCl}}{\text{36}\text{.5}\cancel{\text{ g HCl}}}=0.614\text{ mol HCl}$$

Now we can use the definition of molarity to determine a concentration:

$$\text{M}=\frac{\text{0}\text{.614 mol HCl}}{\text{1}\text{.56 L}}=0.394\text{ M}$$

The definition of molarity can be used to determine the amount of solute or the volume of solution, if the other information is given. Example 4 illustrates this situation.

If you need to determine volume, remember the rule that the unknown quantity must be by itself and in the numerator to determine the correct answer. Thus rearrangement of the definition of molarity is required.

A similar unit of concentration is molalityThe number of moles of solute per kilogram of solvent. (m), which is defined as the number of moles of solute per kilogram of solvent, not per liter of solution:

$$\text{molality}=\frac{\text{moles solute}}{\text{kilograms solvent}}$$

Mathematical manipulation of molality is the same as with molarity.

Another way to specify an amount is percentage composition by massRatio of mass of solute to the total mass of a sample times 100. (or mass percentage, % m/m). It is defined as follows:

$$\%\text{ m/m}=\frac{\text{mass of solute}}{\text{mass of entire sample}}\times 100\%$$

It is not uncommon to see this unit used on commercial products (Figure 11.1 "Concentration in Commercial Applications").

Related concentration units are parts per thousand (ppth)Ratio of mass of solute to total mass of sample times 1,000., parts per million (ppm)Ratio of mass of solute to total mass of sample times 1,000,000., and parts per billion (ppb)Ratio of mass of solute to total mass of sample times 1,000,000,000.. Parts per thousand is defined as follows:

$$\text{ppth}=\frac{\text{mass of solute}}{\text{mass of sample}}\times \mathrm{1,000}$$

There are similar definitions for parts per million and parts per billion:

$$\text{ppm}=\frac{\text{mass of solute}}{\text{mass of sample}}\times \mathrm{1,000,000}$$ and $$\text{ppb}=\frac{\text{mass of solute}}{\text{mass of sample}}\times \mathrm{1,000,000,000}$$

Each unit is used for progressively lower and lower concentrations. The two masses must be expressed in the same unit of mass, so conversions may be necessary.

As with molarity and molality, algebraic rearrangements may be necessary to answer certain questions.

For ionic solutions, we need to differentiate between the concentration of the salt versus the concentration of each individual ion. Because the ions in ionic compounds go their own way when a compound is dissolved in a solution, the resulting concentration of the ion may be different from the concentration of the complete salt. For example, if 1 M NaCl were prepared, the solution could also be described as a solution of 1 M Na⁺(aq) and 1 M Cl⁻(aq) because there is one Na⁺ ion and one Cl⁻ ion per formula unit of the salt. However, if the solution were 1 M CaCl₂, there are two Cl⁻(aq) ions for every formula unit dissolved, so the concentration of Cl⁻(aq) would be 2 M, not 1 M.

In addition, the total ion concentration is the sum of the individual ion concentrations. Thus for the 1 M NaCl, the total ion concentration is 2 M; for the 1 M CaCl₂, the total ion concentration is 3 M.