The Learning Objective of this module is to describe the concentration of a solution in the way that is most appropriate for a particular problem or application. There are several different ways to quantitatively describe the concentration of a solution. For example, molarity was introduced in Chapter 4 as a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions, introduced in Chapter 10, are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example 4 reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.
The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent: \[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \tag{13.5}\] Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of \(H_2O\) under these conditions is very close to 1.0 L, and a 0.50 M solution of \(KBr\) in water, for example, has approximately the same concentration as a 0.50 m solution. Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb): \[\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \tag{13.6}\] \[\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \tag{13.7}\] \[\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \tag{13.8}\] In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution. The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of \(H_2SO_4\) per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature. Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as we saw in Chapter 10. As you will learn in Section 13.5, mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies. Table 13.5 summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example 6. Table 13.5 Different Units for Expressing the Concentrations of Solutions*
The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent.
Key Takeaway
molality \[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \tag{13.5}\] mass percentage \[\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \tag{13.6}\] parts per million \[\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \tag{13.7}\] parts per billion \[\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \tag{13.8}\]
1.
3. 100.0 ml of 0.40 M KI: dissolve 6.64 g of KI in enough water to make 100.0 mL of solution; 100.0 ml of 0.65 M NaCN: dissolve 3.18 g of NaCN in enough water to make 100.0 mL of solution.
5.
9.
13. The molarity is 0.745 M, and the mole fraction is 0.0134.
15. The molarity is 0.0129 M, the molality is 0.0129 m, the mole fraction is \(2.33 \times 10^{-4}, and the solution contains 1830 ppm \(Na_2HPO_4\). Mole fraction is most useful for calculating vapor pressure, because Raoult’s law states that the vapor pressure of a solution containing a non-volatile solute is equal to the mole fraction of solvent times the vapor pressure of the pure solvent. The mole fraction of the solvent is just one minus the mole fraction of solute. 17. \(6.65 \times 10^{-5} mol sodium; \(6.14 \times 10^4\; ppb\)
19. \(1.63 \times 10^{-3} g; \(2.17 \times 10^4\) ppb
21. \(2.22 \times 10^4\; mL\) or 22.2 L
25 0.777 g \(BaSO_4\); \(Na_2SO_4\); 0.0134 M \(Na_2SO_4\) |