Solubility and Concentration Units

The solubility of a substance is the maximum amount of that substance that can be dissolved in a particular solvent at a particular temperature. When this maximum amount of solute has been added, the solution is said to be saturated. In this state, the solution is in a state of dynamic equilibrium: The two opposite processes of dissolution and precipitation (or crystallization) are taking place at the same rate.

If more solute is added to a saturated solution, it will not dissolve. For example, at 18°C, a maximum of 83 g of glucose (C6H12O6) will dissolve in 100 mL of H2O. Thus, we can say that the solubility of glucose is 83 g/100 mL. If more glucose is added, it will remain in solid form, precipitating to the bottom of the container. In some unique temperature conditions more solid can be dissolved than is allowed, in which case the solution is said to be supersaturated. This is, however, an unstable system, and often the slightest disturbance would cause the excess solute to precipitate out.

A solution in which the proportion of solute to solvent is small is said to be dilute, and one in which the proportion is large is said to be concentrated. Of course, very often we need to specify the amount of solute dissolved in a solvent more exactly, and this is expressed by the quantity of concentration. Solubility, then, can be thought of as the maximum possible concentration for the given solute-solvent pair in question. In the example above, we mentioned that the maximum concentration of glucose in water at 18°C is 83 g/100 mL. More commonly, however, the concentration of a solution is expressed as percent composition by mass, parts per million, mole fraction, molarity, molality, or normality.

Percent Composition by Mass

The percent composition by mass of a solution is the mass of the solute divided by the mass of the solution (solute plus solvent), multiplied by 100.

  1. Example: What is the percent composition by mass of a salt water solution if 200 g of the solution contains 0.3 mol of NaCl?
  2. Solution: First we need to convert the amount of NaCl into mass:

Parts Per Million

The parts per million (ppm) of a solution is equal to the grams of solute divided by 1,000,000 g of solution.

  1. Example: A chemical analysis shows that there are 2.2 mg of iron in a 500 g sample of tap water. Convert this to parts per million (ppm).
  2. Solution: First we need to convert the amount of iron to grams.
2.2 mg × 1 g/1000 mg = 0.0022 g Fe
  1. Second, we need to create a proportion between the amount of iron per 500 g of water and the amount of iron per 1,000,000 g of water.
  1. Cross multiply, then solve for x.
  1. The units of grams cancel out, leaving us with units of parts per million.

Mole Fraction

The mole fraction (X) of a compound is equal to the number of moles of the compound divided by the total number of moles of all species within the system. The sum of the mole fractions in a system will always equal 1.

  1. Example: If 92 g of glycerol is mixed with 90 g of water, what will be the mole fractions of the two components? (MW of H2O = 18; MW of C3H8O3 = 92)
  2. Solution: 90 g water = 90 g × (1 mol/18 g) = 5 mol
    92 g glycerol = 92 g × (1 mol/92 g) = 1 mol
    total number of moles = 5 + 1 = 6
    Xwater = 5 mol/6 mol = 0.833
    Xglycerol = 1 mol/6 mol = 0.167

    Since these are the only two components in the system, one can verify that the two mole fractions add up to be 1.

Molarity

The molarity (M) of a solution is the number of moles of solute per liter of solution. Solution concentrations are usually expressed in terms of and in units of mol/L, also abbreviated M molarity. Molarity depends on the volume of the solution, not on the volume of solvent used to prepare the solution. In other words, mixing 1 mol of solute with 1 L of solvent will not in general give a 1 M solution since the final volume after mixing may be different from 1 L. You may recall from your laboratory experience that in order to produce a solution of a particular molarity, you add solvent to a container (volumetric flask) that already has the solute weighed out in it, until the total volume of the mixture reaches a specific value (marked by a ring around the narrow neck of the flask). One generally is not interested in, and does not keep track of, the volume of solvent actually added.

  1. Example: If enough water is added to 11 g of CaCl2 to make 100 mL of solution, what is the molarity of the solution?
  2. Solution: 11 g CaCl2 = 0.10 mol CaCl2
    100 mL = 0.10 L
    ∴ molarity = 0.10 mol/0.10 L = 1.0 M

Molality

The molality (m) of a solution is the number of moles of solute per kilogram of solvent. For dilute aqueous solutions at 25°C the molality is approximately equal to the molarity, because the density of water at this temperature is 1 kilogram per liter and the volume of the solution is presumed to be approximately the same as that of the solvent (i.e., water) added. But note that this is an approximation and true only for dilute aqueous solutions.

  1. Example: If 10 g of NaOH are dissolved in 500 g of water, what is the molality of the solution?
  2. Solution:

Normality

The normality (N) of a solution is equal to the number of equivalents of solute per liter of solution. An equivalent is a measure of the reactive capacity of a molecule, and is defined according to the type of reaction being considered.

To calculate the normality of a solution, then, we must know for what purpose the solution is being used, because it is the concentration of the reactive species with which we are concerned. For example, a 1 molar solution of sulfuric acid would be 2 normal for acid-base reactions (because each mole of sulfuric acid, H2SO4, provides 2 moles of H+ ions) but is only 1 normal for a sulfate precipitation reaction (because each mole of sulfuric acid provides only 1 mole of sulfate ions). Normality is always a whole-number multiple of molarity: for example, a 3 M solution may be 3 N, 6 N, et cetera, but not 2N or 5N.