Solubility

There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the solubility of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table 9.2 “Solubilities of Various Solutes in Water at 25°C “ lists the solubilities of various solutes in water. Solubility varies with temperature, so it is important to include the temperature at which the solubility was determined.  Most solids have a higher solubility in water at higher temperatures. For example, while 91 g of glucose will dissolve in 100 mL of water at 25^oC, 357 g will dissolve at 70^oC.

If a solution contains so much solute that its solubility limit is reached, the solution is said to be saturated. If a solution contains less solute than the solubility limit, it is unsaturated. For some substances, a supersaturated solution can be prepared by dissolving as much solute as possible at a higher temperature, then lowering the temperature.  Less solute should dissolve at the lower temperature, but the excess solute might not precipitate right away.  Thus the solution is supersaturated, but not stable. Eventually the excess solute will crystallize with a release of energy as the mixture achieves a more stable state.  Honey is a supersaturated solution which bees form by using their wings to fan nectar, a dilute sugar solution, to drive off much of the water, forming supersaturated honey.  The excess sugar in honey sometimes recrystallizes, making the honey cloudy or grainy.  It can be restored to a clear state by heating it.

In contrast to increased solubility of most solids at higher temperatures, the solubility of gases in liquid water decreases as temperature increases.  The solubility of O₂ is 8.2 mg O₂/L of H₂O at 25^oC but only 6.0 mg O₂/L of H₂O at 45^oC, both at a pressure of 1 atm.    Pressure must be specified because it also influences the solubility of gases; more gas dissolves when the pressure is higher.   At 0.5 atm, 4.1 mg of O₂ dissolves per L of H₂O, but at 2 atm, the solubility of O₂ is 16.5 mg/L, both at a temperature of 25 ^oC.   When pressure is suddenly changed, such as when you open a can of soda, excess gas can bubble out of solution uncontrollably.

There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:

[latex]\text{%m/m}=\frac{\text{mass solute}}{\text{mass solution}}\times{100}[/latex]

Each mass must be expressed in the same units to determine the proper concentration.  The mass of the solution may be given, or it may be necessary to calculate it by adding the masses of the solute and solvent.

For liquids, volumes are relatively easy to measure, so the concentration of a liquid in solution can be expressed as a volume/volume percent (% v/v): the volume of a solute divided by the volume of a solution times 100:

[latex]\text{%v/v}=\frac{\text{volume solute}}{\text{volume solution}}\times{100}[/latex]

As with mass/mass %, the units of the solute and the solution must be the same.  Both the volume of the solute and the volume of the solution must bespecified because volumes are not additive.  For example, a mixture made from 50 mL of water and 50 mL of ethanol has a volume of about 95 ml because the molecules of the two substances are attracted to each other and enter the spaces between each other’s molecules.

A hybrid concentration unit, mass/volume percent (% m/v), is commonly used for intravenous (IV) fluids (Figure 9.2 “Mass/Volume Percent”). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:

[latex]\text{%m/v}=\frac{\text{grams solute}}{\text{mL solution}}\times{100}[/latex]

Unlike %m/m and %m/v, the units do not cancel and m/v% is not a true percentage.  Never the less, m/v % is widely used in medical settings because it is easy to deliver a certain mass of a medication or other solute by delivering a certain volume of solution.

Each percent concentration can be used to produce a conversion factor between the amount of solute, the amount of solution, and the percent. Furthermore, given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.  While it is possible to substitute known quantities into the percentage equation and solve for the missing quantity, it is easier to interpret percentage and use it as a conversion factor.  x % implies the following pair of conversion factors:

[latex]\frac{\text{x g or mL of solute}}{100\text{ g or mL of solution}}[/latex]   AND   [latex]\frac{100\text{ g or mL of solution}}{\text{ x g or mL of solute}}[/latex]    As always, use the conversion factor that allows the units to cancel.  Make sure to include a descriptor with the units, such as mL solution or g solvent- cancellation must include the unit and the descriptor.  Note that the 100 is placed there by definition of percent, so it does not limit the sig figs of the answer.

In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units can be based on either mass or volume and are defined as follows:

[latex]\text{ppm}=\frac{\text{g or mL solute}}{\text{g or mL solution}}\times1,000,000[/latex]    and     [latex]\text{ppb}=\frac{\text{g or mL solute}}{\text{g or mL solution}}\times1,000,000,000[/latex]

Concentrations of trace elements in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.

Molarity

Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions that take place in solution. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.

Molarity is defined as the number of moles of a solute dissolved per liter of solution:

[latex]\text{molarity}=\frac{\text{number of moles solute}}{\text{number of liters solution}}[/latex]

Notice that the units are complex, mol/L, often abbreviated to just M.  Also note that this is not a percentage so there is no multiplication by 100.

For example, for 1.5 mol of NaCl dissolved in 0.500 L of solution, calculate the molarity as follows:

[latex]\text{molarity}=\frac{\text{number of moles solute}}{\text{number of liters solution}}=\frac{1.5\text{ mol NaCl}}{0.500\text{ L solution}}=3.0\text{ mol/L}=3.0\text{ M NaCl}[/latex]

Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.

When molarity is given, it can be interpreted as a pair of conversion factors that can be used to convert between moles of solute and liters of solution.  A molariy of x :

[latex]\frac{\text{x moles solute}}{1\text{ L solution}}[/latex]    and     [latex]\frac{1\text{ L solution}}{\text{x moles solute}}[/latex]

Note that the 1 L is part of the definition of molarity, so it does not limit the number of sig figs in the answer.

Using Molarity in Stoichiometry Problems

Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit.

The general steps for performing stoichiometry problems using molarities of solutions  are shown in Figure 9.3 “Diagram of Steps for Using Molarity in Stoichiometry Calculations”. You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure 9.3 “Diagram of Steps for Using Molarity in Stoichiometry Calculations” indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.

Figure 9.3 Diagram of Steps for Using Molarity in Stoichiometry Calculations

In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.

Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table 9.3 “Approximate Concentrations of Various Solutes in Some Solutions in the Body*” lists the typical concentrations of some of these solutes.

Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na⁺(aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/L solution of Ca²⁺(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L. (For more information about the ions present in blood plasma, see Chapter 3 “Ionic Bonding and Simple Ionic Compounds”, Section 3.3 “Formulas for Ionic Compounds”.)

Dilution

When solvent is added to dilute a solution, the volume of the solution changes, but the amount of solute does not change. Before dilution, the amount of solute was equal to its original concentration (C_i) times its original volume (V_i).

amount of substance = C_i × V_i

After dilution, the same amount of solute is equal to the final concentration (C_f) times the final volume (V_f):

amount of substance = C_f × V_f

Since the amount of substance does not change during dilution:

C_i × V_i= C_f × V_f

Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit.  Three of the values must be given in the problem so that the fourth value can be determined algebraically.