http://www.ck12.org Chapter 16. Solutions
The aqueous solubility of a given substance is determined experimentally by dissolving increasing amounts into
a known mass of water at a specific temperature until no more solid dissolves. A solution that cannot hold any
more of a given solute is said to besaturated. For example, inFigure16.1 we see that the solubility of sodium
nitrate (NaNO 3 ) is approximately 90 g per 100 g of H 2 O at 20°C. At this temperature, a 100 gram sample of H 2 O
in which 90 grams of sodium nitrate is dissolved would be saturated, and a solution in which less sodium nitrate
is dissolved would beunsaturated. Note that this ratio also holds for samples in which the solvent is present in
different amounts; 50 grams of water would hold 45 grams of sodium nitrate at 20°C, and 300 grams of water would
hold 270 g of NaNO 3. Notice that even relative solubilities of various compounds are temperature dependent. For
example, at 20°C, KCl has a higher solubility than NaCl, but at 50°C, this relationship is reversed.
Solutions can also becomesupersaturated, where the amount of solute dissolved exceeds its solubility. Supersat-
uration most commonly occurs when a saturated solution is slowly cooled. They occur frequently in geological
and meteorological processes. Supersaturated systems are unstable, and eventually, the solute will precipitate until
a saturated solution is regenerated. We can quantify supersaturation by looking at solubility curves. If the ratio of
solute to solvent is above the saturation curve at the given temperature, the solution is supersaturated. If it is on the
curve, the solution is saturated, and if it is below the curve, the solution is unsaturated.
Solubility can be described for any solute-solvent pairing, but because water is such a fundamentally important
solvent, we are mainly focusing on aqueous solutions.
Example 16.1
You dissolve 40 g of KCl in 100 g of water at 40°C. You then cool the solution to 20°C, during which you notice
solid KCl precipitating. How many grams of KCl would you expect to precipitate?
Answer:
ConsultFigure16.1 to find the solubility of KCl at 20°C (approximately 32 g KCl/100 g H 2 O). Therefore, we would
expect approximately 8 g KCl to precipitate out (40 g –32 g = 8 g).
Factors Affecting Solubility
There are three main factors that control solubility.
- Identities of the solute and solvent
- Temperature
- Pressure (for gases only)
Solute and Solvent
Ultimately, the ability of a solute to dissolve in a particular solvent will be dictated by the relative favorability of
solute-solvent interactions compared to solute-solute and solvent-solvent interaction. In particular, the polarity of
these two substances has a major effect on whether a significant amount of solute is able to dissolve. Polar solutes
are typically quite soluble in polar solvents (e.g., ethanol in water), and nonpolar solutes generally dissolve well in
nonpolar solvents (e.g., grease in gasoline). Conversely, polar solutes will have low solubilities in nonpolar solvents
(e.g., NaCl in CCl 4 ), and solubilities will be low for nonpolar solutes in polar solvents (e.g., oil in vinegar).
Temperature
As you can see inFigure16.1, solid and liquid solutes generally become more soluble as the temperature increases.
This is true for solvents other than water as well. This effect varies quite a bit by substance. For example, the
solubility of KNO 3 has a very strong temperature dependence (its solubility curve has a large slope), whereas the