Although some solutions, like one consisting of water and ethyl alcohol, can have any intermediate composition between the pure components, most solutions have an upper limit to the concentration of the solute. That limit is called the solubility of the substance. For example, in a liter of solution, the maximum amount of CaSO 4 dissolved is 0.667 grams, which is 0.0049 moles of that solute. Therefore, the solubility of calcium sulfate may be reported either as 0.667 grams per liter or as 0.0049 M.

A solution containing less solute than the maximum that can dissolve is known as a dilute solution and is said to be unsaturated. A solution containing as much solute as the solubility limit is described as saturated. Adding more of the solute to a saturated solution usually induces some of the solute to separate from the solution; if the separation is by means of the formation of crystals of the solute, these crystals are said to be precipitating from the solution. In some cases, a solution may contain more solute than the solubility limit. But, such a supersaturated state is unstable, and when precipitation begins, it will rapidly lower the concentration of the solute to the saturated level.

Table 1 is a useful summary of the relative solubilities of common chemical compounds. They are classified by their anions and listed from the most soluble at the top to the least soluble at the base. It is also helpful to remember that all compounds with the cation being an alkali metal or ammonium ion are highly soluble.

It is important to realize that temperature markedly affects the solubility of most substances. For almost all salts, which are solid compounds composed of positive and negative ions (most often composed of both metallic and nonmetallic elements), an increase in temperature leads to an increase in the amount of the salt that will dissolve. Figure 1 shows the solubilities of potassium chloride (KCl) and potassium nitrate (KNO 3) as a function of temperature.

Figure 1. Temperature dependence of solubility.


Notice that at 20°C, the KCl is more soluble, but at 40°C the KNO 3 has greater solubility. Although the solubilities of both salts increase with temperature, a given temperature rise enhances the solubility of the nitrate much more than the chloride.

When a compound containing ionic bonds is placed in water, the polar water molecules separate some or all of the substance into its cations and anions. The separation is referred to as ionic dissociation.

The concentrations of the ions may not equal the concentration of the solution. As an example, take 14.2 grams of sodium sulfate (Na 2SO 4) and add it to enough water to make 1 liter of solution. The sodium sulfate is highly soluble and dissolves completely, so the solution is 0.1 molar in Na 2SO 4.


The salt, however, dissociates completely into ions:


In the preceding expression, the (s) denotes a solid and (aq) denotes an aqueous ion. In any reaction, the coefficients are proportional to the number of moles. So each mole of Na 2SO 4 yields two moles of Na + and one mole of equation. The 0.1 M solution of Na 2SO 4 is, consequently, 0.2 M in Na + and 0.1 M in equation.

For ionizing substances that are only slightly soluble, the concentrations of the ions multiply to a constant called the solubility product in a saturated solution. For a hypothetical compound CA, where the single cation is denoted by C and the anion by A, the solubility equation is

[C] [A] = K sp

where the molar concentrations of the two ions are labeled with square brackets and the constant K sp is the solubility product.

Many binary compounds (those with only two elements) contain more than one cation or anion. The general binary compound can be written C xA y, in which the subscripts mean the compound has x cations and y anions. In this case, the solubility equation is

[C] x [A] y = K sp

Practice a solubility calculation using silver carbonate (Ag 2CO 3) as the solute. Dissocation of the salt yields three aqueous ions:


Table 2 states that K sp for Ag 2CO 3 is 8.5 × 10 –12. The solubility equation involves the square of [Ag +] because each formula unit yields two ions of Ag +.


Because the molarity of equation is the same as the overall molarity of Ag 2CO 3 in the solution, call the carbonate concentration x and the silver ion concentration 2 x.


The solution, then, is 0.000129 M Ag 2CO 3, which is identical to the value found for the equation concentration. Because the gram formula mass of Ag 2CO 3 is 275.75, each liter of solution contains 


You can describe the solubility of silver carbonate as 0.0355 gram per liter.

Table 2 gives the solubility products for some important compounds that are sparingly soluble. The values are for 25°C, and each of them will vary with temperature.

You need to use Table 2 to solve the following practice problems:

  • Suppose that you stirred 0.15 gram of cuprous chloride (CuCl) powder into a liter of water. Will the powder entirely dissolve?
  • Determine the solubility in grams per liter for aluminum hydroxide, Al(OH) 3.