Solubility-Product Constants (Chemical Equilibrium)

Solubility-Product Constants

** Most, but not all, sparingly soluble salts are essentially completely dissociated in saturated aqueous solution. 

** For example, when an excess of barium iodate is equilibrated with water, the dissociation process is adequately described by the equation:

Using Equation:

we write:

** The denominator represents the molar concentration of Ba(IO₃)₂ in the solid, which is a phase that is separate from but in contact with the saturated solution. The concentration of a compound in its solid state is, however, constant. In other words, the number of moles of Ba(IO₃)₂ divided by the volume of the solid Ba(IO₃)₂ is constant no matter how much excess solid is present. Therefore, the previous equation can be rewritten in the form:

where the new constant is called the solubility-product constant or the solubility product. It is important to appreciate that Equation above shows that the position of this equilibrium is independent of the amount of Ba(IO₃)₂ as long as some solid is present. In other words, it does not matter whether the amount is a few milligrams or several grams.

Solubility-product constants at 25 °C inorganic salts

** The examples that follow demonstrate some typical uses of solubility product expressions.

The Solubility of a Precipitate in Pure Water

  With the solubility-product expression, we can calculate the solubility of a sparingly soluble substance that ionizes completely in water.

Example (1): What mass (in grams) of Ba(IO₃)₂ (487 g/mol) can be dissolved in 500 mL of water at 25 °C?

Solution

** The solubility-product constant for Ba(IO₃)₂ is 1.57 × 10^-9 (see table above). The equilibrium between the solid and its ions in solution is described by the equation:

** The equation describing the equilibrium reveals that 1 mol of Ba²⁺ is formed for each mole of Ba(IO₃)₂ that dissolves. Therefore,

** Since two moles of iodate are produced for each mole of barium ion, the iodate concentration is twice the barium ion concentration:

** Substituting this last equation into the equilibrium-constant expression gives:

** Since 1 mol Ba²⁺ is produced for every mole of Ba(IO₃)₂,

** To compute the number of millimoles of Ba(IO₃)₂ dissolved in 500 mL of solution, we write

** The mass of Ba(IO₃)₂ in 500 mL is given by:

The Effect of a Common Ion on the Solubility of a Precipitate

  The common-ion effect is a mass-action effect predicted from Le Châtelier’s principle

and is demonstrated by the following examples.

Example (2): Calculate the molar solubility of Ba(IO₃)₂ in a solution that is 0.0200 M in Ba(NO₃)₂.

Solution

** The solubility is not equal to [Ba²⁺] in this case because Ba(NO₃)₂ is also a source of barium ions. We know, however, that the solubility is related to [IO₃^-]:

** There are two sources of barium ions: Ba(NO₃)₂ and Ba(IO₃)₂. The contribution from the nitrate is 0.0200 M, and that from the iodate is equal to the molar solubility,or ½ [IO₃^-]. Thus,

** By substituting these quantities into the solubility-product expression, we find that

** Since this is a cubic equation, we would like to make an assumption that would simplify the algebra required to find [IO₃^-]. The small numerical value of K_sp suggests that the solubility of Ba(IO₃)₂ is quite small, and this finding is confirmed by the result obtained in Example (1). Also, barium ion from Ba(NO₃)₂ will further suppress the limited solubility of Ba(IO₃)₂. Therefore, it seems reasonable to assume that 0.0200 is large with respect to ½ [IO₃^-] in order to find a provisional answer to the problem. That is, we assume that ½ [IO₃^-] << 0.0200, so

** The original equation then simplifies to:

** The assumption that (0.0200 + ½ × 2.80 × 10^-4) ~ 0.0200 causes minimal error because the second term, representing the amount of Ba²⁺ arising from the dissociation of Ba(IO₃)₂, is only about 0.7% of 0.0200. Usually, we consider an assumption of this type to be satisfactory if the discrepancy is less than 10%. Finally, then,

** If we compare this result with the solubility of barium iodate in pure water (Example 1), we see that the presence of a small concentration of the common ion has decreased the molar solubility of Ba(IO₃)₂ by a factor of about 5.

Example (3): Calculate the solubility of Ba(IO₃)₂ in a solution prepared by mixing 200 mL of 0.0100 M Ba(IO₃)₂ with 100 mL of 0.100 M NaIO₃.

Solution

** First, establish whether either reactant is present in excess at equilibrium. The amounts taken are:

** If the formation of Ba(IO₃)₂ is complete:

** Thus,

** As in Example (1),

** In this case, however,

** where 2[Ba²⁺] represents the iodate contributed by the sparingly soluble Ba(IO₃)₂. We find a provisional answer after making the assumption that [IO₃^-] ~ 0.0200. Therefore,

** Since the provisional answer is nearly four orders of magnitude less than 0.0200 M, our approximation is justified, and the solution does not need further refinement.

** Notice that the results from the last two examples demonstrate that an excess of iodate ions is more effective in decreasing the solubility of Ba(IO₃)₂ than is the same excess of barium ions.

Reference: Fundamentals of analytical chemistry / Douglas A. Skoog, Donald M. West, F. James Holler, Stanley R. Crouch. (ninth edition) , 2014 . USA