E - Solubility Equilibria (Lesson)

Solubility Equilibria

The topics covered in this module apply to all manner of equilibria without exception. Attention will now be turned to a specific example of chemical equilibrium involving heterogeneous equilibria wherein the substances involved in the equilibrium are not all in the same phase of matter (e.g. not all gases or all aqueous solutions). The equilibria to be considered here are examples of solubility equilibria which describe the scenario of a substance that is only sparingly soluble in water. Shown below is an equilibrium representing the sparingly soluble salt, BaSO₄.

BaSO₄ (s)  $\rightleftharpoons$  Ba⁺² (aq) + SO₄^-2 (aq)

This is the equilibrium that would exist in a saturated solution where some solid barium sulfate still remains. To be clear, this is also a dynamic equilibrium much like any other equilibrium process, meaning barium sulfate is simultaneously dissolving into its constituent ions while the ions are also precipitating to form barium sulfate as illustrated in the image below.

As with any other equilibrium situation, an expression can be written for the equilibrium above with one notable difference. Because only gases or aqueous solutions appear in equilibrium expressions, the situation arises where a reactant that fits this criteria is not present. As a result, the equilibrium expression will not have a denominator as evidenced below:

K_sp = [Ba⁺²][SO₄^-2]

The same rules for writing equilibrium expressions apply to solubility equilibria: the expression contains products divided by reactants (although the reactants are always solid and therefore omitted) and the concentrations of each product ion are multiplied together with coefficients from the original chemical equations used as exponents.

Calculating K_sp

When discussing solubility in the context of K_sp it is important to distinguish between two different types of solubility:

• Mass solubility (typically expressed in g/L)
• Molar solubility (expressed in mol/L or M)

Consider the solubility equilibria of fluorite, or CaF₂, shown below:

CaF₂ (s)  $\rightleftharpoons$  Ca⁺² (aq) + 2 F^- (aq)

Assume that a saturated solution of fluorite contains a [Ca⁺²] = 2.15 x 10^-4 M. From this information alone it is possible to calculate the value of the solubility product constant, K_sp. Using basic stoichiometry we can determine the [F^-] that is also present at equilibrium:

[F^-] = 2[Ca⁺²] = (2)(2.15 x 10^-4) = 4.30 x 10^-4 M

These values can then be inserted into the equilibrium expression below:

K_sp = [Ca⁺²][F^-]² = (2.15 x 10^-4)(4.30 x 10^-4)² = 3.98 x 10^-11

As with all other equilibrium constants the value for K_sp is a unitless value. Also, it can be inferred from the magnitude of this value that the reactants are highly favored. In other words, sparingly soluble compounds truly are sparingly soluble. Very few ions would exist in these solutions as evidenced by the more quantitative calculations shown below.

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