CHM 1046
General Chemistry II
Dr. Michael Blaber


Additional Aspects of Aqueous Equilibria

Factors that Affect Solubility


There are three effects upon the solubility of a compound that we need to consider:

  1. The presence of common ions
  2. The pH of the solution (i.e. the effect of [H+] or [OH-] on the solubility)
  3. The presence of complexing agents

Common Ion Effect

In the solubility equilibrium of CuCl we have:

CuCl(s) ó Cu+(aq) + Cl-(aq)

Ksp = 1.21 x 10-6

The solubility of a partially soluble salt is reduced by the presence of a second solute that provides a common ion


Calculate the molar solubility of CuCl at 25C in a solution containing 0.01M NaCl

The balanced equation for the dissolution of CuCl is:

CuCl(s) ó Cu+(aq) + Cl-(aq)

Ksp = 1.21 x 10-6 = [Cu+] [Cl-]

The 0.01M of NaCl will dissolve to produce 0.01M of Na+ ions and 0.01M of Cl- ions. The common ion will be the Cl- ions.

If X = the molar amount of CuCl that will dissolve at equilibrium, then at equilibrium we have:

Ksp = 1.21 x 10-6 = (X)*(0.01+X)

X2 +0.01X - 1.21 x 10-6 = 0

This is a quadratic with solutions:

X = 1.20 x 10-4M

X = -1.01 x 10-2M

X can't be negative, so the solution is 1.20 x 10-4M . Thus, this is the amount of CuCl that will dissolve upon equilibrium.

In the absence of the common ion, the solubility of CuCl is 1.10 x 10-3M (see the prior lecture on Solubility Equilibria). Thus, the addition of the common ion reduces the solubility.


Solubility and pH

Magnesium hydroxide is an ionic compound that dissociates to produce magnesium ions and hydroxide ions:

Mg(OH)2(s) ó Mg2+(aq) + 2OH-(aq)

Ksp = [Mg2+] [OH-]2 = 1.8 x 10-11

If X = the concentration of Mg(OH)2 that dissolves at equilibrium, then the solubility product constant, Ksp, can be defined as follows:

Ksp = [X] * [2X]2 = 1.8 x 10-11

4X3 = 1.8 x 10-11

X = 1.04 x 10-4M

If the same Mg(OH)2 solution is made in a buffer at pH 9.0, what is the effect upon the solubility?

Ksp = 1.8 x 10-11 = [Mg2+]*[OH-]2

1.8 x 10-11 = [Mg2+]*[1.0 x 10-5]2

[Mg2+] = 0.18M

In this case, lowering the pH reduces the equilibrium [OH-] and from Le Chatelier's principle, this will drive the reaction to the right (i.e. more salt will dissolve)

Consider the dissolution of CaF2:

CaF2(s) ó Ca2+(aq) + 2F-(aq)

F-(aq) + H+(aq) ó HF(aq)

CaF2(s) + 2H+(aq)ó Ca2+(aq) + 2HF(aq)

In both of the above cases (i.e. Mg(OH)2 and CaF2) we have seen that solubility increases with increasing [H+] (decreasing pH)

The solubility of slightly soluble salts containing basic anions increases as [H+] increases (as pH is reduced)

Formation of Complex Ions

Metal ions characteristically act as Lewis acids towards H2O(l)

AgCl(s) has a low solubility in H2O(l), but can be solubilized in H2O(l) with the addition of ammonia (NH3)

AgCl(s) ó Ag+(aq) + Cl-(aq)

Ag+(aq) + 2NH3(aq) ó Ag(NH3)2+(aq)

AgCl(s) + 2NH3(aq) ó Ag(NH3)2+(aq) + Cl-(aq)

The metal ion is hydrated (surrounded, separated and dispersed) by H2O(l) molecules


What is the concentration of Ag+(aq) in a 0.01M solution of AgNO3(s) at equilibrium if NH3(aq) is added to give an equilibrium concentration of NH3(aq) of 0.20M. Don't worry about any volume change when the ammonia is added. The equilibrium equation for the formation of the complex ion of Ag+(aq) with NH3(aq) is:

Ag+(aq) + 2NH3(aq) ó Ag(NH3)2+(aq)

And

Kf = 1.7 x 107

What about the equilibrium concentration of Ag(NH3)2+(aq)?

6.8 x 105X + X = 0.01

6.8 x 105X ~ 0.01

X = 1.47 x 10-8M

The presence of NH3(aq), and the formation of the complex ion with Ag+(aq), significantly reduces the equilibrium concentration of Ag+(aq) ion, and thus drives the dissolution of AgNO3(s)

If a metal forms a complex with a Lewis base, such as NH3, CN- or OH-, the solubility will increase

Amphoterism

If a metal hydroxide, or metal oxide, is amphoteric (i.e. can act as either an acid or a base), then it may be soluble in both acidic an basic solutions (even though it may not be soluble in neutral solutions)


© 2000 Dr. Michael Blaber