CHM 1046
General Chemistry II
Dr. Michael Blaber

Factors that Affect Solubility

• The solubility product constant, Ksp, varies with temperature, therefore, temperature will influence the solubility of a compound.
• However, the presence of other solutes (i.e. other dissolved ions or compounds) can also influence the solubility - although they do not alter Ksp

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

• From Le Chatelier's principle, addition of Cu+ ion, or Cl- ion, will shift the equilibrium to the left
• This will favor formation of the solid, and reduce the concentration of solvated ions, and the solubility of the compound will decrease

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 25°C 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:

• XM of Cu+ ion
• (0.01 + XM) of Cl- ion

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

• The [Mg2+] at equilibrium equals 1.04 x 10-4
• The [OH-] at equilibrium equals 2.08 x 10-4M
• The pOH therefore equals 3.68, and pH is therefore (14.0 - 3.68) = 10.3

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

• At pH = 9.0, the pOH = (14 - 9.0) = 5.0
• Therefore, the [OH-] = 1.0 x 10-5M
• Since the solution is buffered, the [OH-] at equilibrium will also be 1 x 10-5M

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

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

[Mg2+] = 0.18M

• Note from above that at the pH of 10.6 the concentration of dissolved Mg2+ ion was 2.08 x 10-4. Therefore, more of the Mg(OH)2 salt dissolved at pH 9.0 compared to pH 10.6

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)

• The F-(aq) ion is a weak base and can combine with H+(aq) to produce the weak acid HF:

F-(aq) + H+(aq) ó HF(aq)

• In aqueous solution, therefore, the overall (balanced) equation for the dissolution of CaF2(s) would consist of two consecutive reactions whose net reaction would be:

CaF2(s) + 2H+(aq)ó Ca2+(aq) + 2HF(aq)

• Thus, from Le Chatelier's principle, as the [H+] increases (i.e. as pH decreases) the reaction is driven to the right (more of the solid dissolves)

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

• In both cases the solid dissolves to produce an anion that is basic in nature. Increasing the [H+] essentially removes the free anion from solution by forming the weak acid. This drives the reaction to the right and more solid dissolves.

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

• The more basic the anion, the more pronounced the effect

Formation of Complex Ions

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

• They accept a non-bonding pair of electrons from H2O(l) which behaves as a Lewis base
• Other compounds can act as Lewis bases towards metal ions
• Such interactions can affect the solubility of the metal ion

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 presence of NH3(aq) will drive the dissolution of AgCl(s) because it effectively removes free Ag+(aq) ions from solution (thus, the top reaction above is driven to the right by Le Chatelier's principle)

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

• In order for the NH3(aq) molecules to act as a Lewis base with the metal ions, they must have a greater affinity for the metal ion than do the H2O(l) molecules
• An assembly of a metal ion and the Lewis bases bonded to it, is called a complex ion
• The stability of a complex ion can be judged by the magnitude of the equilibrium constant for its formation 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

• The concentration of Ag+(aq) at equilibrium = XM
• The concentration of NH3(aq) at equilibrium is given as 0.20M

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

• Kf is fairly large. Therefore, NH3(aq) will be quite effective at removing the Ag+(aq) ion. Since we have an excess of NH3(aq) compared to AgNO3(s) we can assume that almost all of the AgNO3(s) will be converted to either Ag+(s) or complex ion
• Thus, at equilibrium the concentration of Ag(NH3)2+(aq) = (0.01 - X)M 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)

• Acids promote dissolving of compounds containing basic anions
• Bases promote dissolving of compounds containing acidic cations