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Solubility (Precipitation Equilibria)

Definition of Solubility

Lists of insoluble salts and their Ksp values can be found in most general and analytical chemistry textbooks. Note that some ions that we consider spectator ions when discussing acid-base equilibria will form insoluble salts.

Solubility is defined as moles/L, g/L, or mg/L of the dissolving species.

Example 1: How is the solubility of lead chloride, PbCl2, related to the solution concentrations of Pb+ and Cl-?

Each formula unit of PbCl2 that dissolves produces one lead ion, Pb2+, and two chloride ions, Cl-. The molar solubility of PbCl2 is equivalent to the concentration of Pb2+, SPbCl2 = [Pb2+].

Since two Cl-s are produced per PbCl2 formula unit that dissolves, the molar solubility of PbCl2 equals one-half the solution concentration of Cl-, SPbCl2 = 0.5*[Cl-].

SPbCl2 = [Pb2+] = 0.5*[Cl-]. The following representation shows solid PbCl2 in the bottom of a beaker and ions in solution above the solid. In this example there are two lead ions and four chloride ions in solution, showing that [Pb2+] = 0.5*[Cl-].

```\                              /
|                              |
| - - - - - - - - - - - - - - -|
|    Cl-               Pb2+    |
|             Cl-              |
|        Pb2+                  |
|                    Cl-       |
|  Cl-                         |
|                              |
|  Pb      Pb      Pb      Pb  |
|Cl  Cl  Cl  Cl  Cl  Cl  Cl  Cl|
|  Pb      Pb      Pb      Pb  |
\Cl  Cl  Cl  Cl  Cl  Cl  Cl  Cl/
```

Example 2: What is the solubility of barium iodate, Ba(IO3)2, in pure water at 25 oC?

The solubility equilibrium is:

Ba(IO3)2 (s) Ba2+(aq) + 2 IO3-(aq)

Ksp = [Ba2+][IO3-]2 = 1.5x10-9

[Ba2+] = 0.5*[IO3-]

Ksp = [Ba2+](2*[Ba2+])2 = 1.5x10-9

4*[Ba2+]3 = 1.5x10-9

[Ba2+]3 = 3.75x10-10 M

S = [Ba2+] = 7.2x10-4 M

Or in mass terms: S = (7.2x10-4 M)(487 g/mol) = 0.35 g/L

There is a reason that I picked an obscure salt like barium iodate for this example. Determining the solubility of Ba(IO3)2 is simple compared to determining the solubility of many insoluble salts. Ba2+ is a spectator ion so we do not worry about it reacting in pH=7 water. What about the iodate anion? It is a base and can react with water:

IO3-(aq) + H2O HIO3(aq) + OH-(aq)

The extent to which this reaction proceeds is given by the value of Kb. Ka for iodic acid is 0.17 (it is a reasonably strong acid).

Kb = Kw/Ka = 1x10-14/0.17 = 5.9x10-14.

This value of Kb is so small that we can conclude that the hydrolysis of water by iodate does not affect the solubility of Ba(IO3)2.

Example 3: What is the solubility of barium iodate, Ba(IO3)2, at 25 oC in a solution of 0.10 M barium nitrate, Ba(NO3)2?

The problem is set up the same as before:

Ba(IO3)2 (s) Ba2+(aq) + 2 IO3-(aq)

Ksp = [Ba2+][IO3-]2 = 1.5x10-9

The difference is that there are two sources of Ba2+, the 0.10 M Ba(NO3)2 and the dissolution of Ba(IO3)2:

[Ba2+] = 0.10 M + 0.5*[IO3-]

Ksp = (0.10 M + 0.5*[IO3-])*[IO3-]2 = 1.5x10-9

We could solve a cubic equation or we could try neglecting the [Ba2+] that comes from dissolution of Ba(IO3)2 compared to 0.10 M barium nitrate.

Ksp = (0.10 M)[IO3-]2 = 1.5x10-9

[IO3-]2 = 1.5x10-8

[IO3-] = 1.2x10-4 M

S = 0.5*[IO3-], so

S = 6.1x10-5 M

Our assumption that S << 0.1 M is justified.

Compare this result to the solubility in pure water, S = 7.2x10-4 M. The solubility of barium iodate is much less when one of the ions involved in the equilibrium is present in solution. This effect is called the common-ion effect, and is explained qualitatively by LeChatelier's principle.

Will addition of other ions such as Na+ and Cl- affect this equilibrium? To a first approximation no. However, adding spectator ions to a solution increases the ionic strength and affects the activity coefficients.

In which of the following examples should we worry about activity?

• solubility of barium iodate in water
• solubility of barium iodate in a solution of 0.10 M barium nitrate
• solubility of barium iodate in a solution of 0.50 M NaCl

Example 4: Will a precipitate form when 100.0 mL of 0.2 M Fe3+ in 0.5 M H2SO4 is mixed with 100.0 mL of 1.0 M NaOH?

Before thinking about equilibrium, look at this problem and decide if anything will happen when the two solutions are mixed. A solution of a strong base is mixed with a solution containing strong acid. The acid and base will react until one of both are consumed. In this case an equal number of moles of H+ and OH- are mixed so the resulting solution will be at pH=7.

The solubility equilibrium is:

Fe(OH)3 (s) Fe3+(aq) + 3OH-(aq)

Ksp = 2x10-39 = [Fe3+] [OH-]3

To determine if a precipitate will form: find Q and compare it to Ksp. (The text calls this quantity the ion product, P, but it is the same expression as Q.)

If Q > Ksp the ion concentrations are greater than their equilibrium concentrations and a precipitate will form.

If Q < Ksp the ion concentrations are below their equilibrium concentrations and no precipitate forms.

Q = CFe3+COH-3 = (0.100 M)(1x10-7 M)3 = 1.00x10-22

1x10-22 > 2x10-39
Q > Ksp
So a precipitate will form.

What will be the [Fe3+] of the resulting solution?

2x10-39 = [Fe3+](1x10-7 M)3
[Fe3+] = 2x10-18 M

This is the predicted [Fe3+] in water. In real water systems the total amount of iron in solution can be much higher since the iron can exist in a variety of forms such as iron complexes and colloids.

Aside: Iron and sulfuric acid are present in mine drainage due to the oxidation of pyrite, FeS2. One reaction that occurs is:
FeS2(s) + 14 Fe3+(aq) + 8 H2O 15 Fe2+(aq) + 2 SO42-(aq) + 16 H+(aq)

A recent result [Katrina J. Edwards, and coworkers, Science 287 (2000) 1796.] has identified a microorganism that oxidizes Fe2+ to Fe3+ to keep this reaction going.

When the mine drainage is diluted sufficiently to raise the pH, the iron precipitates as iron hydroxide and creates an orange coating in streams and rivers.

Competing Equilibria

What is the solubility of CaCO3 in water at 25 oC?

The solubility is equal to [Ca2+]. The dissolution equilibrium is:

CaCO3 (s) Ca2+(aq) + CO32-(aq)

but there are competing equilibria:

CO32-(aq) + H2O HCO3-(aq) + OH-(aq)

We will neglect HCO3-(aq) + H2O H2CO3(aq) + OH-(aq). Solving the problem with this equilibrium included follows the same procedure, just with more equations.

Ksp = [Ca2+][CO32-] = 4.5x10-9

```       [HCO3-][OH-]
Kb  =  ------------  =  2.13x10-4
[CO32-]
```

We can always write mass and charge balance expressions to get more equations to solve for multiple unknowns:

[Ca2+] = [CO32-] + [HCO3-] + [H2CO3]

2[Ca2+] + [H+] = 2[CO32-] + [HCO3-] + [OH-]

We are neglecting H2CO3 and since we know the solution will be basic we can probably neglect [H+], which will be less than 1x10-7 M. We can check this assumption when we get an answer. We then have:

[Ca2+] = [CO32-] + [HCO3-]

2[Ca2+] = 2[CO32-] + [HCO3-] + [OH-]

If we multiply the simplified mass balance expression by two and relate it to the charge balance expression we get:

2[CO32-] + 2[HCO3-] = 2[CO32-] + [HCO3-] + [OH-]

[HCO3-] = [OH-]

We can use this relationship to simplify the Kb expression:

```       [HCO3-]2
Kb  =  --------
[CO32-]
```

or [HCO3-] = (Kb[CO32-])1/2

So we can eliminate [HCO3-] from [Ca2+] = [CO32-] + [HCO3-] to get:

[Ca2+] = [CO32-] + (Kb[CO32-])1/2

Now we can eliminate [CO32-] using the Ksp expression.

[Ca2+] = Ksp/[Ca2+] + (KbKsp/[Ca2+])1/2

Multiplying each side by [Ca2+] and rearranging gives:

[Ca2+]2 - (KbKsp)1/2[Ca2+]1/2 - Ksp = 0

or
[Ca2+]2 - (9.8x10-7)[Ca2+]1/2 - 4.5x10-9 = 0

The easiest way to solve this equation is to put the expression [Ca2+]2 + (9.8x10-7)[Ca2+]1/2 - 4.5x10-9 in a spreadsheet and try a series of values for [Ca2+] to find the one that gives a result closest to zero.

In the absence of the second equilibrium the solubility would have been:

Ksp = 4.5x10-9 = [Ca2+][CO32-]

[Ca2+] = (4.5x10-9)1/2

S = 6.7x10-5 M

Which gives us a starting point for finding [Ca2+].

[Ca2+][Ca2+]2 - (9.8x10-7)[Ca2+]1/2 - 4.5x10-9
6.00x10-5-8.49x10-9
7.00x10-5-7.80x10-9
8.00x10-5-6.87x10-9
9.00x10-5-5.70x10-9
1.00x10-4-4.30x10-9
1.20x10-4-8.35x10-10
1.30x10-41.23x10-9
1.40x10-43.50x10-9

This set of data shows that the solution is between 1.20x10-4 and 1.30x10-4. We can do another set of data at smaller intervals to pin down the solution:

[Ca2+][Ca2+]2 - (9.8x10-7)[Ca2+]1/2 - 4.5x10-9
1.22x10-4-4.40x10-10
1.23x10-4-2.40x10-10
1.24x10-4-3.68x10-11
1.25x10-41.68x10-10
1.26x10-43.76x10-10

[Ca2+] = 1.2x10-4 M

The competing equilibrium for carbonate increases the calcium carbonate solubility by a factor of approximately 2 (LeChatelier strikes again).

The assumption we made that 2[Ca2+] >> [H+] in basic solution was valid.

More Metal Hydroxide Equilibria and Another Intro to Complexation

An example where Ksp is relatively large and we can neglect competing equilibrium:

Mg(OH)2 (s) Mg2+(aq) + 2 OH-(aq)

Ksp = [Mg2+][OH-]2 = 7.1x10-12

The solubility of Mg(OH)2 is equal to [Mg2+].

We need another equation to solve for [Mg2+]. From the stoichiometry of the reaction we know that 2 [Mg2+] = [OH-]. Substituting into the Ksp expression to eliminate [OH-]:

Ksp = [Mg2+](2 [Mg2+])2 = 7.1x10-12

4 [Mg2+]3 = 7.1x10-12

[Mg2+]3 = 1.8x10-12

[Mg2+] = 1.2x10-4

An example where Ksp is very small and we can neglect hydroxide from the metal hydroxide dissolution:

Al(OH)3 (s) Al3+(aq) + 3 OH-(aq)

Ksp = [Al3+][OH-]3 = 3x10-34

The solubility of Al(OH)3 is equal to [Al3+].

Since Ksp is so small we can try solving this problem assuming that the [OH-] from the Al(OH)3 dissolution is negligible compared to the [OH-] from the dissociation of water.

Ksp = [Al3+][OH-]3 = 3x10-34

[Al3+] = 3x10-34 / [OH-]3

pH[OH-][Al3+]
31x10-113x10-1
41x10-103x10-4
51x10-93x10-7
61x10-83x10-10
71x10-73x10-13

At some pH the [Al3+] reaches a minimum and then increases with further increase in pH. The solubility of Al(OH)3 increases because a soluble complex, Al(OH)4-, begins to form at high pH.

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