Chapter 16: Equilibria of Other Reaction Classes
16.1 Precipitation and Dissolution
Learning Outcomes
- Write chemical equations and equilibrium expressions representing solubility equilibria
- Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations
Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced chapter on equilibrium concepts and tools to systems involving dissolution and precipitation.
The Solubility Product
Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride (AgCl) is one in which the equilibrium shown below has been established.
In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous

The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is called the solubility product (Ksp) of the solid. Recall from the chapter on solutions and colloids that we use an ion’s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:
Recall that only gases and solutes are represented in equilibrium constant expressions, so the Ksp does not include a term for the undissolved
Table 16.1.1. Common Solubility Products by Decreasing Equilibrium Constants | |
---|---|
Substance | Ksp at 25 °C |
1.2 × 10–6 | |
6.27 × 10–9 | |
1.5 × 10–16 | |
7 × 10–29 | |
2 × 10–32 | |
4 × 10–38 |
Example 16.1.1: Writing Equations and Solubility Products
Write the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:
, silver iodide, a solid with antiseptic properties , calcium carbonate, the active ingredient in many over-the-counter chewable antacids , magnesium hydroxide, the active ingredient in Milk of Magnesia , magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium , the mineral apatite, a source of phosphate for fertilizers
(Hint: When determining how to break 4 and 5 up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)
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Now we will extend the discussion of Ksp and show how the solubility product constant is determined from the solubility of its ions, as well as how Ksp can be used to determine the molar solubility of a substance.
Ksp and Solubility
The Ksp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:
For cases such as these, one may derive Ksp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.
Example 16.1.2: Calculation of Ksp from Equilibrium Concentrations
Fluorite,
The concentration of
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According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a
Substituting the ion concentrations into the Ksp expression gives
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Example 16.1.3: Determination of Molar Solubility from Ksp
The Ksp of copper(I) bromide,
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The solubility product constant of copper(I) bromide is 6.3 × 10–9. The reaction is:
First, write out the solubility product equilibrium constant expression:
Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the
At equilibrium:
Therefore, the molar solubility of
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Example 16.1.4: Determination of Molar Solubility from Ksp, Part II
The Ksp of calcium hydroxide,
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The dissolution equation and solubility product expression are
Create an ICE table, leaving the
At equilibrium:
Therefore, the molar solubility of
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Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. Example 16.1.5 shows how to perform those unit conversions before determining the solubility product equilibrium.
Example 16.1.5: Determination of Ksp from Gram Solubility
Many of the pigments used by artists in oil-based paints (Figure 16.1.2) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow,

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We are given the solubility of Step 1. Use the molar mass of
Step 2. The chemical equation for the dissolution indicates that 1 mol of
The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both
Step 3. Solve.
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Example 16.1.6: Calculating the Solubility of
Calomel,
Calculate the molar solubility of
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The molar solubility of
- Determine the direction of change. Before any
dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium. - Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:
- Note that the change in the concentration of
(2x) is twice as large as the change in the concentration of (x) because 2 mol of Cl– forms for each 1 mol of that forms. is a pure solid, so it does not appear in the calculation. - Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for Ksp and calculate the value of x:
The molar solubility of is equal to , or 6.5 × 10–7M. - Check the work. At equilibrium, Q = Ksp:
The calculations check.
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Tabulated Ksp values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: Q equals Ksp at equilibrium; if Q is less than Ksp, the solid will dissolve until Q equals Ksp; if Q is greater than Ksp, precipitation will occur at a given temperature until Q equals Ksp.
Predicting Precipitation
The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:
It is important to realize that this equilibrium is established in any aqueous solution containing
Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)
Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)
This predictive strategy and related calculations are demonstrated in the next few example exercises.
Example 16.1.7: Precipitation of
The first step in the preparation of magnesium metal is the precipitation of
The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough
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This problem asks whether the reaction:
shifts to the left and forms solid
Because Q is greater than Ksp (Q = 5.4 × 10–8 is larger than Ksp = 2.1 × 10–13), we can expect the reaction to shift to the left and form solid magnesium hydroxide.
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Example 16.1.8: Precipitation of AgCl upon Mixing Solutions
Does silver chloride precipitate when equal volumes of a 2.0 × 10–4–M solution of
(Note: The solution also contains Na+ and
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The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:
The solubility product is 1.6 × 10–10 (see Solubility Products).
The reaction quotient, Q, is momentarily greater than Ksp for
Since supersaturated solutions are unstable,
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In the previous two examples, we have seen that Mg(OH)2 or
Example 16.1.9: Precipitation of Calcium Oxalate
Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion,

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The equilibrium expression is:
For this reaction:
Solid
A concentration of
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It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example 16.1.9—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.
Example 16.1.10: Concentrations Following Precipitation
Clothing washed in water that has a manganese [
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The dissolution of
We need to calculate the concentration of
so
Now we calculate the pH from the pOH:
(final result rounded to one significant digit, limited by the certainty of the Ksp)
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In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.
Example 16.1.11: Precipitation of Silver Halides
A solution contains 0.00010 mol of
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The two equilibria involved are:
If the solution contained about equal concentrations of
For
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Common Ion Effect
Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le ChÂtelier’s principle. Consider the dissolution of silver iodide:
This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved
This effect may also be explained in terms of mass action as represented in the solubility product expression:
The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion’s concentration must be balanced by a proportional decrease in the other.
Example 16.1.12: Common Ion Effect on Solubility
What is the effect on the amount of solid
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The solubility equilibrium is
- Adding a common ion,
, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide. - Adding a common ion,
, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide. - The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.
- Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.
- Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.
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Key Concepts and Summary
The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, Ksp, of the solid. When we have a heterogeneous equilibrium involving the slightly soluble solid MpXq and its ions Mm+ and Xn–:
We write the solubility product expression as:
The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its Ksp, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.
A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.
Key Equations
Try It
- Complete the changes in concentrations for each of the following reactions:
- How do the concentrations of
and in a saturated solution above 1.0 g of solid change when 100 g of solid is added to the system? Explain. - Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Solubility Products):
- the mineral brucite,
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- In dissolution, one unit of substance produces a quantity of discrete ions or polyatomic ions that equals the number of times that the subunit appears in the formula.
Dissolving AgI(s) must produce the same amount of I– ion as it does Ag+ ion.
Dissolving (s) must produce the same amount of ion as it does ion.
When one unit of dissolves, two ions of are formed for each ion.
One unit of provides two units of ion and three units of ion.
One unit of dissolves into five units of ion, three units of ion, and one unit of ion.
- There is no change. A solid has an activity of 1 whether there is a little or a lot.
-
- In each of the following, allow x to be the molar concentration of the ion occurring only once in the formula.
-
- Ksp = [Ag+][I–] = 1.5 – 10–16 = [x2], [x] = 1.2 – 10–8M, [Ag+] = [I–] = 1.2 × 10–8M
- Ksp =
= 1.18 × 10–5 = [2x]2[x], 4x3 = 1.18 × 10–5, x = 1.43 × 10–2M
As there are 2 Ag+ ions for each ion, [Ag+] = 2.86 × 10–2M, = 1.43 × 10–2M; (c) Ksp = [Mn2+]2[OH–]2 = 4.5 × 10–14 = [x][2x]2, 4x3 = 4.5 × 10–14, x = 2.24 × 10–5M. - Since there are two OH– ions for each Mn2+ ion, multiplication of x by 2 gives 4.48 × 10–5M. If the value of x is rounded to the correct number of significant figures, [Mn2+] = 2.2 × 10–5M. [OH–] = 4.5 × 10–5M. If the value of x is rounded before determining the value of [OH–], the resulting value of [OH–] is 4.4 × 10–5M. We normally maintain one additional figure in the calculator throughout all calculations before rounding.
- Ksp = [Sr2+][OH–]2 = 3.2 × 10–4 = [x][2x]2, 4x3 = 3.2 × 10–4, x = 4.3 × 10–2M.
Substitution gives [Sr2+] = 4.3 × 10–2 M, [OH–] = 8.6 × 10–2M - Ksp = [Mg2+]2[OH–]2 = 1.5 × 10–11 = [x][2x]2, 4x3 = 1.5 × 10–11, x = 1.55 × 10–4M, 2x = 3.1 × 10–4.
Substitution and taking the correct number of significant figures gives [Mg2+] = 1.6 × 10–4M, [OH–] = 3.1 × 10–4M. If the number is rounded first, the first value is still [Mg2+] = 1.6 × 10–4M, but the second is [OH–] = 3.2 × 10–4M.
Glossary
common ion effect: effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base
molar solubility: solubility of a compound expressed in units of moles per liter (mol/L)
selective precipitation: process in which ions are separated using differences in their solubility with a given precipitating reagent
solubility product (Ksp): equilibrium constant for the dissolution of a slightly soluble electrolyte
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equilibrium constant for the dissolution of a slightly soluble electrolyte
solubility of a compound expressed in units of moles per liter (mol/L)
process in which ions are separated using differences in their solubility with a given precipitating reagent
effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base