19.3: Solubility Equilibrium
Lesson Objectives
 Write solubility product constant expressions for nearly insoluble ionic compounds.
 Calculate the value of K_{sp} for a compound from its solubility and determine the solubility of a compound with a known K_{sp}.
 Use the K_{sp} of a compound to predict whether a precipitate will form when two solutions are mixed together.
 Describe the common ion effect and its relationship to solubility.
Lesson Vocabulary
 common ion
 common ion effect
 molar solubility
 solubility product constant
Check Your Understanding
Recalling Prior Knowledge
 What is a saturated solution?
 What is the equilibrium that occurs in a saturated solution?
A saturated aqueous solution is one in which the maximum amount of a solute has been dissolved in a given amount of water. A saturated solution may also have undissolved solute present, in which case an equilibrium exists between the dissolved and undissolved solute. In this lesson, you will learn about that equilibrium and how to calculate and use the solubility product constant.
The Solubility Product Constant
Ionic compounds have widely differing solubilities. Sodium chloride has a solubility of about 360 g per liter of water at 25°C. Salts of alkali metals tend to be quite soluble. On the other end of the spectrum, the solubility of zinc hydroxide is only 4.2 × 10^{−4} g/L of water at the same temperature. Many ionic compounds containing hydroxide are relatively insoluble. The chapter Solutions summarized a set of rules for predicting the relative solubilities of various ionic compounds in water.
Most ionic compounds that are considered to be insoluble will still dissolve to a small extent in water. These “mostly insoluble” compounds are still considered to be strong electrolytes, because essentially any portion of the compound that dissolves will also dissociate into ions. As an example, silver chloride dissociates to a small extent into silver ions and chloride ions upon being added to water.


 \begin{align*}\mathrm{AgCl}(s) \mathrm{\rightleftharpoons Ag^+}(aq) \mathrm{+Cl^}(aq)\end{align*}

The process is written as an equilibrium because the dissociation occurs only to a small extent. Therefore, an equilibrium expression can be written for the process. Keep in mind that the solid silver chloride does not have a variable concentration, so it is not included in the equilibrium expression.


 \begin{align*}\mathrm{K_{sp}=[Ag^+][Cl^]}\end{align*}

This equilibrium constant is called the solubility product constant, (K_{sp}) and is equal to the mathematical product of the ions each raised to the power of the coefficient of the ion in the dissociation equation.
The formula of the ionic compound dictates the form of the K_{sp} expression. For example, the formula of calcium phosphate is Ca_{3}(PO_{4})_{2}. The dissociation equation and K_{sp} expression are shown below:


 \begin{align*}\mathrm{Ca_3(PO_4)_2}(s) \mathrm{\rightleftharpoons 3Ca^{2+}}(aq) \mathrm{+2PO^{3}_4}(aq)\end{align*}



 \begin{align*}\mathrm{K_{sp}=[Ca^{2+}]^3[PO^{3}_4]^2}\end{align*}

Table below lists solubility product constants for some common nearly insoluble ionic compounds.
Compound  K_{sp}  Compound  K_{sp} 

AgBr  5.0 × 10^{−13}  CuS  8.0 × 10^{−37} 
AgCl  1.8 × 10^{−10}  Fe(OH)_{2}  7.9 × 10^{−16} 
Al(OH)_{3}  3.0 × 10^{−34}  Mg(OH)_{2}  7.1 × 10^{−12} 
BaCO_{3}  5.0 × 10^{−9}  PbCl_{2}  1.7 × 10^{5} 
BaSO_{4}  1.1 × 10^{−10}  PbCO_{3}  7.4 × 10^{14} 
CaCO_{3}  4.5 × 10^{−9}  PbI_{2}  7.1 × 10^{9} 
Ca(OH)_{2}  6.5 × 10^{−6}  PbSO_{4}  6.3 × 10^{−7} 
Ca_{3}(PO_{4})_{2}  1.2 × 10^{−26}  Zn(OH)_{2}  3.0 × 10^{−16} 
CaSO_{4}  2.4 × 10^{−5}  ZnS  3.0 × 10^{−23} 
Solubility and K_{sp}
Solubility is normally expressed in grams of solute per liter of saturated solution. However, solubility can also be expressed as moles per liter. Molar solubility is the number of moles of solute in one liter of a saturated solution. In other words, the molar solubility of a given compound represents the highest molarity solution that is possible for that compound. The molar mass of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of Zn(OH)_{2} is 4.2 × 10^{−4} g/L, the molar solubility can be calculated as shown below:


 \begin{align*}\mathrm{\dfrac{4.2 \times 10^{4} \ \cancel{\text{g}}}{1 \ L} \times \dfrac{1 \ mol}{99.41 \ \cancel{\text{g}}}=4.2 \times 10^{6} \ mol/L \ (M)}\end{align*}

Solubility data can be used to calculate the K_{sp} for a given compound. The following steps need to be taken.
 Convert from solubility to molar solubility.
 Use the dissociation equation to determine the concentration of each of the ions in mol/L.
 Apply the K_{sp} equation.
Sample Problem 19.2: Calculating K_{sp} from Solubility
The solubility of lead(II) fluoride is found experimentally to be 0.533 g/L. Calculate the K_{sp} for lead(II) fluoride.
Step 1: List the known quantities and plan the problem.
Known
 solubility of PbF_{2} = 0.533 g/L
 molar mass of PbF_{2} = 245.20 g/mol
Unknown
 K_{sp} of PbF_{2} = ?
The dissociation equation for PbF_{2} and the corresponding K_{sp} expression can be constructed as follows:


 \begin{align*}\mathrm{PbF_2}(s) \mathrm{\rightleftharpoons Pb^{2+}}(aq) \mathrm{+2F^}(aq) \ \ \ \ \text{K}_{\text{sp}} = [\text{Pb}^{2+}][\text{F}^]^2\end{align*}

The steps above will be followed to calculate K_{sp} for PbF_{2}.
Step 2: Solve.


 molar solubility: \begin{align*}\mathrm{\dfrac{0.533 \ \cancel{\text{g}}}{1 \ L} \times \dfrac{1\ mol}{245.20\ \cancel{\text{g}}}=2.17 \times 10^{3} \ M}\end{align*}

The dissociation equation shows that for every mole of PbF_{2} that dissociates, 1 mol of Pb^{2+} and 2 mol of F^{−} are produced. Therefore, at equilibrium, the concentrations of the ions are:


 [Pb^{2+}] = 2.17 × 10^{−3} M and [F^{−}] = 2 × 2.17 × 10^{−3} = 4.35 × 10^{−3} M

Substitute into the equilibrium expression and solve for K_{sp}.


 K_{sp} = (2.17 × 10^{3})(4.35 × 10^{3})^{2} = 4.11 × 10^{8}

Step 3: Think about your result.
The solubility product constant is significantly less than 1 for a nearly insoluble compound such as PbF_{2}.
 From the given solubility data, calculate K_{sp} for each of the following compounds.
 copper(II) iodide, CuI = 4.30 × 10^{−4} g/L
 silver sulfide, Ag_{2}S = 2.84 × 10^{−15} g/L
The known K_{sp} values from Table above can be used to calculate the solubility of a given compound by following the steps listed below.
 Set up an ICE problem (Initial, Change, Equilibrium) in order to use the K_{sp} value to calculate the concentration of each of the ions. Assume that no ions are initially present in the solution.
 The concentrations of the ions can be used to calculate the molar solubility of the compound.
 Use the molar mass to convert from molar solubility to solubility in g/L.
The K_{sp} value of calcium carbonate is 4.5 × 10^{−9}. We begin by setting up an ICE table showing the dissociation of CaCO_{3} into calcium ions and carbonate ions. The variable s will be used to represent the molar solubility of CaCO_{3}. In this case, each formula unit of CaCO_{3} yields one Ca^{2+} ion and one CO_{3}^{2−} ion. Therefore, the equilibrium concentrations of each ion are equal to s.
\begin{align*}\mathrm{CaCO_3}(s) \mathrm{\rightleftharpoons}\end{align*}  \begin{align*}\mathrm{Ca^{2+}}(aq)+\end{align*}  \begin{align*}\mathrm{CO^{2}_3}(aq)\end{align*}  

Initial (M)  0.00  0.00  
Change (M)  +s  +s  
Equilibrium (M)  s  s 
The K_{sp} expression can be written in terms of s and then used to solve for s.


 \begin{align*}\mathrm{K_{sp}=[Ca^{2+}][CO_3^{2}]=(s)(s)=s^2}\end{align*}



 \begin{align*}\mathrm{s=\sqrt{K_{sp}}=\sqrt{4.5 \times 10^{9}}=6.7 \times 10^{5}\:M}\end{align*}

The concentration of each of the ions at equilibrium is 6.7 × 10^{−5} M. We can now use the molar mass to convert from molar solubility to solubility in g/L.


 \begin{align*}\dfrac{6.7 \times 10^{5}\ \cancel{\text{mol}}}{1 \ \text{L}} \times \dfrac{100.09 \ \text{g}}{1 \ \cancel{\text{mol}}}=6.7 \times 10^{3} \ \text{g/L}\end{align*}

So the maximum amount of calcium carbonate that is capable of dissolving in 1 liter of water at 25°C is 6.7 × 10^{−3} grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the K_{sp} being equal to s^{2}. This is referred to as a formula of the type AB, where A is the cation and B is the anion. Now let’s consider a formula of the type AB_{2}, such as Fe(OH)_{2}. In this case, the setup of the ICE table would look like the following:
\begin{align*}\mathrm{Fe(OH)_2}(s) \mathrm{\rightleftharpoons}\end{align*}  \begin{align*}\mathrm{Fe^{2+}}(aq)+\end{align*}  \begin{align*}\mathrm{2OH^}(aq)\end{align*}  

Initial (M)  0.00  0.00  
Change (M)  +s  +2s  
Equilibrium (M)  s  2s 


 \begin{align*}\mathrm{K_{sp}=[Fe^{2+}][OH^]^2=(s)(2s)^2=4s^3}\end{align*}



 \begin{align*}\text{s}=\sqrt[3]{\frac{\text{K}_{\text{sp}}}{4}}=\sqrt[3]{\frac{7.9 \times 10^{16}}{4}}=5.8 \times 10^{6} \ \text{M}\end{align*}

Table below shows the relationship between K_{sp} and molar solubility based on the formula.
Compound Type  Example  K_{sp} Expression  Cation  Anion  K_{sp} in Terms of s 

AB  CuS  [Cu^{2+}][S^{2−}]  s  s  s^{2} 
AB_{2} or A_{2}B  Ag_{2}CrO_{4}  [Ag^{+}]^{2}[CrO_{4}^{2−}]  2s  s  4s^{3} 
AB_{3} or A_{3}B  Al(OH)_{3}  [Al^{3+}][OH^{−}]^{3}  s  3s  27s^{4} 
A_{2}B_{3} or A_{3}B_{2}  Ba_{3}(PO_{4})_{2}  [Ba^{2+}]^{3}[PO_{4}^{3−}]^{2}  3s  2s  108s^{5} 
The K_{sp} expressions in terms of s can be used to solve problems in which the K_{sp} is used to calculate the molar solubility as in the examples above. Molar solubility can then be converted to solubility in g/L.
Predicting Precipitates
Knowledge of K_{sp} values will allow you to be able to predict whether or not a precipitate will form when two solutions are mixed together. For example, suppose that a known solution of barium chloride is mixed with a known solution of sodium sulfate. Barium sulfate (Figure below) is a mostly insoluble compound that could potentially precipitate from the mixture. However, it is first necessary to calculate the ion product, [Ba^{2+}][SO_{4}^{2−}], for the solution. If the value of the ion product is less than the value of the K_{sp}, then the solution will remain unsaturated. No precipitate will form because the concentrations are not high enough to begin the precipitation process. If the value of the ion product is greater than the value of K_{sp}, then a precipitate will form. The formation of the precipitate lowers the concentration of each of the ions until the ion product is exactly equal to K_{sp}, at which point equilibrium is attained.
Barium sulfate is used as a component of white pigments in paints and as an agent in certain xray imaging processes.
Sample Problem 19.3: Predicting Precipitates
Will a precipitate of barium sulfate form when 10.0 mL of 0.0050 M BaCl_{2} is mixed with 20.0 mL of 0.0020 M Na_{2}SO_{4}?
Step 1: List the known quantities and plan the problem.
Known
 concentration of BaCl_{2} solution = 0.0050 M
 volume of BaCl_{2} solution = 10.0 mL
 concentration of Na_{2}SO_{4} solution = 0.0020 M
 volume of Na_{2}SO_{4} solution = 20.0 mL
 K_{sp} of BaSO_{4} = 1.1 × 10^{−10} (Table above)
Unknown
 value of [Ba^{2+}][SO_{4}^{2−}]
 if a precipitate forms
The concentration and volume of each solution that is mixed together must be used to calculate the values of [Ba^{2+}] and [SO_{4}^{2−}]. Each individual solution is diluted when they are mixed together. The ion product is calculated and compared to the K_{sp} to determine whether a precipitate forms.
Step 2: Solve.
The moles of each ion from the original solutions are calculated by multiplying the molarity by the volume in liters.


 mol Ba^{2+} = 0.0050 M × 0.010 L = 5.0 × 10^{5} mol Ba^{2+}



 mol SO_{4}^{2} = 0.0020 M × 0.020 L = 4.0 × 10^{5} mol SO_{4}^{2}

The concentration of each ion after dilution is then calculated by dividing the moles by the final solution volume of 0.030 L.


 \begin{align*}\mathrm{[Ba^{2+}]=\dfrac{5.0 \times 10^{5} \ mol}{0.030 \ L}=1.7 \times 10^{3} \ M}\end{align*}



 \begin{align*}\mathrm{[SO_4^{2}]=\dfrac{4.0 \times 10^{5} \ mol}{0.030 \ L}=1.3 \times 10^{3} \ M}\end{align*}

Now, the ion product is calculated.


 [Ba^{2+}][SO_{4}^{2}] = (1.7 × 10^{3})(1.3 × 10^{3}) = 2.2 × 10^{6}

Since the ion product is greater than the K_{sp}, a precipitate of barium sulfate will form.
Step 3: Think about your result.
Two significant figures are appropriate for the calculated value of the ion product.
 Calculate the ion product for calcium hydroxide when 20.0 mL of 0.010 M CaCl_{2} is mixed with 30.0 mL of 0.0040 M KOH. Decide if a precipitate will form.
The Common Ion Effect
In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in solution.


 \begin{align*}\mathrm{CaSO_4}(s) \mathrm{\rightleftharpoons Ca^{2+}}(aq) \mathrm{+SO^{2}_4}(aq) \ \ \ \ \text{K}_{\text{sp}}=2.4 \times 10^{5}\end{align*}

Suppose that some calcium nitrate were added to this saturated solution. Immediately, the concentration of the calcium ion in the solution would increase. As a result, the ion product [Ca^{2+}][SO_{4}^{2−}] would increase to a value that is greater than the K_{sp}. According to Le Châtelier’s principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to K_{sp}. Note that in the new equilibrium, the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.
This situation describes the common ion effect. A common ion is an ion that is common to more than one salt in a solution. In the above example, the common ion is Ca^{2+}. The common ion effect is a decrease in the solubility of an ionic compound as a result of the addition of a common ion. Adding calcium ions to a saturated solution of calcium sulfate causes additional CaSO_{4} to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.
Sample Problem 19.4: The Common Ion Effect
What is the concentration of zinc ion in 1.00 L of a saturated solution of zinc hydroxide to which 0.040 mol of NaOH has been added?
Step 1: List the known quantities and plan the problem.
Known
 K_{sp} = 3.0 × 10^{−16} (Table above)
 moles of added NaOH = 0.040 mol
 volume of solution = 1.00 L
Unknown
 [Zn^{2+}] = ? M
Express the concentrations of the two ions relative to the variable s. The concentration of the zinc ion will be equal to s, while the concentration of the hydroxide ion will be equal to 0.040 + 2s. Note that the value of s in this case is not equal to the value of s when zinc hydroxide is dissolved in pure water.
Step 2: Solve.
The K_{sp} expression can be written in terms of the variable s.


 K_{sp} = [Zn^{2+}][OH^{}]^{2} = (s)(0.040+2s)^{2}

Because the value of the K_{sp} is so small, we can make the assumption that the value of s will be very small compared to 0.040. This simplifies the mathematics involved in solving for s.


 K_{sp} = (s)(0.040)^{2} = 0.0016s = 3.0 × 10^{16}



 \begin{align*}\mathrm{s=\dfrac{K_{sp}}{[OH^]^2}=\dfrac{3.0 \times 10^{16}}{0.0016}=1.9 \times 10^{13}\ M}\end{align*}

The concentration of the zinc ion is equal to s, so [Zn^{2+}] = 1.9 × 10^{13} M.
Step 3: Think about your result.
The relatively high concentration of the common ion, OH^{}, results in a very low concentration of the zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in pure water.
 Determine the concentration of silver ions in 1.00 L of a saturated solution of silver chloride to which 0.0020 mol of sodium chloride has been added.
Lesson Summary
 In a saturated solution, an equilibrium exists between the dissolved and undissolved solute, at which point the rate of dissolution is equal to the rate of recrystallization. The equilibrium constant expression for this type of equilibrium is called a solubility product constant (K_{sp}).
 The K_{sp} of a compound can be calculated from its solubility (g/L) or molar solubility (mol/L). Known K_{sp} values can be used to calculate the solubility of a compound.
 When two solutions are mixed, a precipitate may be produced. The starting ion concentrations are used to calculate the ion product, which is then compared to the K_{sp}. If the ion product is greater than the value of K_{sp}, a precipitate will form.
 The common ion effect describes a reduction in the solubility of a salt that results from the addition of an ion that is common to both the original solution and the salt being added.
Lesson Review Questions
Reviewing Concepts
 Explain what is meant by the statement that all ionic compounds are strong electrolytes, no matter how soluble.
 What is the relationship between a compound’s solubility and its solubility product constant?
 Write the solubility product constant (K_{sp}) expression for the following compounds.
 NiS
 Ag_{2}CO_{3}
 Fe_{3}(PO_{4})_{2}
 Use Table above to rank the following salts from most to least soluble.
 AgBr
 BaSO_{4}
 ZnS
 PbCO_{3}
 How does the addition of lead(II) ions affect the solubility of lead(II) chloride in water?
Problems
 The molar solubility of copper(I) bromide, CuBr, is 2.0 × 10^{−4} M. Calculate the solubility of CuBr in g/L.
 Calculate K_{sp} for the following compounds from the given solubilities at 25°C.
 SrCO_{3}, 5.9 × 10^{−3} g/L
 Ag_{2}SO_{4}, 4.74 g/L
 Cr(OH)_{3}, 3.4 × 10^{−6} g/L
 What is the concentration of lead(II) ions and iodide ions in a saturated solution of lead(II) iodide at 25°C? (Use Table above for the K_{sp}.)
 Use the K_{sp} values from Table above to calculate the solubility in g/L of the following compounds.
 Ca_{3}(PO_{4})_{2}
 PbSO_{4}
 Ca(OH)_{2}
 Calculate the ion product of silver bromide when 100.0 mL of 0.0020 M AgNO_{3} is mixed with 100.0 mL of 1.0 × 10^{−4} M KBr. Will a precipitate of AgBr form when the solutions are mixed?
 Determine the concentration of lead(II) ions in 1.00 L of a saturated solution of PbF_{2} to which 0.025 mol of fluoride ions has been added. The K_{sp} of PbF_{2} is 4.1 × 10^{−8}.
Further Reading / Supplemental Links
 Solubility Product (K_{sp}), (http://www.kentchemistry.com/links/Kinetics/Ksp.htm)
 Understanding Chemistry: Solubility Products, (http://www.chemguide.co.uk/physical/kspmenu.html#top)
Points to Consider
In the course of an exothermic reaction, heat is released from the system into the surroundings, resulting in a decrease in the enthalpy of the system. This is a favorable reaction because nature prefers a state of lower energy.
 What is meant by the term "driving force" as it relates to chemical reactions?
 What other force is responsible for the occurrence of endothermic reactions, which absorb heat into the system?