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19.3: Solubility Equilibrium

Created by: CK-12

Lesson Objectives

  • Write solubility product constant expressions for nearly insoluble ionic compounds.
  • Calculate the value of Ksp for a compound from its solubility and determine the solubility of a compound with a known Ksp.
  • Use the Ksp 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.

\mathrm{AgCl}(s) \mathrm{\rightleftharpoons Ag^+}(aq) \mathrm{+Cl^-}(aq)

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.

\mathrm{K_{sp}=[Ag^+][Cl^-]}

This equilibrium constant is called the solubility product constant, (Ksp) 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 Ksp expression. For example, the formula of calcium phosphate is Ca3(PO4)2. The dissociation equation and Ksp expression are shown below:

\mathrm{Ca_3(PO_4)_2}(s) \mathrm{\rightleftharpoons 3Ca^{2+}}(aq) \mathrm{+2PO^{3-}_4}(aq)
\mathrm{K_{sp}=[Ca^{2+}]^3[PO^{3-}_4]^2}

Table below lists solubility product constants for some common nearly insoluble ionic compounds.

Solubility Product Constants (25°C)
Compound Ksp Compound Ksp
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
BaCO3 5.0 × 10−9 PbCl2 1.7 × 10-5
BaSO4 1.1 × 10−10 PbCO3 7.4 × 10-14
CaCO3 4.5 × 10−9 PbI2 7.1 × 10-9
Ca(OH)2 6.5 × 10−6 PbSO4 6.3 × 10−7
Ca3(PO4)2 1.2 × 10−26 Zn(OH)2 3.0 × 10−16
CaSO4 2.4 × 10−5 ZnS 3.0 × 10−23

Solubility and Ksp

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:

\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)}

Solubility data can be used to calculate the Ksp for a given compound. The following steps need to be taken.

  1. Convert from solubility to molar solubility.
  2. Use the dissociation equation to determine the concentration of each of the ions in mol/L.
  3. Apply the Ksp equation.

Sample Problem 19.2: Calculating Ksp from Solubility

The solubility of lead(II) fluoride is found experimentally to be 0.533 g/L. Calculate the Ksp for lead(II) fluoride.

Step 1: List the known quantities and plan the problem.

Known

  • solubility of PbF2 = 0.533 g/L
  • molar mass of PbF2 = 245.20 g/mol

Unknown

  • Ksp of PbF2 = ?

The dissociation equation for PbF2 and the corresponding Ksp expression can be constructed as follows:

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

The steps above will be followed to calculate Ksp for PbF2.

Step 2: Solve.

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

The dissociation equation shows that for every mole of PbF2 that dissociates, 1 mol of Pb2+ and 2 mol of F are produced. Therefore, at equilibrium, the concentrations of the ions are:

[Pb2+] = 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 Ksp.

Ksp = (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 PbF2.

Practice Problem
  1. From the given solubility data, calculate Ksp for each of the following compounds.
    1. copper(II) iodide, CuI = 4.30 × 10−4 g/L
    2. silver sulfide, Ag2S = 2.84 × 10−15 g/L

The known Ksp values from Table above can be used to calculate the solubility of a given compound by following the steps listed below.

  1. Set up an ICE problem (Initial, Change, Equilibrium) in order to use the Ksp value to calculate the concentration of each of the ions. Assume that no ions are initially present in the solution.
  2. The concentrations of the ions can be used to calculate the molar solubility of the compound.
  3. Use the molar mass to convert from molar solubility to solubility in g/L.

The Ksp value of calcium carbonate is 4.5 × 10−9. We begin by setting up an ICE table showing the dissociation of CaCO3 into calcium ions and carbonate ions. The variable s will be used to represent the molar solubility of CaCO3. In this case, each formula unit of CaCO3 yields one Ca2+ ion and one CO32− ion. Therefore, the equilibrium concentrations of each ion are equal to s.

\mathrm{CaCO_3}(s) \mathrm{\rightleftharpoons} \mathrm{Ca^{2+}}(aq)+ \mathrm{CO^{2-}_3}(aq)
Initial (M) 0.00 0.00
Change (M) +s +s
Equilibrium (M) s s

The Ksp expression can be written in terms of s and then used to solve for s.

\mathrm{K_{sp}=[Ca^{2+}][CO_3^{2-}]=(s)(s)=s^2}
\mathrm{s=\sqrt{K_{sp}}=\sqrt{4.5 \times 10^{-9}}=6.7 \times 10^{-5}\:M}

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.

\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}

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 Ksp being equal to s2. 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 AB2, such as Fe(OH)2. In this case, the setup of the ICE table would look like the following:

\mathrm{Fe(OH)_2}(s) \mathrm{\rightleftharpoons} \mathrm{Fe^{2+}}(aq)+ \mathrm{2OH^-}(aq)
Initial (M) 0.00 0.00
Change (M) +s +2s
Equilibrium (M) s 2s
\mathrm{K_{sp}=[Fe^{2+}][OH^-]^2=(s)(2s)^2=4s^3}
\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}

Table below shows the relationship between Ksp and molar solubility based on the formula.

Compound Type Example Ksp Expression Cation Anion Ksp in Terms of s
AB CuS [Cu2+][S2−] s s s2
AB2 or A2B Ag2CrO4 [Ag+]2[CrO42−] 2s s 4s3
AB3 or A3B Al(OH)3 [Al3+][OH]3 s 3s 27s4
A2B3 or A3B2 Ba3(PO4)2 [Ba2+]3[PO43−]2 3s 2s 108s5

The Ksp expressions in terms of s can be used to solve problems in which the Ksp 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 Ksp 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, [Ba2+][SO42−], for the solution. If the value of the ion product is less than the value of the Ksp, 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 Ksp, 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 Ksp, at which point equilibrium is attained.

Barium sulfate is used as a component of white pigments in paints and as an agent in certain x-ray imaging processes.

Sample Problem 19.3: Predicting Precipitates

Will a precipitate of barium sulfate form when 10.0 mL of 0.0050 M BaCl2 is mixed with 20.0 mL of 0.0020 M Na2SO4?

Step 1: List the known quantities and plan the problem.

Known

  • concentration of BaCl2 solution = 0.0050 M
  • volume of BaCl2 solution = 10.0 mL
  • concentration of Na2SO4 solution = 0.0020 M
  • volume of Na2SO4 solution = 20.0 mL
  • Ksp of BaSO4 = 1.1 × 10−10 (Table above)

Unknown

  • value of [Ba2+][SO42−]
  • if a precipitate forms

The concentration and volume of each solution that is mixed together must be used to calculate the values of [Ba2+] and [SO42−]. Each individual solution is diluted when they are mixed together. The ion product is calculated and compared to the Ksp 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 Ba2+ = 0.0050 M × 0.010 L = 5.0 × 10-5 mol Ba2+
mol SO42- = 0.0020 M × 0.020 L = 4.0 × 10-5 mol SO42-

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

\mathrm{[Ba^{2+}]=\dfrac{5.0 \times 10^{-5} \ mol}{0.030 \ L}=1.7 \times 10^{-3} \ M}
\mathrm{[SO_4^{2-}]=\dfrac{4.0 \times 10^{-5} \ mol}{0.030 \ L}=1.3 \times 10^{-3} \ M}

Now, the ion product is calculated.

[Ba2+][SO42-] = (1.7 × 10-3)(1.3 × 10-3) = 2.2 × 10-6

Since the ion product is greater than the Ksp, 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.

Practice Problem
  1. Calculate the ion product for calcium hydroxide when 20.0 mL of 0.010 M CaCl2 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.

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

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 [Ca2+][SO42−] would increase to a value that is greater than the Ksp. 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 Ksp. 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 Ca2+. 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 CaSO4 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

  • Ksp = 3.0 × 10−16 (Table above)
  • moles of added NaOH = 0.040 mol
  • volume of solution = 1.00 L

Unknown

  • [Zn2+] = ? 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 Ksp expression can be written in terms of the variable s.

Ksp = [Zn2+][OH-]2 = (s)(0.040+2s)2

Because the value of the Ksp 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.

Ksp = (s)(0.040)2 = 0.0016s = 3.0 × 10-16
\mathrm{s=\dfrac{K_{sp}}{[OH^-]^2}=\dfrac{3.0 \times 10^{-16}}{0.0016}=1.9 \times 10^{-13}\ M}

The concentration of the zinc ion is equal to s, so [Zn2+] = 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.

Practice Problem
  1. 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 (Ksp).
  • The Ksp of a compound can be calculated from its solubility (g/L) or molar solubility (mol/L). Known Ksp 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 Ksp. If the ion product is greater than the value of Ksp, 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

  1. Explain what is meant by the statement that all ionic compounds are strong electrolytes, no matter how soluble.
  2. What is the relationship between a compound’s solubility and its solubility product constant?
  3. Write the solubility product constant (Ksp) expression for the following compounds.
    1. NiS
    2. Ag2CO3
    3. Fe3(PO4)2
  4. Use Table above to rank the following salts from most to least soluble.
    1. AgBr
    2. BaSO4
    3. ZnS
    4. PbCO3
  5. How does the addition of lead(II) ions affect the solubility of lead(II) chloride in water?

Problems

  1. The molar solubility of copper(I) bromide, CuBr, is 2.0 × 10−4 M. Calculate the solubility of CuBr in g/L.
  2. Calculate Ksp for the following compounds from the given solubilities at 25°C.
    1. SrCO3, 5.9 × 10−3 g/L
    2. Ag2SO4, 4.74 g/L
    3. Cr(OH)3, 3.4 × 10−6 g/L
  3. 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 Ksp.)
  4. Use the Ksp values from Table above to calculate the solubility in g/L of the following compounds.
    1. Ca3(PO4)2
    2. PbSO4
    3. Ca(OH)2
  5. Calculate the ion product of silver bromide when 100.0 mL of 0.0020 M AgNO3 is mixed with 100.0 mL of 1.0 × 10−4 M KBr. Will a precipitate of AgBr form when the solutions are mixed?
  6. Determine the concentration of lead(II) ions in 1.00 L of a saturated solution of PbF2 to which 0.025 mol of fluoride ions has been added. The Ksp of PbF2 is 4.1 × 10−8.

Further Reading / Supplemental Links

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?

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Mar 01, 2013

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Aug 22, 2014
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CK.SCI.ENG.SE.1.Chemistry-Intermediate.19.3

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