482 Chapter 7 NEL

7.67.6 The Solubility Product ConstantSalt solutions are very common in nature. Seawater, lakewater, tapwater, tree sap, saliva,and blood plasma all contain a mixture of dissolved salts including NaCl(aq), MgSO4(aq),and NaHCO3(aq). These compounds are all highly soluble electrolytes. However, manyuseful salts are considered to be weak electrolytes that are only slightly soluble in water.These include compounds such as magnesium hydroxide, Mg(OH)2(s), (milk of magnesia,an antacid), calcium sulfate, CaSO4(s) (gypsum, used to make wallboard), and barium sulfate, BaSO4(s) (used in X rays of the gastrointestinal tract).

Sparingly Soluble Solutes in AnimalsWhen an animal digests the proteins and nucleic acids in its food, it produces toxicnitrogenous wastes that must be eliminated from its body. The source of the nitrogen inthese waste products is the amino group (—NH2) that is removed from the amino acidsof proteins and the nitrogenous bases of nucleic acids when these macromolecules aremetabolized for energy or converted into other useful molecules. Different classes ofanimals process the amino groups in different ways.

Most aquatic animals like fish convert the amino groups into ammonia, NH3(aq).Ammonia is highly toxic, but also highly soluble in water. Ammonia never builds up inthe tissues of such organisms because of its solubility and is easily excreted through thegills into the surrounding water as dissolved ammonium ions, NH4

+(aq).

Mammals, amphibians, and sharks, on the other hand, convert amino groups intourea, H2NCONH2(aq), a compound ten thousand times less toxic than ammonia. Ureamay accumulate to relatively high concentrations in the circulatory system of an organismwithout ill effect. Eventually, it is filtered by the kidneys and excreted in liquid urine.

However, birds, insects, and reptiles do not convert the amino groups into urea.Instead, they form uric acid, H2C5N4O3(s), a substance that is thousands of times less sol-uble in water than ammonia or urea. (The solubilitiy of uric acid is 4.2 � 10�8 mol/Lat 25°C.) The white portion of bird droppings is composed of precipitated uric acidcrystals (Figure 2).

Production of uric acid and urea are the only two methods available to terrestrial ani-mals for excreting nitrogen-containing waste products. Ammonia is far too toxic forterrestrial organisms—they cannot excrete nitrogenous wastes by letting them diffuse outof membranes into an external aqueous solution, the way fish do. In land animals,ammonia would accumulate to toxic levels very quickly. Biologists have discovered thatthe method of reproduction of an organism seems to play a role in determining whichmethod a particular group of animals uses to eliminate nitrogenous waste.

Animals that excrete solid uric acid lay eggs with hard shells (birds and reptiles). Theshells of these eggs are permeable to gases, but not liquids. As the embryo develops insidethe egg it produces nitrogenous waste that must be removed, or the growing organismwill be poisoned. Since the shell is impermeable to liquids, dissolved ammonia or ureawould accumulate within the egg to toxic concentrations, killing the developing embryo.Uric acid, with its extremely low solubility in water, poses less of a threat. It precipitatesout of solution even in low concentrations, and can be stored within the egg as solidwaste until the bird hatches.

Another slightly soluble salt has a noticeable effect on the human body. When theconcentrations of calcium ions, Ca2+

(aq), and oxalate ions, C2O42�(aq), become sufficiently high

in the bloodstream, calcium oxalate, CaC2O4(s), “stones” precipitate in the kidneys.

Figure 1Over thousands of years, the actionof water on limestone (calcium car-bonate) rock formations has createdcrevices and caves.

Figure 2Uric acid (white) is present in birddroppings at concentrations farbeyond its solubility.

weak electrolytes salts with relatively low solubility in water

Chemical Systems in Equilibrium 483NEL

What amount of a slightly soluble salt will dissolve in a given volume of water to forma saturated solution? What minimum concentration of aqueous calcium and oxalateions causes the precipitation of the salts that form kidney stones?

These questions involve equilibria between the dissolved ions and solids of slightlysoluble salts (weak electrolytes). Whether we want to determine the degree of dissolu-tion (dissociation), or the conditions that result in precipitation of slightly soluble salts,equilibrium constants and equilibrium law equations are involved.

In previous chemistry courses you learned about solubility. You may have done inves-tigations to find out how much of a solute would dissolve in a solvent under certainconditions. Using solutes with high solubility, you were able to add a visible amount ofthe solute and make it dissolve, forming an unsaturated solution. When you could makeno more solute dissolve, you had prepared a saturated solution, one where there is adynamic equilibrium between undissolved solute and dissolved solute. If the containerwere sealed, and there were no temperature changes, no further changes occurred inthe concentration of the solution or in the quantity of undissolved solute. You had made a system in which the rate of dissolution was equal to the rate of crystallization—a dynamic equilibrium.

Solubility ProductA special case of equilibrium involves any situation where excess solute is in equilibriumwith its aqueous solution. You can establish such an equilibrium either by starting withexcess salt and dissolving it until the solution is saturated and excess solute remains(Figure 3(a)), or by mixing solutions of two salts that results in a product precipitatingout of solution (Figure 3(b)). Once precipitation ends, the remaining solution is saturated,and the dissolved ions form a dynamic equilibrium with the precipitated crystals.

Consider the chemical equation and equilibrium law equation for a weak electrolytesuch as copper(I) chloride. Notice that since copper(I) chloride dissolves very little in water,almost any amount dropped into water will result in the formation of a saturated solu-tion, and there will be a heterogeneous equilibrium between the solid and dissolvedCu+

(aq) and Cl�(aq) ions.

CuCl(s) e Cu�(aq) � Cl�(aq)

The solubility equilibrium law equation is

K � �[Cu

[

+(

Caq

u)

C

][

l

C

(s

l

)

�(

]aq)]

�

which simplifies to

K � [Cu+(aq)][Cl�(aq) ]

as the concentration (density) of the CuCl(s) is constant and so is incorporated into thevalue of the equilibrium constant. The value of this equilibrium constant is

K � 1.7 � 10�7 at 25°C.

For any solute that forms ions in solution, the solubility equilibrium constant is theproduct of the concentrations of the ions in solution raised to the power equal to the coef-ficient of each in the balanced equation, and for that reason is often called the solubilityproduct constant of the substance, symbolized as Ksp. The equilibrium law equationfor the copper(I) chloride equilibrium is written

Section 7.6

Figure 3Saturated solutions.(a) This saturated solution was

formed by adding excesscopper(II) sulfate to water. Thesolid is in equilibrium with theions in solution.

(b) This saturated solution wasformed by adding a solution ofiron(III) nitrate to a solution ofsodium phosphate. The result isa saturated solution of iron(III)phosphate, in which the ions insolution are in equilibrium withthe solid precipitate.

(a)

(b)

solubility the concentration of asaturated solution of a solute in aparticular solvent at a particulartemperature; solubility is a specificmaximum concentration

solubility product constant (Ksp)the value obtained from the equilib-rium law applied to a saturatedsolution

484 Chapter 7 NEL

Ksp � [Cu+(aq)][Cl�(aq)]

Ksp � 1.7 � 10�7 at 25°C

For ionic substances with a more complex formula, like calcium phosphate, the Kspexpression is also more complex. The balanced equation is

Ca3(PO4)2(s) e 3 Ca2+(aq) � 2 PO4

3�(aq)

The equilibrium expression is

Ksp � [Ca2+(aq)]3[PO4

3�(aq)]2

with the value of the constant

Ksp � 2.1 � 10�33 at 25°C

In general, for the dissociation equilibrium equation

BC(s) e b B+(aq) � c C�

(aq)

where BC(s) is a slightly soluble salt, and B+(aq) and C�

(aq) are aqueous ions,

Ksp � [B+(aq)]b [C�

(aq)]c

Ksp values are listed in many chemistry reference sources (see Table 1 and Appendix C8).

Table 1 Some Solubility Product Constants at 25°C

Name Formula Ksp

cobalt(II) hydroxide Co(OH)2(s) 1.1 � 10�15

lithium carbonate Li2CO3(s) 8.2 � 10�4

mercury(I) chloride Hg2Cl2(s)) 1.5 � 10�18

nickel(II) carbonate NiCO3(s) 1.4 � 10�7

tin(II) sulfide SnS(s) 3.2 � 10�28

zinc hydroxide Zn(OH)2(s) 7.7 � 10�17

calcium phosphate Ca3(PO4)2(s) 2.1 � 10�33

magnesium fluoride MgF2(s) 7.4. � 10�11

lead(II) chloride PbCl2(s) 1.2 � 10�5

Reference — CRC Handbook of Chemistry and Physics (76th Ed.)

References typically list Ksp values only for ionic compounds with low solubility,because under ordinary laboratory conditions highly soluble ionic compounds do notform precipitates. Their solutions, as used, are not saturated—no solubility equilibriumis established. Solubilities of highly soluble substances are listed in mol/L or g/100 mLvalues rather than as Ksp values.

Calculating Solubility Using Ksp ValuesA straightforward calculation will convert a solubility value to (or from) a Ksp value, asthe following examples show.

As the units for Ksp vary frommol/L to (mol/L)2 to (mol/L)3,the convention in the scientificcommunity is to omit unitswhen writing Ksp values, as isthe case for other K values.

LEARNING TIP

Do not confuse solubility withsolubility product. The solubilityof a salt is the amount of saltthat dissolves in a given amountof solvent to give a saturatedsolution. Solubility product isthe product of the molar con-centrations of the ions in thesaturated solution.

LEARNING TIP

Chemical Systems in Equilibrium 485NEL

Section 7.6

Magnesium fluoride is a hard, slightly soluble salt that is used to make spectrallenses for technical instruments. Calculate Ksp for magnesium fluoride at 25°C,given a solubility of 0.001 72 g/100 mL.

MgF2(s) e Mg2+(aq) � 2 F�

(aq)

Ksp � [Mg2+(aq)][F�

(aq)]2

From the balanced equation, we know that

[Mg2+(aq)] � [MgF2(aq)]

The [MgF2(aq)] in the above equation refers to the concentration of MgF2 formula unitsthat produce the aqueous ions, Mg2+

(aq) and F�(aq). This is a very small proportion of the

MgF2(s) crystal.Convert the solubility values (g/100mL) to concentration values (mol/L).

[Mg2+(aq)] � [MgF2(aq)] � �

0.100001m72

L�g�

� � �612m.31

olg�

� � �100

10LmL�

�

� 2.8 � 10�4 mol/L

From the balanced reaction equation above,

[F�(aq)] � 2 [Mg2+

(aq) ]

� 2 � 2.8 � 10�4 mol/L

� 5.5 � 10�4 mol/L

Ksp � [Mg2+(aq)][F�

(aq)]2

� (2.8 � 10�4) (5.5 � 10�4)2

Ksp � 8.4 � 10�11

OROnce the concentration of the fraction of MgF2(s) that dissolves is calculated (2.8 � 10�4 mol/L),it is also possible to solve the problem using an ICE table (Table 2).

Calculating the Solubility Product Constant, Given the Solubility SAMPLE problem

Table 2 ICE Table for Calculating Ksp from Solubility

MgF2(s) e Mg 2+(aq) � 2 F–

(aq)

Initial concentration (mol/L) — 0 0

Change in concentration (mol/L) — �x �2x

Equilibrium concentration (mol/L) — x 2x

x � 2.8 � 10�4

2x � 2(2.8 � 10�4)

2x � 5.5 � 10�4

Ksp � [Mg2+(aq)][F–

(aq)]2

� (2.8 � 10�4) (5.5 � 10�4)2

Ksp � 8.4 � 10�11

ExampleCalculate the molar solubility of zinc hydroxide at 25°C, where Ksp is 7.7 � 10�17.

Zn(OH)2(s) e Zn2+(aq) � 2 OH�

(aq)

486 Chapter 7 NEL

Solution

Ksp � [Zn2+(aq)][OH�

(aq)]2

� 7.7 � 10�17

[OH�(aq)] � 2 [Zn2+

(aq)]

Ksp � [Zn2+(aq)](2 [Zn2+

(aq)])2

7.7 � 10�17 � 4[Zn2+(aq)]3

[Zn2+(aq)] � �3 �7.7 �

4�10�17��

� 2.7 � 10�6 mol/L

[Zn(OH)2(aq)] � [Zn2+(aq)]

� 2.7 � 10�6 mol/L

The molar solubility of zinc hydroxide is 2.7 � 10–6 mol/L.

Or, using an ICE table (Table 3):

PracticeUnderstanding Concepts

1. Calculate the solubility of silver iodide at 25°C. The Ksp of AgI(s) is 1.5 � 10�16 at25°C.

2. Calculate the solubility of iron(II) carbonate at 25°C. The Ksp of FeCO3(s) is 3.5 � 10�11 at 25°C.

3. Calculate the solubility of zinc hydroxide at 25°C. The Ksp of Zn(OH)2(s) is 4.5 � 10�17 at 25°C.

4. Given the following solubilities, calculate the value of the solubility product foreach compound:(a) copper (II) sulfide, 8.89 � 10�19 mol/L(b) zinc carbonate, 3.87 � 10�6 mol/L

Table 3 ICE Table for Calculating Ksp from Solubility

Zn(OH)2(s) e Zn2+(aq) � 2 OH�

(aq)

Initial concentration (mol/L) — 0 0

Change in concentration (mol/L) — �x �2x

Equilibrium concentration (mol/L) — x 2x

Ksp � [Zn2+(aq)][OH�

(aq)]2 � 7.7 � 10–17

Ksp � (x)(2x)2

7.7 � 10�17 � 4x3

x � �3

�7.7 �

4�10�17��

x � 2.7 � 10�6 mol/L

The molar solubility of zinc hydroxide is 2.7 � 10�6 mol/L

Answers

1. 1.2 � 10�8 mol/L

2. 5.9 � 10�6 mol/L

3. 2.3 � 10�6 mol/L

4. (a) 7.90 � 10�37

(b) 1.50 � 10�4

Chemical Systems in Equilibrium 487NEL

Predicting PrecipitationIn Section 7.5, you learned that the reaction quotient, Q, is a value that can be used tocalculate whether or not a system is at equilibrium. We can also use the reaction quotientin the context of solubility—to predict whether a precipitate will form when we mixsolutions of metal cations and nonmetal anions. (In previous chemistry courses, youused solubility tables like Table 4 to determine qualitatively whether an ionic compoundhad high or low solubility.) We can now calculate Q to determine whether, after mixing,the ions are present in too high a concentration, in which case a precipitate will form.In this situation, the reaction quotient, Q, is sometimes called the trial ion product.

To predict whether a precipitate will form when solutions containing anions andcations are mixed, we compare the Ksp value for the salt of these ions (from a table likeTable 1, page 484) to Q (as a trial ion product).

As you know, the value of Ksp equals the ion product of a saturated solution in whichdissolved and undissolved solutes are in dynamic equilibrium. For example,

CuCl(s) e Cu+(aq) � Cl�(aq) Ksp � 7.1 � 10�7

Consider a solution containing Cu+(aq) and Cl�(aq) ions, each at a concentration of

4.1 � 10�4 mol/L. Since the trial ion product, Q, is less than Ksp,

Q � [Cu+(aq)][Cl�(aq)]

� (4.1 � 10�4)(4.1 � 10�4)

Q � 1.7 � 10�7

Ksp � 7.1 � 10�7

Q � Ksp

no precipitation occurs—all aqueous ions are able to remain in the dissolved state. Thedynamic equilibrium that exists between dissolved ions and any undissolved soluteensures that there will be no net crystallization. The solution is unsaturated.

Now consider a solution that is saturated. It is not capable of dissolving more ions. Inthis case, the trial ion product is equal to the value of Ksp. A saturated solution will notform a precipitate.

A third possibility is the case where the trial ion product is greater than the Ksp value.In this case, there are more ions in solution than are necessary for saturation. This is asupersaturated solution. A supersaturated solution is unstable; there is a tendency forthe extra solute to precipitate.

We can also use the value of Ksp to determine whether a precipitate will form whentwo solutions are mixed. See the Sample Problem on page 488.

Section 7.6

trial ion product the reaction quotient applied to the ion concentrations of a slightly soluble salt

Table 4 Solubility of Ionic Compounds at SATP

Anions

Cl –, Br –, I – S2– OH– SO42– CO3

2–, PO43–, SO3

2– C2H3O2– NO3

–

high solubility (aq) most Group 1, NH4+ Group 1, NH4

+ most Group 1, NH4+ most all

�0.1 mol/L Group 2 Sr2+, Ba2+, Tl+

(at SATP) All Group 1 compounds, including acids, and all ammonium compounds are assumed to have high solubility in water.

low Solubility (s) Ag+, Pb2+, Tl+, most most Ag+, Pb2+, most Ag+ none<0.1 mol/L Hg2

2+ (Hg+), Ca2+, Ba2+, (at SATP) Cu+ Sr2+, Ra2+

Catio

ns

supersaturated solution a solutionwhose solute concentration exceedsthe equilibrium concentration

488 Chapter 7 NEL

If 100 mL of 0.100 mol/L CaCl2(aq) and 100 mL of 0.0400 mol/L Na2SO4(aq) aremixed at 20°C, determine whether a precipitate will form. For CaSO4(aq) at 20°C,Ksp is 3.6 �10–5.

This problem essentially asks whether a double displacement reaction will occur when asolution of CaCl2(aq) is mixed with a solution of Na2SO4(aq), and, if so, whether CaSO4(s) willprecipitate.

We begin by writing the potential double displacement reaction.

CaCl2(aq) � Na2SO4(aq) → CaSO4(s) � NaCl(aq)

According to Table 4, CaSO4(s) is relatively insoluble. But are the concentrations ofCa2+

(aq) and SO42�(aq) high enough for precipitation to occur?

To determine the concentration of Ca2+(aq) before mixing, we analyze the equation that

describes the calcium chloride solution.

CaCl2(s) e Ca2+(aq) � 2 Cl�(aq)

[Ca2+(aq)] � [CaCl2(aq)] � 0.100 mol/L (before mixing)

Similarly, to determine the concentration of SO42–(aq) before mixing, we analyze the equa-

tion that describes the sodium sulfate solution.

[SO42–(aq)] � [Na2SO4(aq)] � 0.0400 mol/L (before mixing)

Note that mixing two solutions always increases the overall volume, so the initial con-centration of ions in both solutions is always decreased by the act of mixing them.

In this instance, after mixing,

[Ca2+(aq)] � 0.100 mol/L � �

120000

mm

LL

� � 0.0500 mol/L

Similarly, after mixing,

[SO42–(aq)] � 0.0400 mol/L � �

120000

mm

LL

� � 0.0200 mol/L

Now we use these two concentrations to calculate the ion product, Q, for CaSO4(s),which we can write from the dissociation reaction of the salt.

CaSO4(s) e Ca2+(aq) � SO4

2�(aq)

Q � [Ca2+(aq)][SO4

2�(aq)]

� (0.0500)(0.0200)

Q � 1.00 � 10�3

Q gives the reaction quotient for the component ions in CaSO4(s). Q is much greaterthan the Ksp value (3.6 � 10�5), indicating that more ions are present than would bepresent at equilibrium, so the reaction must shift toward the solid to establish equilibrium.Therefore, a precipitate forms.

(Note that, although in this case we are starting with ions and producing a precipitate,the equation is, by convention, written in the form above, with the solid salt on the left.Remember that the reaction under consideration is the backward, or reverse, reaction.)

ExampleWould a precipitate of lead(II) sulfate, PbSO4(s), (Ksp � 1.8 � 10�8) form if 255 mL of0.000 16 mol/L lead(II) nitrate, Pb(NO3)2(aq), is poured into 456 mL of 0.000 23 mol/Lsodium sulfate, Na2SO4(aq)?

Determining Whether a Precipitate Will FormSAMPLE problem

Chemical Systems in Equilibrium 489NEL

Section 7.6

Solution

Pb(NO3)2(aq) � Na2SO4(aq) → PbSO4(s) � 2 NaNO3(aq)

Before mixing:

Pb(NO3)2(aq) e Pb2+(aq) � 2 NO3

�(aq)

[Pb2+(aq)] = [Pb(NO3)2(aq)] = 0.000 16 mol/L

Na2SO4(aq) e 2 Na+(aq) � SO4

2�(aq)

[SO42–(aq)] � [Na2SO4(aq)] � 0.000 23 mol/L

After mixing, there is

255 mL � 456 mL � 711 mL

of the solution. Therefore, the concentrations of the lead(II) and sulfate ions in themixed solution are calculated as

[Pb2+(aq)] � 0.000 16 mol/L � �

275151 m

mLL

� � 5.74 � 10�5 mol/L

[SO42–(aq)] � 0.000 23 mol/L � �

475161 m

mLL

� � 1.48 � 10�4 mol/L

PbSO4(s) e Pb2+(aq) � SO4

2�(aq)

Q � [Pb2+(aq)][SO4

2�(aq)]

� (5.74 � 10�5)(1.48 � 10�4)

Q � 8.46 � 10�9

Ksp � 1.8 � 10�8

Q is smaller than Ksp. Therefore, a precipitate does not form.

PracticeUnderstanding Concepts

5. Refer to Appendix C8 to predict whether a precipitate forms if(a) 25.0 mL of 0.010 mol/L silver nitrate is mixed with 25.0 mL of 0.0050 mol/L

potassium chloride.(b) equal volumes of 0.0010 mol/L calcium nitrate and 0.0020 mol/L potassium

hydroxide are combined.(c) equal volumes of 0.010 mol/L lead(II) nitrate and 0.10 mol/L sodium chloride

are combined.

Making Connections

6. Barium sulfate is a white, insoluble ionic compound that is opaque to X rays. Priorto going for a gastrointestinal X ray, patients are sometimes given a chalky-whitesuspension of barium sulfate to drink. Because barium is opaque to X rays, thepatient’s gastrointestinal tract is clearly visible in the X ray. Barium sulfate has a Kspof 1.1 � 10�10.

BaSO4(s) e Ba2+(aq) � SO4

2�(aq)

Why is it safe for patients to consume barium sulfate even though barium ions areextremely toxic?

Answers

5. (a) Q � 1.25 � 10�5

(b) Q � 5.0 � 10�10

(c) Q � 1.2 � 10�5

6. (a) [Ba2�(aq)] � 1.0 � 10�5 mol/L

490 Chapter 7 NEL

Using Trial Ion ProductWhen determining whether a precipitate will form in a particular solution, rememberthat Ksp is the ion product for a saturated solution—a solution in which there is adynamic equilibrium between dissolved and undissolved solute. However, the reactionquotient Q (trial ion product) may be calculated for any solution, regardless of the extentto which the solute is dissolved.

For example, the Ksp of AgCl is 1.8 � 10�10 at 25°C. In a saturated solution of AgCl,

[Ag+(aq)] � [Cl�(aq)]

and

Ksp � [Ag+(aq)][Cl�(aq)]

Effectively then, the concentration of either of the two ions at equilibrium is the squareroot of the ion product, which (to two significant digits) is 1.3 � 10�5 mol/L. Logicallythen, any AgCl(aq) solution in which the concentration of both ions is less than 1.3 � 10�5 must be unsaturated, and no precipitate will form. The trial ion product forsuch a solution will be less than the Ksp for AgCl.

In an AgCl solution where the [Ag+(aq)] and [Cl�(aq)] are both greater than 1.3 � 10�5

mol/L, the trial ion product will be greater than the Ksp. This indicates that there is moresolute dissolved in the solution than can be maintained in the dissolved state (a super-saturated solution). The excess solute will precipitate out of solution until the concen-trations of Ag+

(aq) and Cl�(aq) equal 1.3 � 10�5 mol/L, at which point the solution issaturated.

By comparing the trial ion product to the Ksp for a particular salt solution, we canpredict whether a precipitate will form.

Ion product, Q > Ksp (supersaturated solution) Precipitate will form.

Ion product, Q � Ksp (saturated solution) Precipitate will not form.

Ion product, Q < Ksp (unsaturated solution) Precipitate will not form.

Using Q to Predict SolubilitySUMMARY

The Common Ion EffectWhen equilibrium exists in a solution involving ions, the equilibrium can be shifted bydissolving into the solution any other compound that adds a common ion, or any com-pound that reacts with one of the ions already in solution.

Consider a saturated solution of sodium chloride, in equilibrium with a small amountof undissolved sodium chloride (Figure 4(a)).

NaCl(s) e Na�(aq) � Cl�(aq)

If a few drops of concentrated hydrochloric acid is added to the equilibrium mixture, addi-tional crystals of sodium chloride will form (Figure 4(b)). How can we explain this?With Le Châtelier’s principle. The hydrochloric acid releases large numbers of chlorideions into the solution.

HCl(aq) e H�(aq) � Cl�(aq)

Cl(aq)�NaCl(s) Na(aq)

HCl(aq)

��

Cl(aq)�NaCl(s) Na(aq)

��shift

Figure 4The common ion effect.(a) A saturated solution of sodium

chloride is in equilibrium withexcess solid.

(b) Adding a common ion, Cl�(aq),from hydrochloric acid, HCl(aq),shifts the equilibrium, causingsodium chloride to precipitateout of solution.

Determining Ksp for CalciumHydroxide (p. 519)Design and carry out your ownversion of this classic experimentto find Ksp.

INVESTIGATION 7.6.2

Determining the Kspof Calcium Oxalate (p. 517)Calcium oxalate may crystallize outof solution in the kidneys and otherparts of the human urinary tract,forming kidney stones. What is theKsp of this low-solubility salt?

INVESTIGATION 7.6.1

(a)

(b)

Chemical Systems in Equilibrium 491NEL

These additional ions increase the concentration of chloride ions in the mixture, shiftingthe sodium chloride equilibrium to the left—i.e., causing NaCl(s) to precipitate out of thesolution. In this example, the chloride ions were common to both solutions that weremixed. We could expect a similar result if we had added a solution containing Na�

(aq)ions instead. In that case the common ion would be the sodium ion. The lowering of thesolubility of an ionic compound by the addition of a common ion is called the commonion effect.

Section 7.6

common ion effect a reduction inthe solubility of a salt caused by thepresence of another salt having acommon ion

What is the molar solubility of PbCl2(s) in a 0.2 mol/L NaCl(aq) solution at SATP?

In this problem, you are asked to determine the amount of PbCl2(s) that will dissolve into asolution that already contains Cl�(aq) ions. The two salts have a common ion, Cl�(aq).

NaCl(s) (which has high solubility) dissolves completely in water to form Na+(aq) and

Cl�(aq) ions.

NaCl(s) → Na+(aq) � Cl�(aq)

Therefore, before the addition of the lead(II) chloride,

[Cl�(aq)] � [NaCl(aq)] � 0.2 mol/L

We can look up the solubility product constant of PbCl2(s) in a solubility table, forexample, the table in Appendix C8. The low Ksp value shows that it is only slightly solublein water. We can therefore say that PbCl2(s) establishes a dynamic equilibrium in solutionaccording to the following equation:

PbCl2(s) e Pb2+(aq) � 2 Cl�(aq) Ksp � 1.7 � 10�5 at 25°C

In this problem, a PbCl2 equilibrium is being established within a solution that alreadycontains Cl�(aq) ions. The result should be that the position of the PbCl2 equilibrium is “left-shifted” from its normal point. In other words, the dissolution should not proceed to theright as much as it would in pure water. More of the lead chloride should stay in solidform (Figure 5).

We can use the PbCl2(s) equilibrium equation and an ICE table to determine the solu-bility of PbCl2(s), noting that an initial [Cl�(aq)] of 0.2 mol/L already exists in the solutionbefore any PbCl2(s) dissolves. When setting up the ICE table (Table 5) in a common ionproblem, treat the overall process as two separate steps. The first step involves the initialsolution into which a salt is going to be dissolved (if this is pure water, then the initialconcentrations of all ions are assumed to be 0 mol/L). In this problem, the first step isdetermining [Cl–(aq)] in the NaCl(aq) solution, which can be entered on the Initial concen-tration line in the table.

The second step in setting up the ICE table is to use the coefficients in the equilibriumequation to give the multiples for x in the Change in concentration line of the ICE table. Asa result, the multiples of x will correspond to the ion coefficients in the balanced equilib-rium equation.

Solubility in Solutions With Common Ions SAMPLE problem

Figure 5Lead(II) chloride is a low solubilitysalt, but its solubility is lower stillin a solution of sodium chloride.

Table 5 ICE Table to Predict the Solubility of PbCl2(s) in a Solution Containing NaCl(aq)

PbCl2(s) e Pb2+(aq) � 2 Cl–(aq)

Initial concentration (mol/L) — 0 0.2

Change in concentration (mol/L) — �x �2x

Equilibrium concentration (mol/L) — x 0.2 � 2x

Note that the coefficients of theequilibrium chemical equation forthe added salt do NOT apply to theoriginal solute. Do NOT multiplyoriginal concentrations by thecoefficients of the equilibriumchemical equation.

LEARNING TIP

492 Chapter 7 NEL

Ksp � [Pb2+(aq)][Cl�(aq)]2 � 1.7 � 10�5

Ksp � (x)(0.2 � 2x)2 � 1.7 � 10�5

We can simplify the math in this question by noting that the Ksp is very small (PbCl2 hasa very low solubility in water). It follows that the value of x and therefore, 2x will beexceedingly small. When this very small value is added to 0.2 (a much larger value), theresult will essentially remain 0.2. We make the simplifying assumption that

0.2 � 2x � 0.2

Therefore,

Ksp � (x)(0.2)2 � 1.7 � 10�5

x � �1.7

(�

0.21)02

�5�

x � 4.2 � 10�4

To determine whether the assumption that 0.2 � 2x � 0.2 is appropriate, we notice that

2x � 2(4.2 � 10�4)

� 8.4 � 10�4

which is indeed much smaller than 0.2. We accept the assumption as valid.The molar solubility of lead(II) chloride in 0.2 mol/L NaCl(aq) solution is 4.2 � 10�4 mol/L.

PracticeUnderstanding Concepts

7. Calculate the solubility of silver chloride in a 0.10 mol/L solution of sodium chlorideat 25°C. At SATP, Ksp AgCl(s) � 1.8 � 10�10.

8. Calculate the solubility of calcium sulfate in 0.010 mol/L calcium nitrate at SATP.

9. The Ksp of Ag2CrO4(s) is 1.12 � 10–12. Calculate the molar solubility of Ag2CrO4(s)(a) in pure water, and(b) in a solution of 0.10 mol/L sodium chromate, Na2CrO4(s).(c) Compare your answers in (a) and (b). Is the difference reasonable? Explain.

10. Name two compounds that will decrease the solubility of barium sulfate, BaSO4(s).

11. Name two compounds that will decrease the solubility of copper(II) carbonate,CuCO3(s).

12. Consider the equilibrium:

AgCl(s) e Ag+(aq) � Cl�(aq)

During the processing of photographic film, unreacted silver compounds areremoved from the film by the addition of thiosulfate ions, S2O3

2�(aq), to produce the

ion Ag(S2O3)23�(aq).

Using Le Châtelier’s principle, explain why the addition of a sodium thiosulfatesolution makes the solid silver compounds dissolve.

Answers

7. 1.8 � 10�9 mol/L

8. 3.6 � 10�3 mol/L

9. (a) 6.54 � 10�5 mol/L

(b) 1.7 � 10�6 mol/L

Chemical Systems in Equilibrium 493NEL

Section 7.6

Section 7.6 QuestionsUnderstanding Concepts

1. Distinguish between solubility and solubility product constant.

2. (a) Define the common ion effect.(b) How does Le Châtelier’s principle explain the common

ion effect?

3. How do the concepts of a saturated solution and a super-saturated solution help explain the formation of the precipi-tate of a slightly soluble salt when certain soluble saltsolutions are mixed?

4. Calculate the molar solubility of barium sulfate at 25°C.

5. Calculate the solubility at 25°C of silver bromide, in g/100 mL.

6. Calculate the molar concentration of fluoride ions in a saturated solution of strontium fluoride at 25°C.

7. Calculate the Ksp of thallium(I) chloride at 100°C. The con-centration of a saturated solution of the salt at this temper-ature is 2.4 g/100 mL.

8. The concentration of a saturated solution of calcium fluo-ride is determined by evaporating a 100-mL sample of satu-rated solution to dryness. A mass of 0.0016 g of solidremains after evaporation. The original solution was formedat 20°C. Calculate the Ksp of the salt at this temperature.

9. Mercury(I) chloride dissolves as shown by the equation

Hg2Cl2(s) e Hg22+(aq) � 2 Cl�(aq)

Calculate the mass of compound required to make 500 mLof a saturated solution of mercury(I) chloride at 25°C.

10. For each of the following mixtures, calculate a trial ionproduct, Q, to predict whether a precipitate forms. All mix-tures are made at SATP.(a) 50 mL of 0.040 mol/L Ca(NO3)2(aq) plus 150 mL of

0.080 mol/L (NH4)2SO4(aq)(b) 50 mL of 2.2 � 10�9 mol/L AgNO3(aq) plus 50 mL of

0.050 mol/L NH4Cl(aq)(c) 100 mL of 2.1 � 10�3 mol/L Pb(NO3)2(aq) plus 50 mL of

0.0060 mol/L NaI(aq)

11. Calculate the molar solubility of PbI2(s) in a 0.10 mol/Lsolution of NaI(aq) at SATP.

Applying Inquiry Skills

12. Lead(II) chloride is a low-solubility salt that dissociatesaccording to this equilibrium:

PbCl2(s) e Pb2+(aq) � 2 Cl�(aq)

Lead ions undergo a single displacement reaction withzinc:

Pb2+(aq) � Zn(s) e Pb(s) � Zn2+

(aq)

Question

What is the Ksp of lead(II) chloride?

Experimental Design

A strip of zinc of known mass is placed into a saturatedsolution of lead(II) chloride overnight. The next day, thestrip is removed from the solution, cleaned, and its newmass measured to determine how much mass was lost.

Evidence

The zinc strip was coated with a black layer of metallic lead.Volume of saturated lead(II) chloride solution: 100 mLMass loss of zinc strip: 0.094 g

Analysis(a) Calculate the amount (in moles) of zinc reacted.(b) Calculate the molar concentration of the lead ions in

the lead(II) chloride solution. (c) Calculate the solubility product for lead(II) chloride.

13. Silver acetate is a low-solubility salt that dissociatesaccording to this equilibrium:

AgCH3COO(s) e Ag+(aq) � CH3COO�

(aq)

Silver ions undergo a single displacement reaction withcopper:

2 Ag+(aq) � Cu(s) e 2 Ag(s) � Cu2+

(aq)

Question

What is the Ksp of silver acetate?

Experimental Design

A coil of copper wire is placed in a saturated solution ofsilver acetate for two days. The copper is removed from thesolution, cleaned, and its mass measured to determine howmuch mass it lost.

Evidence

The copper wire was coated with a silvery-grey layer.

Volume of the saturated silver acetate solution: 100 mL

Mass loss of copper wire: 0.16 g

Analysis(a) Calculate the moles of copper reacted.(b) Calculate the moles of silver that reacted.(c) Calculate the molar concentration of the silver ions in

the silver acetate solution.(d) Calculate the solubility product for silver acetate.

Evaluation(e) In this experiment, it is very difficult to scrape all the

silver off the copper wire. Suppose some silver residueremained on the wire. What effect would this have onthe calculation of the silver acetate Ksp value?