More Equilibria in Aqueous Solutions: Slightly Soluble Salts and Complex Ions

Prentice Hall © 2005Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Chapter Sixteen

1

More Equilibria in Aqueous Solutions:

Slightly Soluble Salts and Complex Ions

Chapter SixteenChapter Sixteen


Chapter Sixteen

2

• Solubility product constant, Ksp: the equilibrium constant expression for the dissolving of a slightly soluble solid.

The Solubility Product Constant, Ksp

Ksp = [ Ba2+ ][ SO42–]

BaSO4(s) Ba2+(aq) + SO42–(aq)

• Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble” – Chapter 4).

• An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution:


Chapter Sixteen

3


Chapter Sixteen

4

Example 16.1Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO4, and (b) chromium(III) hydroxide, Cr(OH)3.


Chapter Sixteen

5Example 16.1

Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO4 and (b) chromium(III)hydroxide, Cr(OH)3.

StrategyWe first write chemical equations for the solubility equilibrium, basing each equation on one mole of solid on the left side. The coefficients on the right side are then the numbers of moles of cations and anions per mole of the compound. As with other equilibrium constant expressions, coefficients in the balanced equation appear as exponents in the Ksp expression.

Write a Ksp expression for equilibrium in a saturated aqueous solution of (a) MgF2, (b) Li2CO3, and (c) Cu3(AsO4)2.

Exercise 16.1A

Write a Ksp expression for equilibrium in a saturated solution of (a) magnesium hydroxide (commonly known as Milk of magnesia), (b) scandium fluoride, ScF3 (used in the preparation of scandium metal), and (c) zinc phosphate (used in dental cements).

Exercise 16.1B

Solution


Chapter Sixteen

6

• Ksp is an equilibrium constant

• Molar solubility is the number of moles of compound that will dissolve per liter of solution.

• Molar solubility is related to the value of Ksp, but molar solubility and Ksp are not the same thing.

• In fact, “smaller Ksp” doesn’t always mean “lower molar solubility.”

• Solubility depends on both Ksp and the form of the equilibrium constant expression.

Ksp and Molar Solubility


Chapter Sixteen

7

Example 16.2At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag2CO3 per liter of solution. Calculate Ksp for Ag2CO3 at 20 °C. The balanced equation is

Ag2CO3(s) 2 Ag+(aq) + CO32–(aq) Ksp = ?

Example 16.3From the Ksp value for silver sulfate, calculate its molar solubility at 25 °C.

Ag2SO4(s) 2 Ag+(aq) + SO42–(aq)

Ksp = 1.4 x 10–5 at 25 °C


Chapter Sixteen

8

StrategyTo evaluate Ksp, we need the molarities of the individual ions in the saturated solution, but we are given only the solubility of the solute. Thus, the heart of the calculation is to obtain ion molarities from a concentration in milligrams of solute per liter. Once we have these ion molarities, we can get a value for Ksp by substituting them into the solubility constant expression.

Example 16.2

At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag2CO3 per liter of solution. Calculate Ksp for Ag2CO3 at 20 °C. The balanced equation is

Solution

From the equation for the solubility equilibrium, we see that 2 mol Ag+ and 1 mol CO32–

appear in solution for every 1 mol Ag2CO3 that dissolves. We can represent this fact by conversion factors (shown in blue).


Chapter Sixteen

9Example 16.2 continued

Now we can write the Ksp expression for Ag2CO3 and substitute in the equilibrium concentrations of the ions.

Ksp = [Ag+]2[CO32–] = (2.3 x 10–4)2(1.2 x 10–4) = 6.3 x 10–12

In Example 15.4, we determined the molar solubility of magnesium hydroxide from the measured pH of its saturated solution. Use data from that example to determine Ksp forMg(OH)2.

Exercise 16.2A

A saturated aqueous solution of silver(I) chromate contains 14 ppm of Ag+ by mass. Determine Ksp for silver(I) chromate. Assume the solution has a density of 1.00 g/mL.

Exercise 16.2B


Chapter Sixteen

10Example 16.3

From the Ksp value for silver sulfate, calculate its molar solubility at 25 °C.

StrategyThe equation shows that when 1 mol Ag2SO4(s) dissolves, 1 mol SO4

2– and 2 mol Ag+ appear in solution. If we let s represent the number of moles of Ag2SO4 that dissolve per liter of solution, the two ion concentrations are

[SO42–] = s and [Ag+] = 2s

We then substitute these concentrations into the Ksp expression and solve for s,

which, because 1 mol of Ag2SO4 produces 1 mol of SO42–, is the molar solubility we are

seeking.

SolutionFirst, we write the Ksp expression for Ag2SO4 based on the solubility equilibrium equation.

Ksp = [Ag+]2[SO42–] = 1.4 x 10–5

Then, we substitute the values [Ag+] = 2s and [SO42–] = s to obtain the equation that must

be solved.


Chapter Sixteen


Now, by recalling that s = [SO42–] and that [SO4

2–] is the same as the molar solubility of Ag2SO4, we obtain our final result.

AssessmentIn substituting ion concentrations into the Ksp expression, it is important to note that (i)

the factor 2 appears in 2s because the concentration of Ag+ is twice that of SO42–, and (ii)

the exponent 2 appears in the term (2s)2 because the concentration of Ag+ (that is, 2s) must be raised to the second power in the Ksp expression. Be alert for similar situations that might arise in other problems.

Using the Ksp value in Table 16.1, determine the quantity of I–, in parts per million, in a saturated aqueous solution of PbI2. Assume the solution has a density of 1.00 g/mL.

Exercise 16.3B

Calculate the molar solubility of silver arsenate, given that

Exercise 16.3A


Chapter Sixteen

12

Example 16.4 A Conceptual Example

Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI.


Chapter Sixteen

13Example 16.4

Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI.

Analysis and ConclusionsThe key to solving this problem is to recognize that all three solutes are of the same type;that is, the ratio of number of cations to anions in each is 1:1. As suggested by the following equation for the solubility equilibrium of AgX, the ion concentrations are equal to each other and to the molar solubility, s.

The order of increasing molar solubility is the same as the order of increasing Ksp values:

Increasing molar solubilities: AgI < AgBr < AgCl

Increasing Ksp: 8.5 x 10–17 < 5.0 x 10–13 < 1.8 x 10–10

This method works only if we are comparing solutes that are all of the same general formula, that is, all MX, all MX2, and so on.


Chapter Sixteen

14

• The common ion effect affects solubility equilibria as it does other aqueous equilibria.

• The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.

The Common Ion Effectin Solubility Equilibria


Chapter Sixteen

15

Common Ion Effect Illustrated

Na2SO4(aq)

SaturatedAg2SO4(aq)

Ag2SO4

precipitates

The added sulfate ion reduces the

solubility of Ag2SO4.


Chapter Sixteen

16

Common Ion Effect Illustrated

When Na2SO4(aq) is added to the

saturated solution of Ag2SO4 …

… [Ag+] attains a new, lower equilibrium

concentration as Ag+ reacts with SO4

2– to

produce Ag2SO4.


Chapter Sixteen

17

Example 16.5 Calculate the molar solubility of Ag2SO4 in 1.00 M Na2SO4(aq).


Chapter Sixteen

18Example 16.5

Calculate the molar solubility of Ag2SO4 in 1.00 M Na2SO4(aq).

StrategyThe molar solubility of Ag2SO4 will be influenced by the concentration of SO4

2– in solution coming from the 1.00 M Na2SO4(aq). We will assume that the Na2SO4 is completely dissociated and that the presence of Ag2SO4 has no effect on this dissociation. As in Example 16.3, we will use s to represent the number of moles of Ag2SO4 dissolved

per liter of saturated solution. We will then represent [Ag+] and [SO42–] in terms of s and

solve the Ksp expressions for s.

SolutionWe can tabulate the relevant data, including the already present 1.00 mol SO4

2–/L, in the ICE format introduced in Chapter 14.


Chapter Sixteen


The equilibrium concentrations from the ICE table must satisfy the Ksp expression.

Ksp = [Ag+]2[SO42–]

1.4 x 10–5 = (2s)2(1.00 + s)To simplify the equation, let us assume that s is much smaller than 1.00 M, so that (1.00 + s) 1.00.

(2s)2(1.00) = 1.4 x 10–5

4s2 = 1.4 x 10–5

We obtain our final result by solving for s. s2 = 3.5 x 10–6

s = molar solubility = (3.5 x 10–6)1/2

= 1.9 x 10–3 mol Ag2SO4/LAssessmentWhen we assumed that (1.00 + s) 1.00, we were saying that essentially all the common ion comes from the added source (1.00 M) and essentially none comes from the slightly soluble solute (s). The assumption is valid here: (1.00 + 1.9 x 10–3) = 1.00 (to two decimal places). This type of assumption will often be valid in calculations of this type.

Notice that the solubility of Ag2SO4 found here is only about one-eighth of that found for Ag2SO4 in pure water in Example 16.3, illustrating, as we had intended, the common ion effect.

Calculate the molar solubility of Ag2SO4 in 1.00 M AgNO3(aq).

Exercise 16.5A

How many grams of silver nitrate must be added to 250.0 mL of a saturated solution of Ag2SO4(aq) to reduce the solubility of the Ag2SO4 to 1.0 x 10–3 M (Ksp of Ag2SO4 =

1.4 x 10–5)?

Exercise 16.5B


Chapter Sixteen

20

• Ions that are not common to the precipitate can also affect solubility.– CaF2 is more soluble in 0.010 M Na2SO4 than it is in

water.

• Increased solubility occurs because of interionic attractions.

• Each Ca2+ and F– is surrounded by ions of opposite charge, which impede the reaction of Ca2+ with F–.

• The effective concentrations, or activities, of Ca2+ and F– are lower than their actual concentrations.

Solubility and Activities


Chapter Sixteen

21

• Qip can then be compared to Ksp.

• Precipitation should occur if Qip > Ksp.

• Precipitation cannot occur if Qip < Ksp.

• A solution is just saturated if Qip = Ksp.

• In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered.

• Qip is the ion product reaction quotient and is based on initial conditions of the reaction.

Qip and Qc: new look, same great taste!

Will Precipitation Occur? Is It Complete?


Chapter Sixteen

22

Example 16.6If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3, will a precipitate form?

Ag2CrO4(s) 2 Ag+(aq) + CrO42–(aq) Ksp = 1.1 x 10–12


Chapter Sixteen

23

StrategyWe can apply the three-step strategy just outlined.

Example 16.6

If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3(aq), will a precipitate form?

SolutionStep 1: Initial concentrations of ions: The initial concentration of Ag+ is simply 1.5 x 10–

4 M. To establish CrO42–, we first need to determine the number of moles of

CrO42– placed in solution.

We then determine [CrO42–] in the 225 mL (0.225 L) of solution that contains

6.17 x 10–6 mol CrO42–.

Step 2: Evaluation of Qip:

Qip = [Ag+]initial2[CrO4

2–]initial = (1.5 x 10–4)2(2.74 x 10–5) = 6.2 x 10–13


Chapter Sixteen


Step 3: Comparison of Qip and Ksp:

Qip = 6.2 x 10–13 is less than Ksp = 1.1 x 10–12

Because Qip < Ksp, we conclude that no precipitation occurs.

If 1.00 g Pb(NO3)2 and 1.00 g MgI2 are both added to 1.50 L of H2O, should a precipitate

form (Ksp for PbI2 = 7.1 x 10–9)?

Exercise 16.6A

Will precipitation occur from a solution that has 2.5 ppm MgCl2 and a pH of 10.35?

Assume the solution density is 1.00 g/mL [Ksp for Mg(OH)2 = 1.8 x 10–11].

Exercise 16.6B


Chapter Sixteen

25

Example 16.7 A Conceptual ExamplePictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2. What is the solid that first appears? Explain why it then disappears.

Example 16.8

If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form?

MgF2(s) Mg2+(aq) + 2 F–(aq) Ksp = 3.7 x 10–8


Chapter Sixteen

26Example 16.7

Figure 16.4 shows the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2(aq). What is the yellow solid that appears at first? Explain why it then disappears.

Describe how you might use observations of the type suggested by the photographs ofFigure 16.4 to estimate the value of Ksp for PbI2.

Exercise 16.7A

Analysis and ConclusionsWhere the drops of concentrated KI(aq) first enter the Pb(NO3) solution, there is a

localized excess of iodide ion. In this region of the solution, [I–] is large enough so that

Qip = [Pb2+][I–]2 > Ksp

Because Qip > Ksp, yellow PbI2(s) precipitates. When the PbI2(s) that initially formed

settles below the point of entry of the KI(aq), however, [I–] is no longer large enough to maintain a saturated solution for the concentration of Pb2+ present, and therefore the solid redissolves. Based on a uniform distribution of I– ion throughout the solution, the following holds true:

Qip = [Pb2+][I–]2 < Ksp

Consider the following situation similar to Example 16.7: Initially, the 100.0 mL of dilutesolution in the beaker is 1 x 10–4 M and the more concentrated solution added dropwise is 0.10 M. (a) If the dilute solution is Pb(NO3)2(aq) and the more concentrated one is KI(aq), will the observations made resemble Figure 16.4? Explain. (b) What should be observed if the dilute solution is KI(aq) and the more concentrated one is Pb(NO3)2? Explain.

Exercise 16.7B


Chapter Sixteen

27Example 16.8

If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form?

StrategyWe can apply the same three-step strategy as in Example 16.6, but now we will incorporate into Step 1 the assumption that the volume of the mixture of solutions is 0.100 L + 0.200 L = 0.300 L.

SolutionStep 1: Initial concentrations of ions: First, we need to determine [Mg2+] and [F–] as they

initially exist in this mixed solution.


Chapter Sixteen



Qip = [Mg2+][F–]2 = (5.0 x 10–4)(1.7 x 10–2)2 = 1.4 x 10–7

Step 3: Comparison of Qip and Ksp: Because Qip > Ksp, we conclude that precipitation should occur.

Qip = 1.4 x 10–7 is greater than Ksp = 3.7 x 10–8.

Should a precipitate of MgF2(s) form when equal volumes of 0.0010 M MgCl2(aq) and 0.020 M NaF(aq) are mixed?

Exercise 16.8A

How many grams of KI(s) should be dissolved in 10.5 L of 0.00200 M Pb(NO3)2(aq) so that precipitation of PbI2(s) will just begin?

Exercise 16.8B


Chapter Sixteen

29

• A slightly soluble solid does not precipitate totally from solution …

• … but we generally consider precipitation to be “complete” if about 99.9% of the target ion is precipitated (0.1% or less left in solution).

• Three conditions generally favor completeness of precipitation:

1. A very small value of Ksp.

2. A high initial concentration of the target ion.

3. A concentration of common ion that greatly exceeds that of the target ion.

To Determine Whether Precipitation Is Complete


Chapter Sixteen

30

Example 16.9 To a solution with [Ca2+] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), to make the initial [C2O4

2–] = 0.0051 M. Will precipitation of Ca2+ as CaC2O4(s) be complete?

CaC2O4(s) Ca2+(aq) + C2O42–(aq) Ksp = 2.7 x 10–9


Chapter Sixteen

31Example 16.9

To a solution with [Ca2+] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), to make the initial [C2O4

2–] = 0.0051 M. Will the precipitation of Ca2+as CaC2O4(s) be complete?

StrategyWe can begin with the three-step approach to determine whether precipitation occurs. If no precipitation occurs, we need not proceed further—precipitation obviously will not be complete. If precipitation does occur, we will need to make further calculations, which we will do in two parts: First, we will treat the situation as if complete precipitation occurred. This will enable us to determine the excess [C2O4

2–] present. Then, we will solve for the solubility of a slightly soluble solute in the presence of a common ion in a calculation similar to that of Example 16.5.

SolutionFirst, we use our usual three-step approach to determine if precipitation occurs.

Step 1: Initial concentrations of ions: These are given.

[Ca2+] = 0.0050 M, and [C2O42–] = 0.0051 M.


Qip = [Ca2+][C2O42–] = (5.0 x 10–3)(5.1 x 10–3) = 2.6 x 10–5


Chapter Sixteen


Step 3: Comparison of Qip and Ksp:

Qip = 2.6 x 10–5 exceeds Ksp = 2.7 x 10–9

Because Qip > Ksp, precipitation should occur.

Next, we follow the two-part approach outlined in our basic strategy.

Assume complete precipitation.We describe the stoichiometry of the precipitation in a format resembling the ICE format for equilibrium calculations (see Example 15.17b).

The reaction: Ca2+(aq) + C2O42–(aq) CaC2O4(s)

Initial concentrations, M: 0.0050 0.0051Changes, M: –0.0050 –0.0050

After precipitation, M: 0 0.0001

The solubility equilibrium.

Here we enter the appropriate data into the ICE format for solubility equilibrium in the presence of a common ion.


Chapter Sixteen


As usual, we assume that s << 0.0001 and that (0.0001 + s) 0.0001. (We test the validity of this assumption in the Assessment.)

Ksp = [Ca2+][C2O42–] = s(0.0001 + s)

s x 0.0001 = 2.7 x 10–9

s = [Ca2+] = 3 x 10–5 M

We calculate the percentage of Ca2+ remaining in solution as the ratio of the remaining Ca2+ concentration to the initial Ca2+ concentration (5 x 10–3 M).

Our rule for complete precipitation requires that less than 0.1% of an ion should remain in solution. We therefore conclude that precipitation is incomplete.

AssessmentIn this problem, we made the assumption that the contribution of s to the term 0.0001 + s was negligible. Such assumptions are usually valid in common ion situations because the concentration of the common ion is generally relatively large, but 0.0001 M is not a high concentration. If we had not made the simplifying assumption and solved a quadratic equation, we would have obtained s = 2 x 10–5 M. From this result, the percent Ca2+ remaining in solution is 0.4%, but our conclusion that precipitation is incomplete is still valid. If our task had been to calculate [Ca2+] remaining in solution, we would not have made the simplifying assumption.


Chapter Sixteen


Consider a solution in which [Ca2+] = 0.0050 M. If we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), so that the solution is also [C2O4

2–] = 0.0100 M, will the

precipitation of Ca2+ as CaC2O4(s) be complete?

Exercise 16.9A

The two members of each pair of ions are brought together in solution. Without doing detailed calculations, arrange the pairs according to how close to completeness the cation precipitation is—farthest from completeness first and closest to completeness last.

(a) [Ca2+] = 0.110 M and [SO42–] = 0.090 M

(b) [Pb2+] = 0.12 M and [Cl–] = 0.25 M

(c) [Mg2+] = 0.12 M and [SO42–] = 0.15 M

(d) [Ag+] = 0.050 M and [Cl–] = 0.055 M

Exercise 16.9B


Chapter Sixteen

35

Selective Precipitation

AgNO3 added to a mixture

containing Cl– and I–


Chapter Sixteen

36

Example 16.10An aqueous solution that is 2.00 M in AgNO3 is slowly

added from a buret to an aqueous solution that is 0.0100 M in Cl– and also 0.0100 M in I–.

a. Which ion, Cl– or I–, is the first to precipitate from solution?

b. When the second ion begins to precipitate, what is the remaining concentration of the first ion?

c. Is separation of the two ions by selective precipitation feasible?

AgCl(s) Ag+(aq) + Cl–(aq) Ksp = 1.8 x 10–10

AgI(s) Ag+(aq) + I–(aq) Ksp = 8.5 x 10–17


Chapter Sixteen

37Example 16.10

An aqueous solution that is 2.00 M in AgNO3 is slowly added from a buret to an aqueous

solution that is 0.0100 M in Cl– and 0.0100 M in I– (Figure 16.5).

(a) Which ion, Cl– or I–, is the first to precipitate from solution?

(b) When the second ion begins to precipitate, what is the remaining concentration of the first ion?

(c) Is separation of the two ions by selective precipitation feasible?


Chapter Sixteen


StrategyThe molarity of the silver nitrate solution is 200 times greater than [Cl–] or [I–], which means that only a small volume of AgNO3(aq) is required in the precipitation. Because of this, we can neglect throughout the calculation the slight dilution that occurs as the AgNO3(aq) is added to the beaker.

Solution(a) We must find the [Ag+] necessary to precipitate each anion. Precipitation will begin

when the solution has just been saturated, a point at which Qip = Ksp.

To precipitate Cl–.When Qip is equal to Ksp for silver chloride, [Cl–] = 0.0100 M. We then solve the Ksp

expression for [Ag+].

To precipitate I–.

When Qip is equal to Ksp for silver chloride, [I–] = 0.0100 M. We then solve the Ksp

expression for [Ag+].


Chapter Sixteen


Precipitation of AgI(s) begins as soon as [Ag+] = 1.8 x 10–15 M, a concentration much too low for AgCl(s) to precipitate. Therefore I– is the first ion to precipitate.

(b) Next, we determine [I–] remaining in solution when [Ag+] = 1.8 x 10–8 M, the [Ag+] at which AgCl(s) begins to precipitate. To determine this quantity, we rearrange the Ksp expression for silver iodide and solve for [I–].

(c) Now, let us see what percent of the I– is still in solution when AgCl(s) begins to precipitate.

The amount of I– remaining is far below the 0.1% we require for completeness of precipitation. Thus, complete precipitation of I– occurs before Cl– precipitation begins. Separation of the two anions is feasible.

AssessmentThis example illustrates a general rule: The greater the difference in magnitude of the relevant Ksp values, the more likely it is that ion separations by selective precipitation can

be achieved. The Ksp values of AgI and AgCl differ by a factor of about 2 x 106.


Chapter Sixteen

40

• If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered:

• If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate.

Added H+ reacts with, and removes, F–; LeChâtelier’s

principle says more F– forms.

H+ does not consume Cl– ; acid does not affect the

equilibrium.

Effect of pH on Solubility

CaF2(s) Ca2+(aq) + 2 F–(aq)

AgCl(s) Ag+(aq) + Cl–(aq)


Chapter Sixteen

41

Example 16.11 What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x 10–5 M, that is, pH = 9.00?

Mg(OH)2(s) Mg2+(aq) + 2 OH–(aq) Ksp = 1.8 x 10–11

Example 16.12 A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH)2(s) is most soluble:(a) 1.00 M NH3

(b) 1.00 M NH3 /1.00 M NH4+

(c) 1.00 M NH4Cl.


Chapter Sixteen

42

StrategyFirst, note that because the solution is a buffer, any OH– coming from the dissolution of Mg(OH)2(s) is neutralized by the acidic component of the buffer. Therefore, the pH will

remain at 9.00 and the [OH–] at 1.0 x 10–5 M. We will use the Ksp expression and [OH–] =

1.0 x 10–5 M to calculate [Mg2+] in the solution.

Example 16.11

What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x

10–5 M, that is, pH = 9.00?

Strategy

For every mole of Mg2+(aq) appearing in solution, 1 mol of Mg(OH)2(s) must have dissolved. Thus the molar solubility of Mg(OH)2 and the equilibrium concentration of

Mg2+ are the same, meaning that the molar solubility of Mg(OH)2 at pH 9.00 is 0.18 molMg(OH)2/L.


Chapter Sixteen

43

Equilibria Involving Complex Ions

Silver chloride becomes more soluble, not less soluble, in high concentrations of chloride ion.


Chapter Sixteen

44

• A complex ion consists of a central metal atom or ion, with other groups called ligands bonded to it.

• The metal ion acts as a Lewis acid (accepts electron pairs).

• Ligands act as Lewis bases (donate electron pairs).

• The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, Kf:

Complex Ion Formation

Ag+(aq) + 2 Cl–(aq) [AgCl2]–(aq)

[AgCl2]–

Kf = –––––––––– = 1.2 x 108 [Ag+][Cl–]2


Chapter Sixteen

45

Complex Ion Formation

Concentrated NH3 added to a solution of

pale-blue Cu2+ …… forms deep-blue

Cu(NH3)42+.


Chapter Sixteen

46


Chapter Sixteen

47

Complex Ion Formationand Solubilities

AgCl is insoluble in water.

But if the concentration of NH3 is made high

enough …

… the AgCl forms the soluble

[Ag(NH3)2]+ ion.


Chapter Sixteen

48

Example 16.13 Calculate the concentration of free silver ion, [Ag+], in an aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3.

Ag+(aq) + 2 NH3(aq) [Ag(NH3)2]+(aq) Kf = 1.6 x 107

Example 16.14If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution?

AgBr(s) Ag+(aq) + Br–(aq) Ksp = 5.0 x 10–13


Chapter Sixteen

49Example 16.13

Calculate the concentration of free silver ion, [Ag+], in an aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3.

StrategyFirst, because Kf is such a large number, we can assume that the formation of [Ag(NH3)2]

+

goes to completion. We can then establish the concentrations of [Ag(NH3)2]+ and free NH3

through a straightforward stoichiometric calculation. Next, we can return to the formation reaction for the complex ion and consider the extent to which it proceeds in the reverse direction. For this calculation, we will use the Kf expression and solve for x = [Ag+].

SolutionWe complete the bottom row of our ICE tabulation to establish the concentrations of complex ion and free ammonia, assuming that the formation reaction goes to completion.


Chapter Sixteen


Now, we restate the question in this way: What is [Ag+] in a solution that is 0.10 M[Ag(NH3)2]

+ and 2.8 M NH3? To answer this question, we need to focus on the reverse

direction in the reaction for [Ag(NH3)2]+ formation.

When we substitute the equilibrium concentrations into the Kf expression, we obtain an

algebraic equation that we must solve for x = [Ag+].

Because the formation constant is so large, practically all the silver will be present as[Ag(NH3)2]

+ and very little as Ag+. We can therefore assume that x is very much smaller than 0.10 and that 2x is very much smaller than 2.8.


Chapter Sixteen

51Example 16.14

If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should anyAgBr(s) precipitate from the solution?

StrategyWe need to compare Qip for the described solution with Ksp for silver bromide. In the Qip

expression, [Ag+] is that calculated in Example 16.13. We can establish [Br–] from datagiven in this example.

SolutionFrom Example 16.13, we know that [Ag+] = 8.0 x 10–10 M. We determine [Br–] when 1.00 g KBr is dissolved in 1.00 L of the solution with the equation

Because Qip > Ksp, we conclude that some AgBr(s) should precipitate from the solution.


Chapter Sixteen

52

Example 16.16 A Conceptual ExampleFigure 16.10 shows that a precipitateforms when HNO3(aq) is added to

the solution in the beaker on the right in Figure 16.9. Write the equation(s) to show what happens.

Example 16.15What is the molar solubility of AgBr(s) in 3.0 M NH3?

AgBr(s) + 2 NH3(aq) [Ag(NH3)2]+(aq) + Br–(aq)

Kc = 8.0 x 10–6


Chapter Sixteen

53Example 16.15

What is the molar solubility of AgBr(s) in 3.0 M NH3?

StrategyThe key relationship in the dissolution equilibrium is that for every formula unit of AgBr(s) that is dissolved, one ion each of [Ag(NH3)2]

+ and Br– is formed. The molar solubility is the same as the molarity of either of these ions at equilibrium. We calculate these ion molarities in an equilibrium calculation based on the Kc expression.

SolutionWe can use the ICE format to establish equilibrium concentrations in terms of the initial concentration of NH3(aq) and the unknown equilibrium concentration of [Ag(NH3)2]

+(aq) = s.

Now we write the equilibrium constant expression for the reaction and substitute the above data.


Chapter Sixteen


At this point we could assume that s << 3.0 and proceed in the usual fashion, but we can simplify our calculation just by taking the square root of each side of the equation.

Then, we can complete the calculation.

s = (3.0 – 2s)(2.8 x 10–3) = (8.4 x 10–3) – 5.6 x 10–3s

1.0056s = 8.4 x 10–3

s = [Ag(NH3)2]+ = [Br–] = 8.4 x 10–3 M

The molar solubility is therefore 8.4 x 10–3 mol AgBr/L.

What is the molar solubility of AgBr(s) in 0.500 M Na2S2O3(aq)?

Exercise 16.15A

Without doing detailed calculations, determine in which of the following 0.100 M solutions AgI(s) should be most soluble: NH3(aq), Na2S2O3(aq), or NaCN(aq). Explain.

Exercise 16.15B


Chapter Sixteen

55

• Water molecules are commonly found as ligands in complex ions (H2O is a Lewis base).

• The electron-withdrawing power of a small, highly charged metal ion can weaken an O—H bond in one of the ligand water molecules.

• The weakened O—H bond can then give up its proton to another water molecule in the solution.

• The complex ion acts as an acid.

Complex Ions in Acid–Base Reactions

[Na(H2O)4]+ [Al(H2O)6]3+ [Fe(H2O)6]3+


Chapter Sixteen

56

Ionization of a Complex Ion

[Fe(H2O)6]3+ + H2O [Fe(H2O)5OH]2+ + H3O+

The highly-charged iron(III) ion withdraws electron density

from the O—H bonds.

Ka = 1 x 10–7


Chapter Sixteen

57

• Certain metal hydroxides, insoluble in water, are amphoteric; they will react with both strong acids and strong bases.

• Al(OH)3, Zn(OH)2, and Cr(OH)3 are amphoteric.

Amphoteric Species


Chapter Sixteen

58

• Acid–base chemistry, precipitation reactions, oxidation–reduction, and complex ion formation all apply to an area of analytical chemistry called classical qualitative inorganic analysis.

• “Qualitative” signifies that the interest is in determining what is present.– Quantitative analyses are those that determine how much of a

particular substance or species is present.

• Although classical qualitative analysis is not used as widely today as are instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.

Qualitative Inorganic Analysis


Chapter Sixteen

59Qualitative Analysis Outline

In acid, H2S produces very little S2–, so only

the most-insoluble sulfides precipitate.

In base, there is more S2–, and the less-insoluble

sulfides also precipitate.

Some hydroxides also precipitate here.


Chapter Sixteen

60

• If aqueous HCl is added to an unknown solution of cations, and a precipitate forms, then the unknown contains one or more of these cations: Pb2+, Hg2

2+, or Ag+.

• These are the only ions to form insoluble chlorides.

• Any precipitate is separated from the mixture and further tests are performed to determine which of the three Group 1 cations are present.

• The supernatant liquid is also saved for further analysis (it contains the rest of the cations).

• If there is no precipitate, then Group 1 ions must be absent from the mixture.

Cation Group 1


Chapter Sixteen

61

• Precipitated PbCl2 is slightly soluble in hot water.

• The precipitate is washed with hot water, then aqueous K2CrO4 is added to the washings.

• If Pb2+ is present, a precipitate of yellow lead chromate forms, which is less soluble than PbCl2.

• (If all of the precipitate dissolves in the hot water, what does that mean?)

Cation Group 1 (cont’d)

Analyzing for Pb2+


Chapter Sixteen

62

• Next, any undissolved precipitate is treated with aqueous ammonia.

• If AgCl is present, it will dissolve, forming Ag(NH3)2+ (the

dissolution may not be visually apparent).

• If Hg22+ is present, the precipitate will turn dark gray/

black, due to a disproportionation reaction that forms Hg metal and HgNH2Cl.

• The supernatant liquid (which contains the Ag+, if present) is then treated with aqueous nitric acid.

• If a precipitate reforms, then Ag+ was present in the solution.

Cation Group 1 (cont’d)

Analyzing for Ag+ and Hg22+


Chapter Sixteen

63

PbCl2 precipitates when HCl is added.

Hg2Cl2 reacts with NH3 to form black Hg metal and HgNH2Cl.

The presence of lead is confirmed by adding chromate ion; yellow PbCrO4 precipitates.

Group 1 Cation Precipitates


Chapter Sixteen

64

• Once the Group 1 cations have been precipitated, hydrogen sulfide is used as the next reagent in the qualitative analysis scheme.

• H2S is a weak diprotic acid; there is very little ionization of the HS– ion and it is the precipitating agent.

• Hydrogen sulfide has the familiar rotten egg odor that is very noticeable around volcanic areas.

• Because of its toxicity, H2S is generally produced only in small quantities and directly in the solution where it is to be used.

Hydrogen Sulfide in theQualitative Analysis Scheme


Chapter Sixteen

65

• The concentration of HS– is so low in a strongly acidic solution, that only the most insoluble sulfides precipitate.

• These include the eight metal sulfides of Group 2.

• Five of the Group 3 cations form sulfides that are soluble in acidic solution but insoluble in alkaline NH3/NH4

+.

• The other three Group 3 cations form insoluble hydroxides in the alkaline solution.

• The cations of Groups 4 and 5 are soluble.

• Group 4 ions are precipitated as carbonates.

• Group 5 does not precipitate; these must be determined by flame test.

Cation Groups 2, 3, 4, and 5


Chapter Sixteen

66

Cumulative Example A solid mixture containing 1.00 g of ammonium chloride and 2.00 g of barium hydroxide is heated to expel ammonia. The liberated NH3 is then dissolved in 0.500 L of water containing 225 ppm Ca2+ as calcium chloride. Will a precipitate form in this water?


Chapter Sixteen

67Cumulative Example

A solid mixture containing 1.00 g of ammonium chloride and 2.00 g of barium hydroxide is heated to expel ammonia. The liberated NH3(g) is then dissolved in 0.500 L of water

containing 225 ppm Ca2+ as calcium chloride. Will a precipitate form in this water?

StrategyFirst, the precipitate, if there is one, will be Ca(OH)2(s). The Ca2+ is present in the water,

and after the liberated NH3(g) dissolves in the water, the OH– is produced by ionization of the weak base NH3(aq):

If a precipitate is to form, the final ion product in the water, Qip = [Ca2+][OH–]2, must exceed Ksp for Ca(OH)2:

To determine the amount of NH3(g) liberated, we must perform a limiting-reactant calculation (recall Example 3.20) for the reaction

The final steps in the calculation involve obtaining [OH–] from equation (a), converting ppm Ca2+ to [Ca2+], and comparing Qip and Ksp for reaction (b).


Chapter Sixteen

68Cumulative Example continued

SolutionTo determine the amount of NH3(g) available, we complete two stoichiometric calculations based on reaction (c).

Assuming NH4Cl is the limiting reactant:

Assuming Ba(OH)2 is the limiting reactant:

Based on the smaller of the two values just calculated, we determine the molarity of NH3.

Now we use the method of Example 15.8 to determine [OH–], where [OH–] = x and [NH3] = 0.0374 – x.

We assume that x << 0.0374 and solve for x = [OH–].


Chapter Sixteen

69Cumulative Example continued

The quantity of liquid into which we inject the NH3(g) is 0.500 L. Because this liquid is nearly pure water with d = 1.00 g/mL, its mass is 500 g. We can then calculate the mass of Ca2+ in this 500 g water, based on the literal meaning of 225 ppm.

By converting from mass to moles of Ca2+ and dividing by the solution volume, we determine [Ca2+].

Now we can evaluate the ion product, Qip, and compare it with Ksp for Ca(OH)2.

Qip = [Ca2+][OH–]2 = 5.64 x 10–3 x (8.2 x 10–4)2 = 3.8 x 10–9

When we compare Qip with Ksp, we conclude that no precipitate will form.

Qip = 3.8 x 10–9 is considerably smaller than Ksp = 5.5 x 10–6.

AssessmentIn determining [OH–], we made the assumption that it would be much smaller than [NH3],

and this assumption is valid: x = 8.2 x 10–4 is only about 2% of 0.0374.Note that even if a precipitate had formed, its quantity would have been very small.

The maximum yield from 0.113 g Ca2+ would be 0.209 g Ca(OH)2 [that is, 0.113 g Ca2+ x

(74.1 g Ca(OH)2/ 40.1 g 2 Ca2+]. With incomplete precipitation, the quantity would have been smaller still. Yet even a small amount of precipitate may be discernible as turbidity in a solution.

Date post:	14-Jan-2016
Category:	Documents
Upload:	isra
View:	41 times
Download:	2 times

More Equilibria in Aqueous Solutions: Slightly Soluble Salts and Complex Ions

Documents