19-1 Chapter 19 Ionic Equilibria in Aqueous Systems

19-1

Chapter 19

Ionic Equilibria in Aqueous Systems

19-2

Ionic Equilibria in Aqueous Systems

19.1 Equilibria of acid-base buffer systems

19.2 Acid-base titration curves

19.3 Equilibria of slightly soluble ionic compounds

19.4 Equilibria involving complex ions

19.5 Application of ionic equilibria to chemical analysis

19-3

Figure 19.1

The effect of addition of acid or base to …

an unbuffered solution.

a buffered solution.

acid added base added

acid added base added

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

19-4

Acid-Base Buffer Systems

Buffers function by reducing changes in [H3O+]

that result from additions of acid or base to the solution.

Buffers are composed of the conjugateacid-base pair of a weak acid.

Buffers function via the common ion effect.

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

The common ion effect occurs when a reactant containinga given ion is added to an equilibrium mixture that alreadycontains that ion and the position of the equilibrium shiftsaway from forming more of it.

19-5

Table 19.1

The Effect of Added Acetate Ion on the Dissociation of Acetic Acid

[CH3COOH]initial [CH3COO-]added% dissociation* pH

* % dissociation = [CH3COOH]dissoc

[CH3COOH]initial

x 100

0.10 0.00

0.10 0.050

0.10

0.10 0.10

0.15

1.3

0.036

0.018

0.012

2.89

4.44

4.74

4.92


19-6

Figure 19.2

How an acetic acid/acetate buffer works

buffer with equal concentrations of weak acid and its conjugate

base

OH-H3O+

buffer after addition of H3O+

H2O + CH3COOH H3O+ + CH3COO-

buffer after addition of OH-

CH3COOH + OH- H2O + CH3COO-


HA scavenges OH-, A- scavenges H+

19-7

Some Details

Ka = [CH3COO-][H3O+]/[CH3COOH]

[H3O+] = Ka x [CH3COOH]/[CH3COO-]

Since Ka is constant, [H3O+] depends directly on the ratio

of the concentrations of HA and its conjugate base, A-.

19-8

Sample Problem 19.1 Calculating the effect of added H3O+ and OH-

on buffer pH

PROBLEM: Calculate the pH of the following solutions.

(a) A buffer solution consisting of 0.50 M CH3COOH and 0.50 M CH3COONa

(b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution in part (a)

(c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in part (a)

Ka of CH3COOH = 1.8 x 10-5 (assume the additions cause negligible volume

changes)

PLAN:PLAN: We know Ka and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed stepwise through changes in the system.

initial

change

equilibrium

0.50+ x

0.50 - x

--

-

0.50 0+ x

0.50 + x x

- x

SOLUTION:

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration (M)

(a)

___________________________________________________________

19-9

Sample Problem 19.1 (continued)

[CH3COOH]eq ≈ 0.50 M [CH3COO-]eq ≈ 0.50 M[H3O+] = x

Ka =[H3O

+][CH3COO-]

[CH3COOH][H3O

+] = x = Ka

[CH3COO-]

[CH3COOH]= 1.8 x 10-5 M

Check the assumption: 1.8 x 10-5/0.50 X 100 = 3.6 x 10-3 %

CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l)concentration (M)

before addition

addition

after addition

(b)[OH-]added =

0.020 mol

1.0 L soln= 0.020 M NaOH

0.50 - 0.50 -

-

-

- 0.020 -

0.48 0 0.52___________________________________________________________

pH = 4.74

19-10


Set up a reaction table with the new values.


initialchange

equilibrium

0.48 -- x

0.48 - x

-

-

0.52 0

x

+ x + x

0.52 + x

[H3O+] = 1.8 x 10-5 x

0.48

0.52= 1.7 x 10-5 pH = 4.77

CH3COO-(aq) + H3O+(aq) CH3COOH(aq) + H2O

(l)concentration (M)

before addition

addition

after addition

(c) [H3O+]added =

0.020 mol

1.0 L soln= 0.020 M H3O

+

0.50 - 0.50 -

-

-

- 0.020 -

0.48 0 0.52

___________________________________________________________

___________________________________________________________

19-11


Set up a reaction table with the new values.


initialchange

equilibrium

0.52 -- x

0.52 - x

-

-

0.48 0

x

+ x + x

0.48 + x

[H3O+] = 1.8 x 10-5 x

0.48

0.52= 2.0 x 10-5 pH = 4.70

___________________________________________________________

Note that this buffer resists changes in pH fromadditions of either strong acid or strong base!

(What is the solution pH of 0.020 M HCl or 0.020 M NaOH?)

19-12

The Henderson-Hasselbalch Equation

An equation that relates pH, pKa and the ratio ofweak acid to its conjugate base for a buffer solution.

HA + H2O H3O+ + A-

Ka = [H3O+][A-]/[HA]

[H3O+] = Ka x [HA]/[A-]

-log [H3O+] = -log Ka - log ([HA]/[A-])

pH = pKa + log ([A-]/[HA])

pH = pKa + log ([base]/[acid])

Special case: when [base] = [acid], the pH of the buffersolution equals the pKa of the weak acid.

or

19-13

Figure 19.3

The relationship between buffer capacity and pH change

Buffer capacity refers to theability of a buffer to resist

pH change; buffer capacity increases as the concentrations

of its components (i.e., theweak acid and its conjugate

base) increase.

Buffer pH and buffer capacityare different concepts.

For an acetic acid/acetate buffer

19-14

A buffer has the highest capacity when the

concentrations of HA and A- are equal.

For a given addition of acid or base, the concentration ratio

([A-]/[HA]) changes less for similar buffer component concentrations than it does for different concentrations.

Key Concepts

Buffer range: the pH range over which the buffer acts effectively;defined as pKa +/- 1 pH unit

A buffer whose pH is equal to or near the pKa of its acid component has the highest buffer capacity.

19-15

Sample Problem 19.2 Preparing a buffer

SOLUTION:

PROBLEM: An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M

NaHCO3 to prepare the buffer? Ka of HCO3- is 4.7 x 10-11.

PLAN: We know the Ka and the conjugate acid-base pair. Convert pH to

[H3O+], find the number of moles of carbonate and convert to mass.

HCO3-(aq) + H2O(l) CO3

2-(aq) + H3O+(aq)

Ka =[CO3

2-][H3O+]

[HCO3-]

pH = 10.00; [H3O+] = 1.0 x 10-10 4.7 x 10-11 =

[CO32-][10-10]

[0.20][CO3

2-] = 0.094 M

moles of Na2CO3 = (1.5 L)(0.094 mol/L) = 0.14 moles

0.14 moles 105.99 g

mol= 15 g Na2CO3

x

19-16

Acid-Base Titration Curves

Acid-Base Indicator: a weak organic acid (HIn) that has a

different color than its conjugate base (In-); small amountsare used in acid-base titrations; change color over differentpH ranges

19-17

Figure 19.4

Colors and approximate pH ranges of some common acid-base indicators

19-18

Figure 19.5

The color change of the indicator bromthymol blue

acidic

basic

change occurs over ~2 pH

units


pH 6

pH 7.5

19-19 Figure 19.6

Curve for a strong acid-strong base titration


40 mL of 0.1000 M HCl

19-20

Figure 19.7

Curve for a weak acid-

strong base titration

Titration of 40.00 mL of 0.1000 M HPr with 0.1000 M NaOH

[HPr] = [Pr-]

pH = 8.80 at equivalence pointpKa of HPr =

4.89

methyl red

HPr = CH3CH2COOH

19-21

Sample Problem 19.3 Calculating the pH during a weak acid-strong base titration

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M

propanoic acid (HPr; Ka = 1.3 x 10-5) after adding the following volumes of 0.1000 M NaOH:

(a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL

PLAN: The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition.

SOLUTION: (a) Find the starting pH using the methods of Chapter 18.

Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O

+]

€

x= (1.3x10−5)(0.10) x = 1.1 x 10-3 ; pH = 2.96

(b)

before addition

addition

after addition

0.004000

0.003000

0.0030000.001000

0 -

-

-0

-

- -

HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)amount (mol)

_________________________________________________

19-22


[H3O+] = 1.3 x 10-5 x

0.001000 mol

0.003000 mol = 4.3 x 10-6 M pH = 5.37

(c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr-]will be: 0.004000 mol

0.04000 L + 0.04000 L= 0.05000 M

Ka x Kb = Kw Kb = Kw/Ka = 1.0 x 10-14/1.3 x 10-5 = 7.7 x 10-10

[H3O+] = Kw / = 1.6 x 10-9 M

€

Kbx[Pr−] pH = 8.80

(d) 50.00 mL of NaOH will produce an excess of OH-.

mol excess OH- = (0.1000 M)(0.05000 L - 0.04000 L) = 0.00100 mol

M = 0.00100 mol0.0900 L

M = 0.01111

[H3O+] = 1.0 x 10-14/0.01111 = 9.0 x 10-13 M

pH = 12.05

19-23

Figure 19.8

Curve for a weak base-strong acid

titration

Titration of 40.00 mL of 0.1000 M NH3 with 0.1000 M HCl

pH = 5.27 at equivalence

point

pKa of NH4+

= 9.25

19-24

pKa2 = 7.19

pKa1 = 1.85

Figure 19.9

Curve for the titration of a weak polyprotic

acid.

Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH

sulfurous acid: a diprotic weak acid

19-25

Figure 19.10

Sickle shape of red blood cells in sickle cell anemia

Single-site mutationsin the hemoglobin

molecule can change thenet charge on the

protein, which causesprotein aggregationand a consequent

change in cellmorphology

19-26

Dealing with Solubility Equilibria

Slightly soluble ionic compounds

Ion-Product Expressions (Qsp); Solubility-Product Constants (Ksp)

PbSO4(s) Pb2+(aq) + SO42-(aq)

Qc = [Pb2+][SO4 2-]/[PbSO4]

Qsp = [Pb2+][SO42-]

At saturation, Qsp attains a constant value (equilibrium

has been established); thus, Qsp = Ksp

Generally, for a saturated solution of a slightly soluble ionic compound

with formula MpXq: Qsp = [Mn+]p[Xz-]q = Ksp

19-27

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

A slightly solubleionic compound

PbCl2

19-28

Sample Problem 19.4 Writing ion-product expressions for slightly soluble ionic compounds

SOLUTION:

PROBLEM: Write the ion-product expression for each of the following:

(a) magnesium carbonate (b) iron(II) hydroxide(c) calcium phosphate (d) silver sulfide

PLAN: Write an equation which describes a saturated solution. Take note of the unusual behavior of the sulfide ion produced in (d).

Ksp = [Mg2+][CO32-]

(a) MgCO3(s) Mg2+(aq) + CO32-(aq)

Ksp = [Fe2+][OH-]2

(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)

Ksp = [Ca2+]3[PO4

3-]2

(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

(d) Ag2S(s) 2Ag+(aq) + S2-(aq)

S2-(aq) + H2O(l) HS-(aq) + OH-(aq)

Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq) Ksp = [Ag+]2[HS-][OH-]

19-29

Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 25 oC

Name and Formula Ksp

aluminum hydroxide, Al(OH)3

cobalt(II) carbonate, CoCO3

iron(II) hydroxide, Fe(OH)2

lead(II) fluoride, PbF2

lead(II) sulfate, PbSO4

silver sulfide, Ag2S

zinc iodate, Zn(IO3)2

3 x 10-34

1.0 x 10-10

4.1 x 10-15

3.6 x 10-8

1.6 x 10-8

4.7 x 10-29

8 x 10-48

mercury(I) iodide, Hg2I2

3.9 x 10-6


The magnitude of Ksp is a measure of how far to the right the dissolutionproceeds at equilibrium (i.e., at saturation).

19-30

Sample Problem 19.5 Determining Ksp from solubility data

PROBLEM: (a) Lead(II) sulfate is a key component in car batteries. Its solubility

in water at 25 oC is 4.25 x 10-3 g/100 mL solution. What is the Ksp of

PbSO4?(b) When lead(II) fluoride (PbF2) is shaken with pure water at 25 oC,

the solubility is found to be 0.64 g/L. Calculate the Ksp of PbF2.

PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into the solubility product constant expression.

Ksp = [Pb2+][SO42-]

4.25 x 10-3 g100 mL soln

1000 mL

L 303.3 g PbSO4

mol PbSO4= 1.40 x 10-4 M PbSO4

Ksp = [Pb2+][SO42-] = (1.40 x 10-4)2 = 1.96 x 10-8

SOLUTION: PbSO4(s) Pb2+(aq) + SO42-(aq)(a)

x x

19-31


(b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2

0.64 g

L soln 245.2 g PbF2

mol PbF2

= 2.6 x 10-3 M PbF2

Ksp = (2.6 x 10-3)(5.2 x 10-3)2 = 7.0 x 10-8

x

[Pb2+] [F-]

19-32

Sample Problem 19.6 Determining solubility from Ksp

PROBLEM: Calcium hydroxide is a major component of mortar, plaster and cement, and solutions of Ca(OH)2 are used in industry as a

cheap, strong base. Calculate the solubility of Ca(OH)2 in water

at 25 oC if its Ksp is 6.5 x 10-6.PLAN: Write a dissociation equation and Ksp expression; find the molar

solubility (S) using a table.

SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2

-initial

change

equilibrium

-

-

0 0

+S + 2S

S 2S

Ksp = (S)(2S)2 = 4S3 S =

€

6.5x10−6

43

= 1.2 x 10-2 M

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)concentration (M)

____________________________________________________

19-33

Ksp and Solubilities

Ksp values are used to determine relative solubilities provided that comparisonsare made between compounds whose formulas contain the same totalnumber of ions. The compound having the higher Ksp is more soluble.

19-34

Table 19.3 Relationship Between Ksp and Solubility at 25 oC

no. of ions formula cation:anion Ksp solubility (M)

2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4

2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4

2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5

3 Ca(OH)21:2 6.5 x 10-6 1.2 x 10-2

3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3

3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4

3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5


19-35

Common Ion Effects on Solubility

The presence of a common ion decreases the solubility of aslightly soluble ionic compound.

PbCrO4(s) Pb2+(aq) + CrO42-(aq)

Add Na2CrO4 (a very soluble salt; a strong electrolyte)

Result: equilibrium shifts to the left (LeChâtelier’s principle)

19-36

Figure 19.11

The effect of a common ion on solubility

PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4

2-(aq)

CrO42- added


19-37

Sample Problem 19.7Calculating the effect of a common ion on

solubility

PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2

in water. What is its solubility in 0.10 M Ca(NO3)2? Ksp of

Ca(OH)2 is 6.5 x 10-6.PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2.

The Ca(NO3)2 will supply extra [Ca2+], which will influence the solubility of the Ca(OH)2 through the common ion effect.

SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)concentration (M)

initial

change

equilibrium

-

-

-

0.10 0

+S +2S

0.10 + S 2S

Ksp = 6.5 x 10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2 (S << 0.10)

S ≈ = 4.0 x 10-3 M Check the assumption:

4.0 x 10-3 M 0.10 M

x 100 = 4.0 %

_____________________________________________

(6.5 x 10-5/4)0.5

= 1.2 x 10-2 M

19-38

The Effect of pH on Solubility

If the compound contains the anion of a weak acid, addition

of H3O+ (from a strong acid) increases its solubility

(LeChâtelier’s principle)

CaCO3(s) Ca2+(aq) + CO32-(aq)

CO32-(aq) + H3O

+(aq) HCO3- (aq) + H2O(l)

HCO3-(aq) + H3O

+(aq) H2CO3(aq) + H2O(l)

H2CO3(aq) 2H2O(l) + CO2(g)

Adding H3O+ shifts the equilibrium to the right.

19-39

Figure 19.12

Test for the presence of a

carbonate

Effect of the additionof a strong acid

(release of carbondioxide)

19-40

Sample Problem 19.8Predicting the effect of adding a strong acid

on solubility

PROBLEM: Write balanced equations to explain whether addition of H3O+ from a

strong acid will affect the solubility of the following ionic compounds:

(a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) sulfide

PLAN: Write dissolution equations and consider how strong acid would affect the anion component.

Br- is the anion of a strong acid. No effect.

SOLUTION: (a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)

The OH- reacts with the added hydronium ion to form water. The equilibrium will shift to the right and thus solubility will increase.

(c) FeS(s) Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will

react with water to produce OH-.

Both weak acids serve to increase the solubility of FeS.

FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)

19-41

Predicting Whether a Precipitate Will Form

Compare Qsp with Ksp.

When Qsp = Ksp, the solution is saturated and no change occurs.

When Qsp > Ksp, a precipitate forms until the solution is saturated.

When Qsp < Ksp, the solution is unsaturated and no precipitate forms.

19-42

Sample Problem 19.9 Predicting whether a precipitate will form

PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

PLAN: Write out a reaction equation to see which salt could form. Look up the Ksp values in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.

SOLUTION: CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2 x 10-11

mol Ca2+ = 0.100 L(0.30 mol/L) = 0.030 mol [Ca2+] = 0.030 mol/0.300 L = 0.10 M

mol F- = 0.200 L(0.060 mol/L) = 0.012 mol [F-] = 0.012 mol/0.300 L = 0.040 M

Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6 x 10-4

Q is >> Ksp, thus CaF2 will precipitate.

19-43

Equilibria Involving Complex Ions

A complex ion consists of a central metal ion covalently bondedto two or more anions (or molecules) called ligands. Ionic ligands include

hydroxide (OH-), chloride (Cl-) and cyanide (CN-) anions. Water, COand NH3 are examples of molecular ligands.

All complex ions are Lewis adducts. The metal acts as a Lewis acidand the ligand acts as a Lewis base.

We will consider equilibria of hydrated ions with ligands other than water.

19-44

Figure 19.13

Cr(NH3)63+

is a typical complex ion.

Cr3+ is the central metal,and is surrounded by six

NH3 ligands

19-45

Formation Constants, Kf, of Complex Ions

M(H2O)42+(aq) + 4NH3(aq) M(NH3)4

2+(aq) + 4H2O(l)

Kc = [M(NH3)42+][H2O]4/[M(H2O)4

2+][NH3]4

Kf = Kc/[H2O]4 = [M(NH3)42+]/[M(H2O)4

2+][NH3]4

In fact, four sequential reactions occur, defined by four Kf values,

for the systematic substitution of H2O ligands by NH3 ligands.

Thus, Kf = Kf1 x Kf2 x Kf3 x Kf4

19-46

Figure 19.14

The stepwise exchange of NH3 for H2O in

M(H2O)42+

M(H2O)42+

M(H2O)3(NH3)2+

M(NH3)42+

NH3

3NH3


19-47

19-48

Sample Problem 19.10 Calculating the concentration of a complex ion

SOLUTION:

PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more stable

Zn(NH3)42+ by mixing 50.0 L of 0.0020 M Zn(H2O)4

2+ and 25.0 L

of 0.15 M NH3. What is the final [Zn(H2O)42+]? Kf of Zn(NH3)4

2+ is

7.8 x 108.PLAN: Write the reaction equation and Kf expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. The large excess of NH3 will drive the reaction to completion.

Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4

2+(aq) + 4H2O(l)

Kf =[Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]

4

[Zn(H2O)42+]initial = (50.0 L)(0.0020 M)

75.0 L= 1.3 x 10-3 M

[NH3]initial = (25.0 L)(0.15 M)

75.0 L= 5.0 x 10-2 M

19-49


Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4

2+(aq) + 4H2O(l)concentration (M)

initial

change

equilibrium

1.3 x 10-3 5.0 x 10-2 0 -

~(-1.3 x 10-3) ~(-5.2 x 10-3)

Since we assume that all of the Zn(H2O)42+ has reacted, it would consume

four times its amount in NH3..

[NH3]used = 4 (1.3 x 10-3 M) = 5.2 x 10-3 M

~(+1.3 x 10-3) -

x 4.5 x 10-2 1.3 x 10-3

[Zn(H2O)42+]remaining = x (a very small amount)

-

Kf =[Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]

4 7.8 x 108 = (1.3 x 10-3)

x (4.5 x 10-2)4x = 4.1 x 10-7 M

_____________________________________________________________

A result of thevery large Kf!

19-50

Solubility Issues

A ligand increases the solubility of a slightly soluble ionic compoundif it forms a complex ion with the cation.

ZnS(s) + H2O(l) Zn2+(aq) + HS-(aq) + OH-(aq) Ksp = 2.0 x 10-22

Upon addition of some 1.0 M NaCN:

Zn2+(aq) + 4CN-(aq) Zn(CN)42+(aq) Kf = 4.2 x 1019

The overall equation is:

ZnS(s) + 4CN-(aq) + H2O(l) Zn(CN)42+(aq) + HS-(aq) + OH-(aq)

Koverall = Ksp x Kf = 8.4 x 10-3

19-51

Sample Problem 19.11 Calculating the effect of complex ion formation on solubility

SOLUTION:

PROBLEM: In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion

Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0 M

hypo. Kf of Ag(S2O3)23- is 4.7 x 1013 and Ksp AgBr is 5.0 x 10-13.

PLAN: Write equations for the reactions involved. Use Ksp to find S, the molar solubility. Consider the shift in the equilibrium upon the addition of the complexing agent.

AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0 x 10-13

S = [AgBr]dissolved = [Ag+] = [Br-] Ksp = S2 = 5.0 x 10-13; S = 7.1 x 10-7 M(a)

(b) AgBr(s) Ag+(aq) + Br-(aq)

Ag+(aq) + 2S2O32-(aq) Ag(S2O3)2

3-(aq)

AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2

3-(aq)

19-52


Koverall = Ksp x Kf = [Br-][Ag(S2O3)2

3- ]

[S2O32-]2

= (5.0 x 10-13)(4.7 x 1013) = 24

AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2

3-(aq)concentration (M)

initial

change

equilibrium

-

-

-

1.0

-2S

1.0 - 2S

0 0

+S +S

S S

Koverall =S2

(1.0 - 2S)2

= 24S

1.0 - 2S= (24)1/2

S = [Ag(S2O3)23-] = 0.45 M

__________________________________________________________

19-53

Complex Ions of Amphoteric Hydroxides

Amphoteric hydroxides: compounds that dissolve very little in waterbut to a much greater extent in acidic and basic solutions

Al(OH)3(s) Al3+(aq) + 3OH-(aq) Ksp = 3 x 10-34

Al(OH)3 dissolves in acid: 3H3O+(aq) + 3OH- (aq) 6H2O(l)

Al(OH)3 dissolves in base through the formation of a complex ion:

Al(OH)3(s) + OH-(aq) Al(OH)4-(aq)

Al(OH)3(s) + 3H3O+(aq) Al3+(aq) + 6H2O(l)

19-54 Figure 19.15

The amphoteric behavior of aluminum hydroxide

19-55

Selective Precipitation

Selection of an ion in solution by precipitation; achieved byadding a precipitating agent (ion) to the solution until theQsp of the more soluble compound is almost equal to its

Ksp (but Qsp < Ksp). This ion remains in solution, whereas

the other ion(s) having Qsp > Ksp precipitate.

19-56

Sample Problem 19.12 Separating ions by selective precipitation

SOLUTION:

PROBLEM: A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2. Calculate

the [OH-] that would separate the metal ions as their hydroxides.

Ksp of Mg(OH)2 = is 6.3 x 10-10; Ksp of Cu(OH)2 = 2.2 x 10-20.PLAN: Both precipitates have the same ion ratio, 1:2, so we can compare

their Ksp values to determine which has the greater solubility.

Cu(OH)2 will precipitate first (it has the smaller Ksp) so we calculate

the [OH-] needed for a saturated solution of Mg(OH)2. This will

ensure that we do not precipitate Mg(OH)2. We can then check how

much Cu2+ remains in solution.Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3 x 10-10

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2 x 10-20

[OH-] needed for a saturated Mg(OH)2 solution =

€

Ksp

[Mg2+]

€

=6.3x10−10

0.20

= 5.6 x 10-5 M

19-57


Use the Ksp for Cu(OH)2 to find the amount of Cu2+ remaining in solution.

[Cu2+] = Ksp/[OH-]2 = 2.2 x 10-20/(5.6 x 10-5)2 = 7.0 x 10-12 M

Since the solution was 0.10 M CuCl2, virtually none of the

Cu2+ remains in solution.

19-58

Qualitative Analysis

Self-Study: pp. 841-845 in textbook

Only an overview of this materialwill be provided in class. You will be responsible,

however, for this material for the Final Exam.

19-59 Figure 19.16

General procedure for separating ions in qualitative analysis

add precipitating

ion

cent

rifug

e

add precipitating

ion

cent

rifug

e


19-60

Figure 19.17

A qualitative analysis scheme for separating cations into five ion groups

add 6 M HCl

cent

rifug

eacidify to pH 0.5;

add H2S

cent

rifug

e

add

NH3/NH4+

buffer(pH 8)

cent

rifug

e

add (NH4)2HPO4

cent

rifug

e


19-61

19-62

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

Tests to determine thepresence of cations in

ion group 5

(a) flame test for Na+: yellow-orange

(b) flame test for K+: violet

(c) NH4+ + OH- : pH test for NH3 gas

Figure 19.18

19-63

Figure 19.18

Step 1: Add

NH3(aq)

cent

rifug

e

cent

rifug

e

Step 2: Add HCl

Step 3: Add NaOH

cent

rifug

eStep 4:

Add HCl, Na2HPO4

Step 5: Dissolve in

HCl and add KSCN

A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+

19-64

End of Assigned Material

19-65

Figure B19.1

A view inside Carlsbad Caverns, New Mexico

19-66

Figure B19.2

Formation of acidic precipitation

19-67

Figure B19.4

The effect of acid rain on statuary

Location: New York City

19-68

A forest damaged by acid rainFigure B19.3

A forest damaged by

acid rain

Documents

19-1 Chapter 19 Ionic Equilibria in Aqueous Systems