87
19-1 CHEM 1B: GENERAL CHEMISTRY Chapter 19: Ionic Equilibria in Aqueous Systems Instructor: Dr. Orlando E. Raola Santa Rosa Junior College

CHEM 1B: Chapter 19: GENERAL CHEMISTRY Ionic Equilibria …oraola/CHEM1BLECT/Ch. 19/ch19_lecture_6e.pdf · 19-3 Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffers

Embed Size (px)

Citation preview

19-1

CHEM 1B:

GENERAL CHEMISTRY Chapter 19:

Ionic Equilibria in Aqueous

Systems

Instructor: Dr. Orlando E. Raola

Santa Rosa Junior College

19-2

Chapter 19

Ionic Equilibria in Aqueous Systems

19-3

Ionic Equilibria in Aqueous Systems

19.1 Equilibria of Acid-Base Buffers

19.2 Acid-Base Titration Curves

19.3 Equilibria of Slightly Soluble Ionic Compounds

19.4 Equilibria Involving Complex Ions

19-4

Acid-Base Buffers

An acid-base buffer is a solution that lessens the impact of

pH from the addition of acid or base.

An acid-base buffer usually consists of a conjugate acid-

base pair where both species are present in appreciable

quantities in solution.

An acid-base buffer is therefore a solution of a weak acid

and its conjugate base, or a weak base and its

conjugate acid.

19-5

Figure 19.1 The effect of adding acid or base to an unbuffered

solution.

A 100-mL sample of

dilute HCl is adjusted

to pH 5.00.

The addition of 1 mL of strong acid (left)

or strong base (right) changes the pH by

several units.

19-6

Figure 19.2 The effect of adding acid or base to a buffered

solution.

A 100-mL sample of

an acetate buffer is

adjusted to pH 5.00.

The addition of 1 mL of strong acid (left)

or strong base (right) changes the pH very

little.

The acetate buffer is made by mixing 1 mol/L CH3COOH ( a weak acid)

with 1 mol/L CH3COONa (which provides the conjugate base, CH3COO-).

19-7

Buffers and the Common-ion Effect

A buffer works through the common-ion effect.

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

acetic acid acetate ion

Acetic acid in water dissociates slightly to produce some

acetate ion:

If NaCH3COO is added, it provides a source of

CH3COO- ion, and the equilibrium shifts to the left.

CH3COO- is common to both solutions.

The addition of CH3COO- reduces the % dissociation of

the acid.

19-8

Table 19.1 The Effect of Added Acetate Ion on the Dissociation

of Acetic Acid

[CH3COOH]init [CH3COO-]added % Dissociation* [H3O+] pH

0.10 0.00 1.3 1.3x10-3 2.89

0.10 0.050 0.036 3.6x10-5 4.44

0.10 0.10 0.018 1.8x10-5 4.74

0.10 0.15 0.012 1.2x1015 4.92

* % Dissociation = [CH3COOH]dissoc

[CH3COOH]init

x 100

19-9

How a Buffer Works

The buffer components (HA and A-) are able to consume

small amounts of added OH- or H3O+ by a shift in

equilibrium position.

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

Added H3O+ reacts with

CH3COO-, causing a

shift to the left.

Added OH- reacts with

CH3COOH, causing a shift to

the right.

The shift in equilibrium position absorbs the change in

[H3O+] or [OH-], and the pH changes only slightly.

19-10

Figure 19.3 How a buffer works.

Buffer has equal

concentrations of A- and HA.

H3O+

Buffer has more HA after

addition of H3O+.

H2O + CH3COOH ← H3O+ + CH3COO-

OH-

Buffer has more A- after

addition of OH-.

CH3COOH + OH- → CH3COO- + H2O

19-11

Relative Concentrations of Buffer Components

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

Ka = [CH3COO-][H3O

+]

[CH3COOH]

If the ratio increases, [H3O+] decreases.

[HA]

[A-]

Since Ka is constant, the [H3O+] of the solution depends

on the ratio of buffer component concentrations.

[H3O+] = Ka x

[CH3COOH]

[CH3COO-]

If the ratio decreases, [H3O+] increases.

[HA]

[A-]

19-12

Sample Problem 19.1 Calculating the Effect of Added H3O+ or

OH- on Buffer pH

Calculate the pH:

PLAN: We can calculate [CH3COOH]init and [CH3COO-]init from the

given information. From this we can find the starting pH. For (b) and (c)

we assume that the added OH- or H3O+ reacts completely with the

buffer components. We write a balanced equation in each case, set up

a reaction table, and calculate the new [H3O+].

PROBLEM:

(a) Of a buffer solution consisting of 0.50 mol/L CH3COOH and 0.50

mol/L CH3COONa

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

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

Ka of CH3COOH = 1.8 x 10-5. (Assume the additions cause a negligible

change in volume.)

19-13

Sample Problem 19.1

SOLUTION: (a)

Initial 0.50 - 0.50 0

Change - −x +x +x

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

(mol/L)

Equilibrium - 0.50 - x x 0.50 + x

1.8x10-5 mol/L

0.50 mol/L

x 100 = 3.6x10-3% (< 5%; assumption is justified.)

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.50 – x ≈ 0.50 mol/L and [CH3COO-] = 0.50 + x ≈ 0.50 mol/L

x = [H3O+] = Ka x

[CH3COOH]

[CH3COO-] ≈ 1.8x10-5 x 0.50

0.50

= 1.8x10-5 mol/L

Checking the assumption: pH = -log(1.8x10-5) = 4.74

19-14

Sample Problem 19.1

Initial 0.50 0.020 0.50 -

Change -0.020 -0.020 +0.020 -

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Concentration

(mol/L)

(b) [OH−]added = 0.020 mol

1.0 L soln = 0.020 mol/L

OH−

Equilibrium 0 0.48 - 0.52

Setting up a reaction table for the stoichiometry:

Initial 0.48 - 0.52 0

Change - −x +x +x

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

(mol/L)

Equilibrium - 0.48 - x x 0.52 + x

Setting up a reaction table for the acid dissociation, using new initial [ ]:

19-15

Sample Problem 19.1

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.48 – x ≈ 0.48 mol/L and [CH3COO-] = 0.52 + x ≈ 0.52 mol/L

x = [H3O+] = Ka x

[CH3COOH]

[CH3COO-]

pH = -log(1.7x10-5) = 4.77

≈ 1.8x10-5 x 0.48

0.52 = 1.7x10-5 mol/L

19-16

Sample Problem 19.1

Initial 0.50 0.020 0.50 -

Change -0.020 -0.020 +0.020 -

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

(mol/L)

(c) [H3O+]added =

0.020 mol

1.0 L soln = 0.020 mol/L

H3O+

Equilibrium 0 0.48 - 0.52

Setting up a reaction table for the stoichiometry:

Initial 0.52 - 0.48 0

Change - −x +x +x

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

(mol/L)

Equilibrium - 0.52 - x x 0.48 + x

Setting up a reaction table for the acid dissociation, using new initial [ ]:

19-17

Sample Problem 19.1

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.52 – x ≈ 0.52 mol/L and [CH3COO-] = 0.48 + x ≈ 0.48 mol/L

x = [H3O+] = Ka x

[CH3COOH]

[CH3COO-]

pH = -log(2.0x10-5) = 4.70

≈ 1.8x10-5 x 0.52

0.48 = 2.0x10-5 mol/L

19-18

The Henderson-Hasselbalch Equation

HA(aq) + H2O(l) A-(aq) + H3O+(aq)

Ka = [H3O

+][A-]

[HA] [H3O

+] = Ka x [HA]

[A-]

-log[H3O+] = -logKa – log

[HA]

[A-]

pH = pKa + log [base]

[acid]

19-19

Buffer Capacity

The buffer capacity is a measure of the “strength” of the

buffer, its ability to maintain the pH following addition of

strong acid or base.

The greater the concentrations of the buffer

components, the greater its capacity to resist pH

changes.

The closer the component concentrations are to each

other, the greater the buffer capacity.

19-20

Figure 19.4 The relation between buffer capacity and pH change.

When strong base is

added, the pH increases

least for the most

concentrated buffer.

This graph shows the final pH values for four different buffer solutions after

the addition of strong base.

19-21

Buffer Range

The buffer range is the pH range over which the buffer is

effective.

Buffer range is related to the ratio of buffer component

concentrations.

[HA]

[A-] The closer is to 1, the more effective the buffer.

If one component is more than 10 times the other, buffering

action is poor. Since log10 = 1, buffers have a usable

range within ± 1 pH unit of the pKa of the acid

component.

19-22

Sample Problem 19.2 Using Molecular Scenes to Examine Buffers

PROBLEM: The molecular scenes below represent samples of four

HA/A- buffers. (HA is blue and green, A- is green, and

other ions and water are not shown.)

(a) Which buffer has the highest pH?

(b) Which buffer has the greatest capacity?

(c) Should we add a small amount of concentrated strong acid or

strong base to convert sample 1 to sample 2 (assuming no volume

changes)?

19-23

Sample Problem 19.2

PLAN: Since the volumes of the solutions are equal, the scenes

represent molarities as well as numbers. We count the particles

of each species present in each scene and calculate the ratio

of the buffer components.

SOLUTION:

[A-]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3,

4/4 = 1; and sample 4, 4/2 = 2.

19-24

Sample Problem 19.2

(c) Sample 2 has a lower [A-]/[HA] ratio than sample 1, so we need to

increase the [A-] and decrease the [HA]. This is achieved by

adding strong acid to sample 1.

(a) As the pH rises, more HA will be converted to A-. The scene with

the highest [A-]/[HA] ratio is at the highest pH. Sample 4 has the

highest pH because it has the highest ratio.

(b) The buffer with the greatest capacity is the one with the [A-]/[HA]

closest to 1. Sample 3 has the greatest buffer capacity.

19-25

Preparing a Buffer

• Choose the conjugate acid-base pair.

– The pKa of the weak acid component should be close to the

desired pH.

• Calculate the ratio of buffer component concentrations.

• Determine the buffer concentration, and calculate the

required volume of stock solutions and/or masses of

components.

• Mix the solution and correct the pH.

pH = pKa + log [base]

[acid]

19-26

Sample Problem 19.3 Preparing a Buffer

SOLUTION:

PROBLEM: An environmental chemist needs a carbonate buffer of

pH 10.00 to study the effects of the acid rain on

limsetone-rich soils. How many grams of Na2CO3 must

she add to 1.5 L of freshly prepared 0.20 mol/L NaHCO3

to make the buffer? Ka of HCO3- is 4.7x10-11.

PLAN: The conjugate pair is HCO3- (acid) and CO3

2- (base), and we

know both the buffer volume and the concentration of HCO3-.

We can calculate the ratio of components that gives a pH of

10.00, and hence the mass of Na2CO3 that must be added to

make 1.5 L of solution.

[H3O+] = 10-pH = 10-10.00

= 1.0x10-10 mol/L

HCO3-(aq) + H2O(l) H3O

+(aq) + CO32-(aq) Ka =

[CO32-][H3O

+]

[HCO3-]

19-27

Sample Problem 19.3 Preparing a Buffer

[CO32-] =

Ka[HCO3-]

[H3O+]

= (4.7x10-11)(0.20)

1.0x10-10 = 0.094 mol/L

Amount (mol) of CO32- needed = 1.5 L soln x 0.094 mol CO3

2-

1 L soln

= 0.14 mol CO32-

105.99 g Na2CO3

1 mol Na2CO3

0.14 mol Na2CO3 x = 15 g Na2CO3

The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 mol/L

NaHCO3 and add more 0.20 mol/L NaHCO3 to make 1.5 L. Using a

pH meter, she can then adjust the pH to 10.00 by dropwise addition of

concentrated strong acid or base.

19-28

Acid-Base Indicators

An acid-base indicator is a weak organic acid (HIn)

whose color differs from that of its conjugate base (In-).

The color of an indicator changes over a specific,

narrow pH range, a range of about 2 pH units.

The ratio [HIn]/[In-] is governed by the [H3O+] of the

solution. Indicators can therefore be used to monitor the

pH change during an acid-base reaction.

19-29

pH

Figure 19.5 Colors and approximate pH range of some

common acid-base indicators.

19-30

Figure 19.6 The color change of the indicator bromthymol blue.

pH < 6.0 pH > 7.5 pH = 6.0-7.5

19-31

Acid-Base Titrations

In an acid-base titration, the concentration of an acid (or a

base) is determined by neutralizing the acid (or base) with

a solution of base (or acid) of known concentration.

The equivalence point of the reaction occurs when the

number of moles of OH- added equals the number of

moles of H3O+ originally present, or vice versa.

The end point occurs when the indicator changes color.

- The indicator should be selected so that its color change occurs at a

pH close to that of the equivalence point.

19-32

Figure 19.7 Curve for a strong acid–strong base titration.

The initial pH is low.

The pH rises very rapidly

at the equivalence point,

which occurs at pH = 7.00.

The pH increases gradually

when excess base has been

added.

19-33

Calculating the pH during a

strong acid–strong base titration

Initial pH

[H3O+] = [HA]init

pH = -log[H3O+]

pH before equivalence point

initial mol H3O+ = Vacid x Macid

mol OH- added = Vbase x Mbase

mol H3O+

remaining = (mol H3O+

init) – (mol OH-added)

[H3O+] = pH = -log[H3O

+] mol H3O

+remaining

Vacid + Vbase

19-34

pH at the equivalence point

pH = 7.00 for a strong acid-strong base titration.

Calculating the pH during a

strong acid–strong base titration

pH beyond the equivalence point

initial mol H3O+ = Vacid x Macid

mol OH- added = Vbase x Mbase

mol OH-excess = (mol OH-

added) – (mol H3O+

init)

[OH-] =

pOH = -log[OH-] and pH = 14.00 - pOH

mol OH-excess

Vacid + Vbase

19-35

Example:

40.00 mL of 0.1000 mol/L HCl is titrated with 0.1000 mol/L NaOH.

The initial pH is simply the pH of the HCl solution:

[H3O+] = [HCl]init = 0.1000 mol/L and pH = -log(0.1000) = 1.00

To calculate the pH after 20.00 mL of NaOH solution has been added:

OH- added = 0.02000 L NaOH x 0.1000 mol

1 L = 2.000x10-3 mol OH-

Initial mol of H3O+ = 0.04000 L HCl x 0.1000 mol

1 L = 4.000x10-3 mol H3O

+

H3O+ remaining = 4.000x10-3 – 2.000x10-3 = 2.000x10-3 mol H3O

+

The OH- ions react with an equal amount of H3O+ ions, so

19-36

[H3O+] =

2.000x10-3 mol

0.04000 L + 0.02000 L = 0.03333 mol/L

pH = -log(0.03333) = 1.48

To calculate the pH after 50.00 mL of NaOH solution has been added:

The equivalence point occurs when mol of OH- added = initial mol of

HCl, so when 40.00 mL of NaOH has been added.

OH- added = 0.05000 L NaOH x 0.1000 mol

1 L = 5.000x10-3 mol OH-

OH- in excess = 5.000x10-3 – 4.000x10-3 = 1.000x10-3 mol OH-

[OH-] = 1.000x10-3 mol

0.04000 L + 0.05000 L = 0.01111 mol/L

pOH = -log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05

19-37

Figure 19.8 Curve for a weak acid–strong base titration.

The initial pH is higher than

for the strong acid solution.

The curve rises

gradually in the buffer

region. The weak acid

and its conjugate

base are both present

in solution.

The pH increases slowly

beyond the equivalence

point.

The pH at the equivalent

point is > 7.00 due to the

reaction of the conjugate

base with H2O.

19-38

Calculating the pH during a

weak acid–strong base titration

Initial pH

[H3O+] =

pH = -log[H3O+]

[H3O+][A-]

[HA] Ka =

pH before equivalence point

[H3O+] = Ka x or

pH = pKa + log [base]

[acid]

[HA]

[A-]

19-39

Calculating the pH during a

weak acid–strong base titration

A-(aq) + H2O(l) HA(aq) + OH-(aq)

pH at the equivalence point

[OH- ] =

where [A-] = and Kw

Ka

Kb = mol HAinit

Vacid + Vbase

[H3O+] ≈ and pH = -log[H3O

+] Kw

pH beyond the equivalence point

[OH-] =

pH = -log[H3O+]

mol OH-excess

Vacid + Vbase

[H3O+] =

Kw

[OH-]

19-40

Sample Problem 19.4 Finding the pH During a Weak Acid–

Strong Base Titration

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

mol/L propanoic acid (HPr; Ka = 1.3x10-5) after adding the

following volumes of 0.1000 mol/L NaOH:

PLAN: The initial pH must be calculated using the Ka value for the weak

acid. We then calculate the number of moles of HPr present

initially and the number of moles of OH- added. Once we know

the volume of base required to reach the equivalence point we

can calculate the pH based on the species present in solution.

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

SOLUTION:

(a) [H3O+] = = 1.1x10-3 mol/L

pH = -log(1.1x10-3) = 2.96

19-41

Sample Problem 19.4

(b) 30.00 mL of 0.1000 mol/L NaOH has been added.

Initial amount of HPr = 0.04000 L x 0.1000 mol/L = 4.000x10-3 mol HPr

Amount of NaOH added = 0.03000 L x 0.1000 mol/L = 3.000x10-3 mol OH-

Each mol of OH- reacts to form 1 mol of Pr-, so

HPr(aq) + OH-(aq) → Pr-(aq) + H2O(l) Concentration

(mol/L) Initial 0.004000 0.003000 0 -

Change −0.003000 -0.003000 +0.003000 -

Equilibrium 0.001000 0 - 0.003000

[H3O+] = Ka x [HPr]

[Pr-]

pH = -log(4.3x10-6) = 5.37

0.001000

0.003000

= (1.3x10-5) x = 4.3x10-6 mol/L

19-42

Sample Problem 19.4

(c) 40.00 mL of 0.1000 mol/L NaOH has been added.

This is the equivalence point because mol of OH- added = 0.004000

= mol of HAinit.

All the OH- added has reacted with HA to form 0.004000 mol of Pr-.

0.004000 mol

0.04000 L + 0.04000 L [Pr-] = = 0.05000 mol/L

Pr- is a weak base, so we calculate Kb = Kw

Ka

= 1.0x10-14

1.3x10-5 = 7.7x10-10

[H3O+] ≈

Kw = 1.0x10-14

= 1.6x10-9 mol/L

pH = -log(1.6x10-9) = 8.80

19-43

Sample Problem 19.4

(d) 50.00 mL of 0.1000 mol/L NaOH has been added.

Amount of OH- added = 0.05000 L x 0.1000 mol/L = 0.005000 mol

Excess OH- = OH-added – HAinit = 0.005000 – 0.004000 = 0.001000 mol

[OH-] = mol OH-

excess

total volume

= 0.001000 mol

0.09000 L = 0.01111 mol/L

[H3O+] =

Kw

[OH-] =

1x10-14

0.01111 = 9.0x10-13 mol/L

pH = -log(9.0x10-13) = 12.05

19-44

Figure 19.9 Curve for a weak base–strong acid titration.

The pH decreases

gradually in the buffer

region. The weak base

and its conjugate acid

are both present in

solution. The pH at the equivalence

point is < 7.00 due to the

reaction of the conjugate

acid with H2O.

19-45

Figure 19.10 Curve for the titration of a weak polyprotic acid.

Titration of 40.00 mL of 0.1000

mol/L H2SO3 with 0.1000 mol/L

NaOH

pKa2 = 7.19

pKa1 = 1.85

19-46

Amino Acids as Polyprotic Acids

An amino acid contains a weak base (-NH2) and a weak

acid (-COOH) in the same molecule.

Both groups are protonated at low pH and the amino

acid behaves like a polyprotic acid.

19-47

Figure 19.11 Abnormal shape of red blood cells in sickle cell

anemia.

In sickle cell anemia, the

hemoglobin has two amino

acids with neutral R groups

instead of charged groups. The

abnormal hemoglobin causes

the red blood cells to have a

sickle shape, as seen here.

Several amino acids have

charged R groups in addition

to the NH2 and COOH group.

These are essential to the

normal structure of many

proteins.

19-48

Equilibria of Slightly Soluble Ionic Compounds

Any “insoluble” ionic compound is actually slightly

soluble in aqueous solution.

We assume that the very small amount of such a compound that

dissolves will dissociate completely.

For a slightly soluble ionic compound in water, equilibrium

exists between solid solute and aqueous ions.

PbF2(s) Pb2+(aq) + 2F-(aq)

Qc = [Pb2+][F-]2

[PbF2] Qsp = Qc[PbF2] = [Pb2+][F-]2

19-49

Qsp and Ksp

Qsp is called the ion-product expression for a slightly

soluble ionic compound.

For any slightly soluble compound MpXq, which consists

of ions Mn+ and Xz-,

Qsp = [Mn+]p[Xz-]q

When the solution is saturated, the system is at equilibrium,

and Qsp = Ksp, the solubility product constant.

The Ksp value of a salt indicates how far the dissolution

proceeds at equilibrium (saturation).

19-50

Metal Sulfides

Metal sulfides behave differently from most other slightly

soluble ionic compounds, since the S2- ion is strongly

basic.

We can think of the dissolution of a metal sulfide as a

two-step process:

MnS(s) Mn2+(aq) + S2-(aq)

S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)

MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq)

Ksp = [Mn2+][HS-][OH-]

19-51

Sample Problem 19.5 Writing Ion-Product Expressions

SOLUTION:

PROBLEM: Write the ion-product expression at equilibrium for each

compound:

(a) magnesium carbonate (b) iron(II) hydroxide

(c) calcium phosphate (d) silver sulfide

PLAN: We write an equation for a saturated solution of each

compound, and then write the ion-product expression at

equilibrium, Ksp. Note the sulfide in part (d).

(a) MgCO3(s) Mg2+(aq) + CO32-(aq) Ksp = [Mg2+][CO3

2-]

(b) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) Ksp = [Fe2+][OH-]2

(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO4

3-]2

19-52

Sample Problem 19.5

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

S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)

Ksp = [Ag+]2[HS-][OH-]

Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq)

19-53

Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic

Compounds at 25°C

Name, 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

3x10-34

1.0x10-10

4.1x10-15

3.6x10-8

1.6x10-8

4.7x10-29

8x10-48

Mercury(I) iodide, Hg2I2

3.9x10-6

19-54

Sample Problem 19.6 Determining Ksp from Solubility

PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in lead-

acid car batteries. Its solubility in water at 25°C is

4.25x10-3 g/100 mL solution. What is the Ksp of

PbSO4?

(b) When lead(II) fluoride (PbF2) is shaken with pure

water at 25°C, the solubility is found to be 0.64 g/L.

Calculate the Ksp of PbF2.

PLAN: We write the dissolution equation and the ion-product

expression for each compound. This tells us the number of

moles of each ion formed. We use the molar mass to convert

the solubility of the compound to molar solubility (molarity),

then use it to find the molarity of each ion, which we can

substitute into the Ksp expression.

19-55

Sample Problem 19.6

Ksp = [Pb2+][SO42-]

Ksp = [Pb2+][SO42-] = (1.40x10-4)2

SOLUTION:

= 1.96x10-8

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

Converting from g/mL to mol/L:

4.25x10-3g PbSO4

100 mL soln x 1000 mL

1 L

x 1 mol PbSO4

303.3 g PbSO4 = 1.40x10-4 mol/L PbSO4

Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO42-, so

[Pb2+] = [SO42-] = 1.40x10-4 mol/L

19-56

Sample Problem 19.6

Ksp = [Pb2+][F-]2

Ksp = [Pb2+][F-]2 = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

(b) PbF2(s) Pb2+(aq) + F-(aq)

Converting from g/L to mol/L:

0.64 g PbF2

1 L soln

x 1 mol PbF2

245.2 g PbF2 = 2.6x10-3 mol/L PbF2

Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F-, so

[Pb2+] = 2.6x10-3 mol/L and [F-] = 2(2.6x10-3) = 5.2x10-3 mol/L

19-57

Sample Problem 19.7 Determining Solubility from Ksp

PROBLEM: Calcium hydroxide (slaked lime) is a major component of

mortar, plaster, and cement, and solutions of Ca(OH)2

are used in industry as a strong, inexpensive base.

Calculate the molar solubility of Ca(OH)2 in water if the

Ksp is 6.5x10-6.

PLAN: We write the dissolution equation and the expression for Ksp.

We know the value of Ksp, so we set up a reaction table that

expresses [Ca2+] and [OH-] in terms of S, the molar solubility.

We then substitute these expressions into the Ksp expression

and solve for S.

SOLUTION:

Ksp = [Ca2+][OH-]2 = 6.5x10-6 Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)

19-58

Sample Problem 19.7

- Initial 0 0

Change - +S + 2S

Ksp = [Ca2+][OH-]2 = (S)(2S)2 = 4S3 = 6.5x10-6

= 1.2x10-2

mol/L

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Concentration

(mol/L)

Equilibrium - S 2S

S = =

19-59

Table 19.3 Relationship Between Ksp and Solubility at 25°C

No. of Ions Formula Cation/Anion Ksp Solubility

(mol/L)

2 MgCO3 1/1 3.5x10-8 1.9x10-4

2 PbSO4 1/1 1.6x10-8 1.3x10-4

2 BaCrO4 1/1 2.1x10-10 1.4x10-5

3 Ca(OH)2 1/2 6.5x10-6 1.2x10-2

3 BaF2 1/2 1.5x10-6 7.2x10-3

3 CaF2 1/2 3.2x10-11 2.0x10-4

3 Ag2CrO4 2/1 2.6x10-12 8.7x10-5

The higher the Ksp value, the greater the solubility, as long as we

compare compounds that have the same total number of ions in

their formulas.

19-60

Figure 19.12 The effect of a common ion on solubility.

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

If Na2CrO4 solution is added to a saturated solution of PbCrO4, it

provides the common ion CrO42-, causing the equilibrium to shift to

the left. Solubility decreases and solid PbCrO4 precipitates.

19-61

Sample Problem 19.8 Calculating the Effect of a Common Ion

on Solubility

PROBLEM: In Sample Problem 19.7, we calculated the solubility of

Ca(OH)2 in water. What is its solubility in 0.10 mol/L

Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

PLAN: The addition of Ca2+, an ion common to both solutions, should

lower the solubility of Ca(OH)2. We write the equation and Ksp

expression for the dissolution and set up a reaction table in

terms of S, the molar solubility of Ca(OH)2. We make the

assumption that S is small relative to [Ca2+]init because Ksp is

low. We can then solve for S and check the assumption.

SOLUTION:

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

19-62

Sample Problem 19.8

Change - +S + 2S

Ksp = [Ca2+][OH-]2 = 6.5x10-6 ≈ (0.10)(2S)2 = (0.10)(4S2)

= 4.0x10-3 mol/L

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Concentration

(mol/L)

Equilibrium - 0.10 + S 2S

[Ca2+]init = 0.10 mol/L because Ca(NO3)2 is a soluble salt, and

dissociates completely in solution.

4S2 ≈ 6.5x10-6

0.10 so S ≈

Checking the assumption: 4.0x10-3 mol/L

0.10 mol/L

x 100 = 4.0% < 5%

- Initial 0.10 0

19-63

Effect of pH on Solubility

Changes in pH affects the solubility of many slightly

soluble ionic compounds.

The addition of H3O+ will increase the solubility of a salt

that contains the anion of a weak acid.

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

CO32-(aq) + H3O

+(aq) → HCO3-(aq) + H2O(l)

HCO3-(aq) + H3O

+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)

The net effect of adding H3O+ to CaCO3 is the removal

of CO32- ions, which causes an equilibrium shift to the

right. More CaCO3 will dissolve.

19-64

Figure 19.13 Test for the presence of a carbonate.

When a carbonate mineral is treated with HCl, bubbles of CO2 form.

19-65

Sample Problem 19.9 Predicting the Effect on Solubility of

Adding Strong Acid

PROBLEM: Write balanced equations to explain whether addition of

H3O+ from a strong acid affects the solubility of each ionic

compound:

PLAN: We write the balanced dissolution equation for each compound

and note the anion. The anion of a weak acid reacts with H3O+,

causing an increase in solubility.

SOLUTION:

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

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

Br- is the anion of HBr, a strong acid, so it does not react with H3O+. The

addition of strong acid has no effect on its solubility.

19-66

Sample Problem 19.9

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

OH- is the anion of H2O, a very weak acid, and is in fact a strong base. It

will react with H3O+:

The addition of strong acid will cause an increase in solubility.

OH-(aq) + H3O+(aq) → 2H2O(l)

(c) FeS(s) Fe2+(aq) + S2-(aq)

S2- is the anion of HS-, a weak acid, and is a strong base. It will react

completely with water to form HS- and OH-. Both these ions will react

with added H3O+:

The addition of strong acid will cause an increase in solubility.

HS-(aq) + H3O+(aq) → H2S(aq) + H2O(l)

OH-(aq) + H3O+(aq) → 2H2O(l)

19-67

Figure 19.14 Limestone cave in Nerja, Málaga, Spain.

Limestone is mostly CaCO3 (Ksp = 3.3x10-9).

CO2(g) CO2(aq)

CO2(aq) + 2H2O(l) H3O+(aq) + HCO3

-(aq)

Ground water rich in CO2 trickles over

CaCO3, causing it to dissolve. This

gradually carves out a cave.

CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)

Water containing HCO3- and Ca2+ ions

drips from the cave ceiling. The air has

a lower P than the soil, causing CO2

to come out of solution. A shift in

equilibrium results in the precipitation

of CaCO3 to form stalagmites and

stalactites.

CO2

19-68

Predicting the Formation of a Precipitate

For a saturated solution of a slightly soluble ionic salt,

Qsp = Ksp.

When two solutions containing the ions of slightly soluble

salts are mixed,

If Qsp = Ksp,

the solution is saturated and no change will occur.

If Qsp > Ksp,

a precipitate will form until the remaining solution is saturated.

If Qsp =< Ksp,

no precipitate will form because the solution is unsaturated.

19-69

Sample Problem 19.10 Predicting Whether a Precipitate Will

Form

PROBLEM: A common laboratory method for preparing a precipitate is

to mix solutions containing the component ions. Does a

precipitate form when 0.100 L of 0.30 mol/L Ca(NO3)2 is

mixed with 0.200 L of 0.060 mol/L NaF?

PLAN: First we need to decide which slightly soluble salt could form,

look up its Ksp value in Appendix C, and write the dissolution

equation and Ksp expression. We find the initial ion

concentrations from the given volumes and molarities of the

two solutions, calculate the value for Qsp and compare it to Ksp.

SOLUTION:

The ions present are Ca2+, NO3-, Na+, and F-. All Na+ and NO3

- salts are

soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10-11).

CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2

19-70

Sample Problem 19.10

Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.

We need to calculate [Ca2+] and [F-] in the final solution.

Amount (mol) of Ca2+ = 0.030 mol/L Ca2+ x 0.100 L = 0.030 mol Ca2+.

[Ca2+]init = 0.030 mol Ca2+

0.100 L + 0.200 L = 0.10 mol/L Ca2+

Amount (mol) of F- = 0.060 mol/L F- x 0.200 L = 0.012 mol F-.

[F-]init = 0.012 mol F-

0.100 L + 0.200 L = 0.040 mol/L F-

Qsp = [Ca2+]init[F-]2init = (0.10)(0.040)2 = 1.6x10-4

Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10-11.

19-71

Sample Problem 19.11 Using Molecular Scenes to Predict

Whether a Precipitate Will Form

PROBLEM: These four scenes represent solutions of silver (gray) and

carbonate (black and red) ions above solid silver

carbonate. (The solid, other ions, and water are not

shown.)

(a) Which scene best represents the solution in equilibrium with the

solid?

(b) In which, if any, other scene(s) will additional solid silver

carbonate form?

(c) Explain how, if at all, addition of a small volume of concentrated

strong acid affects the [Ag+] in scene 4 and the mass of solid

present.

19-72

Sample Problem 19.11

PLAN: We need to determine the ratio of the different types of ion in

each solution. A saturated solution of Ag2CO3 should have

2Ag+ ions for every 1 CO32- ion. For (b) we need to compare

Qsp to Ksp. For (c) we recall that CO32- reacts with H3O

+.

SOLUTION:

First we determine the Ag+/CO32- ratios for each scene.

Scene 1: 2/4 or 1/2 Scene 2: 3/3 or 1/1

Scene 3: 4/2 or 2/1 Scene 4: 3/4

(a) Scene 3 is the only one that has an Ag+/CO32- ratio of 2/1, so this

scene represents the solution in equilibrium with the solid.

19-73

(b) We use the ion count for each solution to determine Qsp for each

one. Since Scene 3 is at equilibrium, its Qsp value = Ksp.

Sample Problem 19.11

Scene 1: Qsp = (2)2(4) = 16

Scene 3: Qsp = (4)2(2) = 32

Scene 2: Qsp = (3)2(3) = 27

Scene 4: Qsp = (3)2(4) = 36

Scene 4 is the only one that has Qsp > Ksp, so a precipitate forms in

this solution.

(c) Ag2CO3(s) 2Ag+(aq) + CO32-(aq)

CO32-(aq) + 2H3O

+(aq) → [H2CO3(aq)] + 2H2O(l) → 3H2O(l) + CO2(g)

The CO2 leaves as a gas, so adding H3O+ decreases the [CO3

2-] in

solution, causing more Ag2CO3 to dissolve.

[Ag+] increases and the mass of Ag2CO3 decreases.

19-74

Selective Precipitation

Selective precipitation is used to separate a solution

containing a mixture of ions.

A precipitating ion is added to the solution until the Qsp

of the more soluble compound is almost equal to its Ksp.

The less soluble compound will precipitate in as large a

quantity as possible, leaving behind the ion of the more

soluble compound.

19-75

Sample Problem 19.12 Separating Ions by Selective Precipitation

PROBLEM: A solution consists of 0.20 mol/L MgCl2 and 0.10 mol/L

CuCl2. Calculate the [OH-] that would separate the metal

ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10;

Ksp of Cu(OH)2 is 2.2x10-20.

PLAN: Both compounds have 1/2 ratios of cation/anion, so we can

compare their solubilities by comparing their Ksp values.

Mg(OH)2 is 1010 times more soluble than Cu(OH)2, so

Cu(OH)2 will precipitate first. We write the dissolution

equations and Ksp expressions. Using the given cation

concentrations, we solve for the [OH-] that gives a saturated

solution of Mg(OH)2. Then we calculate the [Cu2+] remaining

to see if the separation was successful.

19-76

Sample Problem 19.12

SOLUTION:

Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+][OH-]2 = 6.3x10-10

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = [Cu2+][OH-]2 = 2.2x10-20

[OH-] = = = 5.6x10-5 mol/L

This is the maximum [OH-] that will not precipitate Mg2+ ion.

Calculating the [Cu2+] remaining in solution with this [OH-]

Ksp

[OH-]2 [Cu2+] = =

2.2x10-20

(5.6x10-5)2 = 7.0x10-12 mol/L

Since the initial [Cu2+] is 0.10 mol/L, virtually all the Cu2+ ion is

precipitated.

19-77

Figure B19.1 Formation of acidic precipitation.

Chemical Connections

Since pH affects the solubility of many slightly soluble ionic compounds, acid

rain has far-reaching effects on many aspects of our environment.

19-78

Figure 19.15 Cr(NH3)63+, a typical complex ion.

A complex ion consists of a central metal ion covalently bonded to

two or more anions or molecules, called ligands.

19-79

Figure 19.16 The stepwise exchange of NH3 for H2O in M(H2O)42+.

The overall formation constant is given by

Kf = [M(NH3)4

2+]

[M(H2O)42+][NH3]

4

19-80

Table 19.4 Formation Constants (Kf) of Some Complex Ions at

25°C

19-81

Sample Problem 19.13 Calculating the Concentration of a

Complex Ion

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

stable Zn(NH3)42+ by mixing 50.0 L of 0.0020 mol/L

Zn(H2O)42+ and 25.0 L of 0.15 mol/L NH3. What is the

final [Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)4

2+ is

7.8x108.

PLAN: We write the reaction equation and the Kf expression, and use

a reaction table to calculate equilibrium concentrations. To set

up the table, we must first find [Zn(H2O)42+]init and [NH3]init

using the given volumes and molarities. With a large excess

of NH3 and a high Kf, we assume that almost all the

Zn(H2O)42+ is converted to Zn(NH3)4

2+.

SOLUTION:

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

2+(aq) + 4H2O(l)

Kf = [Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]

4

19-82

Sample Problem 19.13

[Zn(H2O)42+]initial = = 1.3x10-3

mol/L

50.0 L x 0.0020 mol/L

50.0 L + 25.0 L

[NH3]initial = = 5.0x10-2

mol/L

25.0 L x 0.15 mol/L

50.0 L + 25.0 L

Initial 1.3x10-3 5.0x10-2 0 -

Change ~(-1.3x10-3) ~(-5.2x10-3) ~(+1.3x10-3) -

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

2+(aq) + 4H2O(l) Concentration

(mol/L)

Equilibrium x 4.5x10-2 1.3x10-3 -

4 mol of NH3 is needed per mol of Zn(H2O4)2+, so

[NH3]reacted = 4(1.3x10-3 mol/L) = 5.2x10-3 mol/L and

[Zn(NH3)42+] ≈ 1.3x10-3 mol/L

19-83

Kf = [Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]

4 = 7.8x108 =

(1.3x10-3)

x(4.5x10-2)4

x = [Zn(H2O)42+ = 4.1x10-7

mol/L

Sample Problem 19.13

19-84

Sample Problem 19.14 Calculating the Effect of Complex-Ion

Formation on Solubility

PROBLEM: In black-and-white film developing, excess AgBr is

removed from the film negative by “hypo”, an aqueous

solution of sodium thiosulfate (Na2S2O3), which forms the

complex ion Ag(S2O3)23-. Calculate the solubility of AgBr

in (a) H2O; (b) 1.0 mol/L hypo.

Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13.

PLAN: After writing the equation and the Ksp expression, we use the

given Ksp value to solve for S, the molar solubility of AgBr. For

(b) we note that AgBr forms a complex ion with S2O32-, which

shifts the equilibrium and dissolves more AgBr. We write an

overall equation for the process and set up a reaction table to

solve for S.

19-85

Sample Problem 19.14

SOLUTION:

S = [AgBr]dissolved = [Ag+] = [Br-]

Ksp = [Ag+][Br-] = S2 = 5.0x10-13

(b) Write the overall equation:

(a) AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13

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

3-(aq)

S = 7.1x10-7

mol/L

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

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

3-(aq)

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

3-]

[S2O32-]2

= (5.0x10-13)(4.7x1013) = 24

19-86

Sample Problem 19.14

Initial - 1.0 0 0

Change - -2S +S +S

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

3-(aq) Concentration

(mol/L)

Equilibrium - 1.0 - 2S S S

Koverall = S2

(1.0 - 2S)2 = 24

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

mol/L

S

1.0 - 2S = = 4.9 so

S = 4.9 mol/L – 0.9S and 10.9S = 4.9 mol/L

19-87

Figure 19.17 The amphoteric behavior of aluminum hydroxide.

When solid Al(OH)3 is treated with H3O+ (left) or with OH- (right), it dissolves

as a result of the formation of soluble complex ions.