Lecture PowerPoint
Chemistry
The Molecular Nature of
Matter and Change
Seventh Edition
Martin S. Silberberg
and Patricia G. Amateis
19-1Copyright  McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19
Ionic Equilibria in Aqueous Systems
19-2
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
19-3
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 acidbase 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.
19-4
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.
19-5
The addition of 1 mL of strong acid (left)
or strong base (right) changes the pH by
several units.
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 M CH3COOH ( a weak acid) with
1 M CH3COONa (which provides the conjugate base, CH3COO-).
19-6
Buffers and the Common-ion Effect
A buffer works through the common-ion effect.
Acetic acid in water dissociates slightly to produce some
acetate ion:
CH3COOH(aq) + H2O(l)
acetic acid
CH3COO-(aq) + H3O+(aq)
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.
19-7
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
1.2x10-5
4.92
*
% Dissociation =
19-8
[CH3COOH]dissoc
[CH3COOH]init
x 100
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)
Added OH– reacts with
CH3COOH, causing a shift to
the right.
CH3COO–(aq) + H3O+(aq)
Added H3O+ reacts with
CH3COO–, causing a
shift to the left.
The shift in equilibrium position absorbs the change in
[H3O+] or [OH–], and the pH changes only slightly.
19-9
Figure 19.3
How a buffer works.
Buffer has more HA after
Buffer has equal
addition of H3O+.
concentrations of A– and HA.
H3O+
Buffer has more A- after
addition of OH–.
OH-
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
H2O + CH3COOH ← H3O+ + CH3COO-
19-10
CH3COOH + OH- → CH3COO- + H2O
Relative Concentrations of Buffer Components
CH3COOH(aq) + H2O(l)
[CH3COO–][H3O+]
Ka =
[CH3COOH]
CH3COO–(aq) + H3O+(aq)
[H3
O +]
[CH3COOH]
= Ka x
[CH3COO–]
Since Ka is constant, the [H3O+] of the solution depends
on the ratio of buffer component concentrations.
[HA]
If the ratio – increases, [H3O+] increases.
[A ]
[HA]
+] decreases.
If the ratio
decreases,
[H
O
3
[A–]
19-11
Sample Problem 19.1
Calculating the Effect of Added H3O+ or
OH- on Buffer pH
PROBLEM: Calculate the pH:
(a) Of 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 (a).
(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.)
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+].
19-12
Sample Problem 19.1
SOLUTION: (a)
Concentration (M) CH3COOH(aq) + H2O(l)
Initial
Change
Equilibrium
0.50
−x
0.50 − x
CH3COO–(aq) + H3O+(aq)
-
0.50
+x
0
+x
-
0.50 + x
x
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.50 – x ≈ 0.50 M and [CH3COO-] = 0.50 + x ≈ 0.50 M
x = [H3O+] = Ka x
[CH3COOH]
[CH3COO–]
≈ 1.8x10–5 x 0.50 = 1.8x10–5 M
0.50
pH = -log(1.8x10–5) = 4.74
Checking the assumption:
1.8x10–5 M x 100 = 3.6x10-3% (< 5%; assumption is justified.)
0.50 M
19-13
Sample Problem 19.1
0.020 mol
(b) [OH−]added =
= 0.020 M OH–
1.0 L soln
Setting up a reaction table for the stoichiometry:
Concentration (M) CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)
Initial
Change
Equilibrium
0.50
−0.020
0.020
−0.020
0.50
+0.020
-
0.48
0
0.52
-
Setting up a reaction table for the acid dissociation, using new initial [ ]:
Concentration (M) CH3COOH(aq) + H2O(l)
Initial
Change
Equilibrium
19-14
0.48
−x
0.48 − x
-
CH3COO–(aq) + H3O+(aq)
0.52
+x
0.52 + x
0
+x
x
Sample Problem 19.1
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.48 – x ≈ 0.48 M and [CH3COO–] = 0.52 + x ≈ 0.52 M
x = [H3O+] = Ka x
[CH3COOH]
[CH3COO–]
≈ 1.8x10–5 x 0.48 = 1.7x10–5 M
0.52
pH = −log(1.7x10-5) = 4.77
Addition of a small amount of base caused the pH to rise only
slightly, from 4.74 to 4.77.
19-15
Sample Problem 19.1
0.020 mol
(c) [H3O+]added =
= 0.020 M H3O+
1.0 L soln
Setting up a reaction table for the stoichiometry:
Concentration (M) CH3COO–(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)
Initial
Change
Equilibrium
0.50
−0.020
0.020
−0.020
0.50
+0.020
-
0.48
0
0.52
-
Setting up a reaction table for the acid dissociation, using new initial [ ]:
Concentration (M) CH3COOH(aq) + H2O(l)
Initial
Change
Equilibrium
19-16
0.52
−x
0.52 - x
-
CH3COO–(aq) + H3O+(aq)
0.48
+x
0.48 + x
0
+x
x
Sample Problem 19.1
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.52 – x ≈ 0.52 M and [CH3COO–] = 0.48 + x ≈ 0.48 M
[CH3COOH]
x = [H3O+] = Ka x
≈ 1.8x10–5 x 0.52 = 2.0x10–5 M
[CH3COO–]
0.48
pH = −log(2.0x10–5) = 4.70
Addition of a small amount of acid caused the pH to drop only
slightly, from 4.74 to 4.70.
19-17
Figure 19.4
19-18
Effect of added acid or base on concentrations of
buffer components
The Henderson-Hasselbalch Equation
HA(aq) + H2O(l)
[H3O+][A–]
Ka =
[HA]
-log[H3
O+]
A-(aq) + H3O+(aq)
[H3O+] = Ka x
[HA]
= -logKa – log
[A–]
[base]
pH = pKa + log
[acid]
19-19
[HA]
[A–]
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 19.5
The relation between buffer capacity and pH change.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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]
The closer – is to 1, the more effective the buffer.
[A ]
If the concentration of one component is more than 10 times
the concentration of 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
(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.
(c)
19-25
Sample 2 has a lower [A–]/[HA] ratio than sample 1, so to convert
sample 1 to sample 2, we need to decrease [A–] and increase
[HA]. This is achieved by adding strong acid to sample 1.
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.
[base]
pH = pKa + log
[acid]
• Determine the buffer concentration, and calculate the
required volume of stock solutions and/or masses of
components.
• Mix the solution and correct the pH.
19-26
Sample Problem 19.3
Preparing a Buffer
PROBLEM: An environmental chemist needs a carbonate buffer of
pH 10.00 to study the effects of acid rain on limestonerich soils. How many grams of Na2CO3 must she add to
1.5 L of freshly prepared 0.20 M NaHCO3 to make the
buffer? Ka of HCO3– is 4.7x10–11.
PLAN: The conjugate pair is HCO3– (acid) and CO32– (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.
SOLUTION:
[H3O+] = 10–pH = 10–10.00 = 1.0x10–10 M
HCO3
19-27
– (aq)
+ H2O(l)
H3
O+(aq)
+ CO3
2– (aq)
Ka =
[CO32–][H3O+]
[HCO3–]
Sample Problem 19.3
[CO3
2–]
=
Ka[HCO3–]
[H3
O+ ]
Preparing a Buffer
=
(4.7x10–11)(0.20)
1.0x10–10
= 0.094 M
2–
Amount (mol) of CO32– needed = 1.5 L soln x 0.094 mol CO3
1 L soln
= 0.14 mol CO32–
0.14 mol Na2CO3 x 105.99 g Na2CO3
1 mol Na2CO3
= 15 g Na2CO3
The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 M
NaHCO3 and add more 0.20 M 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 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.
The color of an indicator changes over a specific,
narrow pH range, a range of about 2 pH units.
19-29
Figure 19.6
Colors and approximate pH range of some
common acid-base indicators.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
pH
19-30
Figure 19.7
The color change of the indicator bromthymol blue.
pH < 6.0
19-31
pH = 6.0-7.5
pH > 7.5
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 19.8
Curve for a strong acid–strong base titration.
The pH increases gradually
when excess base has been
added.
The pH rises very rapidly at
the equivalence point, which
occurs at pH = 7.00.
The initial pH is low.
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)
[H3
O+]
=
mol H3O+remaining
Vacid + Vbase
19-34
pH = –log[H3O+]
Calculating the pH during a
strong acid–strong base titration
pH at the equivalence point
pH = 7.00 for 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–]
=
mol OH–excess
Vacid + Vbase
pOH = –log[OH–] and pH = 14.00 - pOH
19-35
Example:
40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH.
The initial pH is simply the pH of the HCl solution:
[H3O+] = [HCl]init = 0.1000 M and pH = –log(0.1000) = 1.00
To calculate the pH after 20.00 mL of NaOH solution has been added:
Initial mol of H3O+ = 0.04000 L HCl x 0.1000 mol = 4.000x10–3 mol H3O+
1L
OH– added = 0.02000 L NaOH x 0.1000 mol = 2.000x10–3 mol OH–
1L
The OH– ions react with an equal amount of H3O+ ions, so
H3O+ remaining = 4.000x10–3 – 2.000x10–3 = 2.000x10–3 mol H3O+
19-36
[H3
O +]
2.000x10–3 mol
=
= 0.03333 M
0.04000 L + 0.02000 L
pH = –log(0.03333) = 1.48
The equivalence point occurs when mol of OH– added = initial mol of
HCl, so when 40.00 mL of NaOH has been added.
To calculate the pH after 50.00 mL of NaOH solution has been added:
OH– added = 0.05000 L NaOH x 0.1000 mol = 5.000x10–3 mol OH–
1L
OH– in excess = 5.000x10–3 – 4.000x10–3 = 1.000x10–3 mol OH–
[OH–]
=
1.000x10–3 mol
= 0.01111 M
0.04000 L + 0.05000 L
pOH = –log(0.01111) = 1.95
19-37
pH = 14.00 – 1.95 = 12.05
Figure 19.9
Curve for a weak acid–strong base titration.
The pH increases slowly
beyond the equivalence
point.
The curve rises gradually
in the buffer region. The
weak acid and its
conjugate base are both
present in solution.
19-38
The pH at the equivalence
point is > 7.00 due to the
reaction of the conjugate
base with H2O.
The initial pH is higher than
for the strong acid solution.
Calculating the pH during a
weak acid–strong base titration
Initial pH
Ka =
[H3O+][A–]
[HA]
[H3O+] = √Ka x [HA]init
pH = –log[H3O+]
pH before equivalence point
[H3O+] = Ka x [HA]
[A–]
pH = pKa + log
19-39
[base]
[acid]
or
Calculating the pH during a
weak acid–strong base titration
pH at the equivalence point
A–(aq) + H2O(l)
HA(aq) + OH–(aq)
[OH–] = √Kb x [A–]
mol HAinit
where [A-] =
Vacid + Vbase
[H3O+] ≈
Kw
√Kb x [A–]
and Kb =
Kw
Ka
and pH = -log[H3O+]
pH beyond the equivalence point
[OH–]
=
mol OH–excess
[H3O+] =
Vacid + Vbase
pH = –log[H3O+]
19-40
Kw
[OH–]
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 M propanoic acid (HPr; Ka = 1.3x10–5) 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 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.
SOLUTION:
(a) [H3O+] = √Ka x [HPr]init = √(1.3 x 10–5) (0.100) = 1.1x10–3 M
pH = –log(1.1x10–3) = 2.96
19-41
Sample Problem 19.4
(b) 30.00 mL of 0.1000 M NaOH has been added.
Initial amount of HPr = 0.04000 L x 0.1000 M = 4.000x10–3 mol HPr
Amount of NaOH added = 0.03000 L x 0.1000 M = 3.000x10–3 mol OH–
Each mol of OH– reacts to form 1 mol of Pr–, so
Concentration (M)
HPr(aq) +
OH–(aq) → Pr-(aq) + H2O(l)
Initial
Change
0.004000
−0.003000
0.003000
−0.003000
0
+0.003000
-
Equilibrium
0.001000
0
0.003000
-
[H3O+] = Ka x [HPr]
[Pr-]
= (1.3x10–5) x 0.001000 = 4.3x10–6 M
0.003000
pH = −log(4.3x10–6) = 5.37
19-42
Sample Problem 19.4
(c) 40.00 mL of 0.1000 M 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
= 0.05000 M
0.04000 L + 0.04000 L
Kw
1.0x10−14
−
Pr is a weak base, so we calculate Kb =
=
= 7.7x10−10
Ka
1.3x10−5
Kw
1.0x10-14
+
-9 M
=
[H3O ] ≈
=
1.6x10
√Kb x [Pr−] √(7.7x10-10)(0.05000)
[Pr−] =
pH = –log(1.6x10-9) = 8.80
19-43
Sample Problem 19.4
(d) 50.00 mL of 0.1000 M NaOH has been added.
Amount of OH– added = 0.05000 L x 0.1000 M = 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 M
–14
1x10
K
w
= 9.0x10–13 M
[H3O+] =
=
0.01111
[OH–]
pH = –log(9.0x10–13) = 12.05
19-44
Figure 19.10
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.
19-45
The pH at the equivalence
point is < 7.00 due to the
reaction of the conjugate
acid with H2O.
Figure 19.11
19-46
Curve for the titration of a weak polyprotic acid.
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 19.12
Abnormal shape of red blood cells in sickle cell
anemia.
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.
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.
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+][F–]2
Qc =
[PbF2]
19-49
Pb2+(aq) + 2F–(aq)
Qsp = Qc[PbF2] = [Pb2+][F–]2
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
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).
SOLUTION:
(a)
MgCO3(s)
(b) Fe(OH)2(s)
(c)
19-52
Ca3(PO4)2(s)
Mg2+(aq) + CO32–(aq)
Fe2+(aq) + 2OH–(aq)
3Ca2+(aq) + 2PO43–(aq)
Ksp = [Mg2+][CO32–]
Ksp = [Fe2+][OH–]2
Ksp = [Ca2+]3[PO43–]2
Sample Problem 19.5
(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–]
19-53
Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 25°C
Name, Formula
19-54
Ksp
Aluminum hydroxide, Al(OH)3
3x10–34
Cobalt(II) carbonate, CoCO3
1.0x10–10
Iron(II) hydroxide, Fe(OH)2
4.1x10–15
Lead(II) fluoride, PbF2
3.6x10–8
Lead(II) sulfate, PbSO4
1.6x10–8
Mercury(I) iodide, Hg2I2
4.7x10–29
Silver sulfide, Ag2S
8x10–48
Zinc iodate, Zn(IO3)2
3.9x10–6
Sample Problem 19.6
Determining Ksp from Solubility
PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in leadacid 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
SOLUTION:
(a)
PbSO4(s)
Pb2+(aq) + SO42(aq)
Ksp = [Pb2+][SO42–]
Converting from g/mL to mol/L:
4.25x10–3 g PbSO4 x 1000 mL x 1 mol PbSO4
100 mL soln
1L
303.3 g PbSO4
= 1.40x10–4 M PbSO4
Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO42–, so
[Pb2+] = [SO42–] = 1.40x10–4 M
Ksp = [Pb2+][SO42–] = (1.40x10–4)2 = 1.96x10–8
19-56
Sample Problem 19.6
(b)
PbF2(s)
Pb2+(aq) + F–(aq)
Ksp = [Pb2+][F–]2
Converting from g/L to mol/L:
0.64 g PbF2 x
1 mol PbF2
1 L soln
245.2 g PbF2
= 2.6x10–3 M PbF2
Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F–, so
[Pb2+] = 2.6x10–3 M and [F–] = 2(2.6x10–3) = 5.2x10–3 M
Ksp = [Pb2+][F–]2 = (2.6x10–3)(5.2x10–3)2 = 7.0x10–8
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:
Ca(OH)2(s)
19-58
Ca2+(aq) + 2OH–(aq)
Ksp = [Ca2+][OH–]2 = 6.5x10–6
Sample Problem 19.7
Concentration (M)
Ca(OH)2(s)
Ca2+(aq) + 2OH–(aq)
Initial
-
0
0
Change
-
+S
+ 2S
Equilibrium
-
S
2S
Ksp = [Ca2+][OH–]2 = (S)(2S)2 = 4S3 = 6.5x10–6
S = 3√Ksp/4
19-59
=
3√ (6.5x10–6)/4
= 1.2x10–2 M
Table 19.3 Relationship Between Ksp and Solubility at 25°C
No. of Ions Formula Cation/Anion
Ksp
Solubility (M)
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 19.13
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 M
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)
19-62
Ca2+(aq) + 2OH–(aq)
Ksp = [Ca2+][OH–]2 = 6.5x10–6
Sample Problem 19.8
[Ca2+]init = 0.10 M because Ca(NO3)2 is a soluble salt, and dissociates
completely in solution.
Concentration (M)
Ca(OH)2(s)
Ca2+(aq) + 2OH–(aq)
Initial
-
0.10
0
Change
-
+S
+ 2S
Equilibrium
-
0.10 + S
2S
Ksp = [Ca2+][OH–]2 = 6.5x10–6 ≈ (0.10)(2S)2 = (0.10)(4S2)
4S2
≈
6.5x10–6
0.10
so S ≈
√(6.5x10-5)/4
= 4.0x10–3 M
-3
Checking the assumption: 4.0x10 M x 100 = 4.0% < 5%
0.10 M
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 19.14
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:
(a) lead(II) bromide (b) copper(II) hydroxide
(c) iron(II) sulfide
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)
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+:
OH–(aq) + H3O+(aq) → 2H2O(l)
The addition of strong acid will cause an increase in solubility.
(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+:
HS–(aq) + H3O+(aq) → H2S(aq) + H2O(l)
OH–(aq) + H3O+(aq) → 2H2O(l)
The addition of strong acid will cause an increase in solubility.
19-67
Figure 19.15
Limestone cave in Nerja, Málaga, Spain.
Limestone is mostly CaCO3 (Ksp = 3.3x10–9).
Ground water rich in CO2 trickles over
CaCO3, causing it to dissolve. This
gradually carves out a cave.
Water containing HCO3- and Ca2+ ions
drips from the cave ceiling. The air has
a lower PCO than the soil, causing CO2
2
to come out of solution. A shift in
equilibrium results in the precipitation
of CaCO3 to form stalagmites and
stalactites.
CO2(g)
CO2(aq)
CO2(aq) + 2H2O(l)
H3O+(aq) + HCO3–(aq)
CaCO3(s) + CO2(aq) + H2O(l)
19-68
Ca2+(aq) + 2HCO3– (aq)
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 M Ca(NO3)2 is mixed
with 0.200 L of 0.060 M 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)
19-70
Ca2+(aq) + 2F–(aq)
Ksp = [Ca2+][F–]2
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 M Ca2+ x 0.100 L = 0.030 mol Ca2+.
2+
0.030
mol
Ca
2+
[Ca ]init =
= 0.10 M Ca2+
0.100 L + 0.200 L
Amount (mol) of F– = 0.060 M F– x 0.200 L = 0.012 mol F–.
[F–]init
=
0.012 mol F–
= 0.040 M F–
0.100 L + 0.200 L
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
Sample Problem 19.11
(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.
Scene 1: Qsp = (2)2(4) = 16
Scene 2: Qsp = (3)2(3) = 27
Scene 3: Qsp = (4)2(2) = 32
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 [CO32–] 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 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 6.3x10–10; 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–] = √Ksp/[Mg2+] = √6.3x10-10/0.20 = 5.6x10–5 M
This is the maximum [OH–] that will not precipitate Mg2+ ion.
Calculating the [Cu2+] remaining in solution with this [OH–]
[Cu2+]
Ksp
2.2x10–20
=
=
–
2
[OH ]
(5.6x10–5)2
= 7.0x10–12 M
Since the initial [Cu2+] is 0.10 M, virtually all the Cu2+ ion is precipitated.
19-77
Chemical Connections
Figure B19.1
Formation of acidic precipitation.
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 19.16
Cr(NH3)63+, a typical complex ion.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A complex ion consists of a central metal ion covalently bonded to
two or more anions or molecules, called ligands.
19-79
Figure 19.17
The stepwise exchange of NH3 for H2O in M(H2O)42+.
The overall formation constant is given by
[M(NH3)42+]
Kf =
[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 M
Zn(H2O)42+ and 25.0 L of 0.15 M NH3. What is the final
[Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)42+ 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)42+.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq)
Kf =
Zn(NH3)42+(aq) + 4H2O(l)
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
19-82
Sample Problem 19.13
[Zn(H2O)42+]initial = 50.0 L x 0.0020 M = 1.3x10–3M
50.0 L + 25.0 L
25.0 L x 0.15 M
= 5.0x10–2 M
[NH3]initial =
50.0 L + 25.0 L
4 mol of NH3 is needed per mol of Zn(H2O4)2+, so
[NH3]reacted ≈ 4(1.3x10–3 M) = 5.2x10–3 M and
[Zn(NH3)42+] ≈ 1.3x10–3 M
Concentration (M)
Initial
Change
Equilibrium
19-83
Zn(H2O)42+(aq) + 4NH3(aq)
1.3x10–3
~(−1.3x10–3)
x
Zn(NH3)42+(aq) + 4H2O(l)
5.0x10–2
0
-
~(−5.2x10–3)
~(+1.3x10–3)
-
4.5x10–2
1.3x10–3
-
Sample Problem 19.13
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
=
7.8x108
x = [Zn(H2O)42+] ≈ 4.1x10–7 M
19-84
≈
(1.3x10–3)
x(4.5x10–2)4
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 M 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:
(a) AgBr(s)
Ag+(aq) + Br– (aq)
Ksp = [Ag+][Br–] = 5.0x10–13
S = [AgBr]dissolved = [Ag+] = [Br–]
Ksp = [Ag+][Br–] = S2 = 5.0x10–13
S = 7.1x10–7 M
(b) Write the overall equation:
AgBr(s)
Ag+(aq) + Br–(aq)
Ag+(aq) + 2S2O32–(aq)
AgBr(s) + 2S2O32–(aq)
Koverall = Ksp x Kf =
19-86
Ag(S2O3)23–(aq)
Br–(aq) + Ag(S2O3)23–(aq)
[Br-][Ag(S2O3)23–]
[S2O32–]2
= (5.0x10–13)(4.7x1013) = 24
Sample Problem 19.14
Concentration (M)
Initial
Change
Equilibrium
AgBr(s) + 2S2O32–(aq)
-
1.0
–2S
1.0 – 2S
S2
Koverall =
= 24
2
(1.0 – 2S)
so
S = 4.9 M – 0.9S and 10.9S = 4.9 M
S = [Ag(S2O3)23–] = 0.45 M
19-87
Br–(aq) + Ag(S2O3)23– (aq)
0
+S
S
0
+S
S
S
= √24 = 4.9
1.0 – 2S
Figure 19.18 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.
19-88