Ionic Equilibria in Aqueous Systems

1
Chapter 19
2
Ionic Equilibria in Aqueous Systems
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 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.
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.
The addition of 1 mL of strong acid (left) or
strong base (right) changes the pH by several
units.
5
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-).
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
The addition of CH3COO- reduces the
dissociation of the acid.
7
Table 19.1 The Effect of Added Acetate Ion on
the Dissociation of Acetic Acid
CH3COOHinit 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
8
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.
The shift in equilibrium position absorbs the
change in H3O or OH-, and the pH changes
only slightly.
9
How a buffer works.
Figure 19.3
10
Relative Concentrations of Buffer Components
Since Ka is constant, the H3O of the solution
depends on the ratio of buffer component
concentrations.
11
Sample Problem 19.1
Calculating the Effect of Added H3O or OH- on
Buffer pH
Calculate the pH
12
Sample Problem 19.1
SOLUTION (a)
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
1.8x10-5 M
pH -log(1.8x10-5) 4.74
Checking the assumption
13
Sample Problem 19.1
Setting up a reaction table for the stoichiometry
Setting up a reaction table for the acid
dissociation, using new initial
14
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
pH -log(1.7x10-5) 4.77
15
Sample Problem 19.1
Setting up a reaction table for the stoichiometry
Setting up a reaction table for the acid
dissociation, using new initial
16
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
pH -log(2.0x10-5) 4.70
17
The Henderson-Hasselbalch Equation
18
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
Figure 19.4
The relation between buffer capacity and pH
change.
This graph shows the final pH values for four
different buffer solutions after the addition of
strong base.
20
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.
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.
21
Sample Problem 19.2
Using Molecular Scenes to Examine Buffers
(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)?
22
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.
23
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) 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.
24
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.

25
Sample Problem 19.3
Preparing a Buffer
SOLUTION
H3O 10-pH 10-10.00 1.0x10-10 M
26
Sample Problem 19.3
Preparing a Buffer
0.094 M
Amount (mol) of CO32- needed 1.5 L soln x
0.14 mol CO32-
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.
27
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.
28
Figure 19.5
Colors and approximate pH range of some common
acid-base indicators.
29
Figure 19.6
The color change of the indicator bromthymol blue.
pH lt 6.0
pH gt 7.5
pH 6.0-7.5
30
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.
31
Figure 19.7
Curve for a strong acidstrong 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.
32
Calculating the pH during a strong acidstrong
base titration
Initial pH H3O HAinit pH -logH3O
33
Calculating the pH during a strong acidstrong
base titration
pH at the equivalence point pH 7.00 for a
strong acid-strong base titration.
34
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 HClinit 0.1000 M and pH
-log(0.1000) 1.00
To calculate the pH after 20.00 mL of NaOH
solution has been added
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
35
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- in excess 5.000x10-3 4.000x10-3
1.000x10-3 mol OH-
pOH -log(0.01111) 1.95
pH 14.00 1.95 12.05
36
Figure 19.8
Curve for a weak acidstrong base titration.
The pH increases slowly beyond the equivalence
point.
The pH at the equivalent point is gt 7.00 due to
the reaction of the conjugate base with H2O.
The curve rises gradually in the buffer region.
The weak acid and its conjugate base are both
present in solution.
The initial pH is higher than for the strong acid
solution.
37
Calculating the pH during a weak acidstrong base
titration
38
Calculating the pH during a weak acidstrong base
titration
39
Sample Problem 19.4
Finding the pH During a Weak AcidStrong Base
Titration
(a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d)
50.00 mL
SOLUTION
1.1x10-3 M
pH -log(1.1x10-3) 2.96
40
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
4.3x10-6 M
pH -log(4.3x10-6) 5.37
41
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-.
0.05000 M
7.7x10-10
1.6x10-9 M
pH -log(1.6x10-9) 8.80
42
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
0.01111 M
pH -log(9.0x10-13) 12.05
43
Figure 19.9
Curve for a weak basestrong 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 lt 7.00 due to
the reaction of the conjugate acid with H2O.
44
Figure 19.10
Curve for the titration of a weak polyprotic acid.
45
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.
46
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.
47
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.
Qsp QcPbF2 Pb2F-2
48
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 MnpXz-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).
49
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
S2-(aq) H2O(l) ? HS-(aq) OH-(aq)
Ksp Mn2HS-OH-
50
Sample Problem 19.5
Writing Ion-Product Expressions
SOLUTION
Ksp Mg2CO32-
Ksp Fe2OH-2
Ksp Ca23PO43-2
51
Sample Problem 19.5
Ksp Ag2HS-OH-
52
Table 19.2 Solubility-Product Constants (Ksp) of
Selected Ionic Compounds at
25C
53
Sample Problem 19.6
Determining Ksp from Solubility
54
Sample Problem 19.6
SOLUTION
Ksp Pb2SO42-
Converting from g/mL to mol/L
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 Pb2SO42- (1.40x10-4)2
1.96x10-8
55
Sample Problem 19.6
Ksp Pb2F-2
Converting from g/L to mol/L
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 Pb2F-2 (2.6x10-3)(5.2x10-3)2
7.0x10-8
56
Sample Problem 19.7
Determining Solubility from Ksp
SOLUTION
Ksp Ca2OH-2 6.5x10-6
57
Sample Problem 19.7
Ksp Ca2OH-2 (S)(2S)2 4S3 6.5x10-6
1.2x10-2 M
58
Table 19.3 Relationship Between Ksp and
Solubility at 25C
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.
59
Figure 19.12
The effect of a common ion on solubility.
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.
60
Sample Problem 19.8
Calculating the Effect of a Common Ion on
Solubility
SOLUTION
Ksp Ca2OH-2
61
Sample Problem 19.8
Ca2init 0.10 M because Ca(NO3)2 is a soluble
salt, and dissociates completely in solution.
Ksp Ca2OH-2 6.5x10-6 (0.10)(2S)2
(0.10)(4S2)
4.0x10-3 M
62
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.
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.
63
Figure 19.13
Test for the presence of a carbonate.
When a carbonate mineral is treated with HCl,
bubbles of CO2 form.
64
Sample Problem 19.9
Predicting the Effect on Solubility of Adding
Strong Acid
(a) lead(II) bromide (b) copper(II)
hydroxide (c) iron(II) sulfide
SOLUTION
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.
65
Sample Problem 19.9
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.
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.
66
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.
67
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 gt Ksp, a precipitate will form until the
remaining solution is saturated.
If Qsp lt Ksp, no precipitate will form because
the solution is unsaturated.
68
Sample Problem 19.10
Predicting Whether a Precipitate Will Form
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).
69
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.
Amount (mol) of F- 0.060 M F- x 0.200 L 0.012
mol F-.
Qsp Ca2initF-2init (0.10)(0.040)2
1.6x10-4
Since Qsp gt Ksp, CaF2 will precipitate until Qsp
3.2x10-11.
70
Sample Problem 19.11
Using Molecular Scenes to Predict Whether a
Precipitate Will Form
(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.
71
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.
72
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 gt Ksp, so a
precipitate forms in this solution.
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.
73
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.
74
Sample Problem 19.12
Separating Ions by Selective Precipitation
75
Sample Problem 19.12
SOLUTION
5.6x10-5 M
This is the maximum OH- that will not
precipitate Mg2 ion.
Calculating the Cu2 remaining in solution with
this OH-
Since the initial Cu2 is 0.10 M, virtually all
the Cu2 ion is precipitated.
76
Chemical Connections
Formation of acidic precipitation.
Figure B19.1
Since pH affects the solubility of many slightly
soluble ionic compounds, acid rain has
far-reaching effects on many aspects of our
environment.
77
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.
78
Figure 19.16
The stepwise exchange of NH3 for H2O in M(H2O)42.
79
Table 19.4 Formation Constants (Kf) of Some
Complex Ions at 25C
80
Sample Problem 19.13
Calculating the Concentration of a Complex Ion
SOLUTION
81
Sample Problem 19.13
4 mol of NH3 is needed per mol of Zn(H2O4)2,
so NH3reacted 4(1.3x10-3 M) 5.2x10-3 M and
Zn(NH3)42 1.3x10-3 M
82
Sample Problem 19.13
x Zn(H2O)42 4.1x10-7 M
83
Sample Problem 19.14
Calculating the Effect of Complex-Ion Formation
on Solubility
84
Sample Problem 19.14
SOLUTION
S AgBrdissolved Ag Br-
Ksp AgBr- S2 5.0x10-13
S 7.1x10-7 M
(b) Write the overall equation
Koverall Ksp x Kf
(5.0x10-13)(4.7x1013)
24
85
Sample Problem 19.14
S 4.9 M 0.9S and 10.9S 4.9 M
S Ag(S2O3)23- 0.45 M
86
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.