Chapter 16 Aqueous Equilibria

1
Chapter 16 Aqueous Equilibria

• The intravenous (IV) solution in this bag might
  save a life. The concentration of each solute in
  the solution is carefully selected to keep the
  total solute concentration within an optimal
  range and to maintain the pH of the blood.
  Because even slight deviations in blood pH can be
  fatal, the first treatment administered to an
  injured person is usually an intravenous solution.

Single acid (base) ? Salts and mixed salts
(anions and cations). Central topic How pH
values can be affected and controlled by ions.
2
Assignment

• 16.5,16.14,16.26,16.31,16.38,
• 16.47,16.52,16.61,16.71,16.76.

3
Acidity/Basicity of Various Solutions
CuSO4
NaCl
KBr
Characteristic colors of 10 mM water solutions of
K2FeO4, KMnO4 and Fe2 (SO4)3. 1 Permanganate 2
Potassium ferrate 3 Ferric chloride. Why?
4
How to modify the pH value of a solution?
5
Indicators tell you how acidic or basic your
samples are by looking at it.
6
Salts in Water

• Ions as acids and bases
• The pH of a salt solution
• The pH of mixed solutions

7
Figure 16.1 Solutions of salts in water give
rise to acidic, neutral, and basic solutions, as
shown here by the color of the indicator
bromothymol blue (see Table 16.3). From left to
right are solutions of ammonium chloride, sodium
chloride, and sodium acetate. Throughout this
chapter, we use the Brønsted definitions of acids
and bases.
Salts that contain the conjugate acids of weak
bases produce acidic aqueous solutions.
All anions that are conjugate bases of weak
acids act as proton acceptors, giving basic
solutions.
8
Figure 16.2 A solution of titanium(III) sulfate
is so acidic that it can releaseH2S from some
sulfides.
Small, highly charged metal ions have strong
polarizing effect on water so that they act as a
proton donator, an acid.
9
Figure 16.3 In water, Al3? cations exist as
hydrated ions that can act as Brønsted acids.
Although, for clarity, only four water molecules
are shown here, metal cations typically have six
H2O molecules attached to them.
Small, highly charged metal ions exert the
greatest pull on the electrons. They look like a
proton donator, an acid.
10
(A cation cannot be a base)
11
However,

• the cations of Group 1 and 2 metals and those of
  charge 1 from other groups are such weak Lewis
  acids that they DO NOT act as acids when
  hydrated. These metal ions are too large or have
  too low a charge to have an appreciable
  polarizing effect on water molecules that
  surround them. As a result, the hydrating water
  molecules do not readily release their protons.

12
Questions

• Is Na an acid in water?
• Is Ca2 an acid in water?

No and No
13
Figure 16.4 The relative strengths of conjugate
acids and bases have a reciprocal relation. (a)
When a species has a high tendency to donate a
proton, the resulting conjugate base has a low
tendency to accept one. (b) When a species has a
low tendency to donate a proton, the resulting
conjugate base has a strong tendency to accept
one. In the insets, a solid blue color represents
the water molecules.
Ka10-12
Kb10-13
Ka0.1
Kb10-2
Strong Acid (Base) ? Very Weak Conjugate Base
(Acid)
Kb10-10 10-4
Ka10-4 10-10
most weak acids ? weak conjugate bases most
weak bases ? weak conjugate acids
Kb10-4 10-10
Ka10-10 10-4
14
Only very few anions are acids
(Very weak bases? Kb? 0 ? neutral)
All anions that are conjugate bases of weak
acids act as proton acceptors, giving basic
solutions.
15
Exercise

• Decide whether aqueous solutions of (a) Ba(NO2)2
  (b) CrCl3 (c) NH4NO3 are acidic, basic or
  neutral.

• Basic, for the NO2- is the conjugate base of a
  weak acid.
• (b) Acidic, for Cr3 is small and highly charged
  so that it can
• polarize water and release proton(s) from water.
• (c) Acidic, for the NH4 is the conjugate acid of
  a weak base
• (and NO3- is the conjugate base of a strong
  acid?neutral)

• (a) Ba(NO2)2 (b) CrCl3 (c) NH4NO3

16
Classroom Exercise

• Decide whether aqueous solutions of (a) Na2CO3
  (b) AlCl3 (c) KNO3 are acidic, basic or neutral.

• Basic for the CO32- is the conjugate base of a
  weak acid.
• (b) Acidic for Al3 is small and highly charged
  so that it can
• polarize water and release proton(s) from water.
• (c) Neutral for the K is large and charge 1 so
  that it cannot
• polarize water.

• (a) Na2CO3 (b) AlCl3 (c) KNO3

17
Figure 16.5 The initial (left) and equilibrium
(right) composition of a solution of a salt
composed of the cation HB? and the anion A?,
where HB? is a weak acid and A? is neutral.
(left) The hypothetical initial situation that we
imagine before deprotonation. (right) At
equilibrium, the acidic cation is partially
deprotonated, and the solution is acidic.
How to calculate the pH of an acidic solution
18
The pH of a Salt Solution
Acidic

1. The initial molarity of the acidic ion is the
   molarity of the initially completely protonated
   ion. We assume that the initial molarities of its
   conjugate base and H3O are 0.
2. Write the increase in molarity of H3O as x mol/L
   and use the reaction stoichiometry to write the
   corresponding changes for the acidic ion and its
   conjugate base. Ignore H3O from the
   autoprotolysis of water this approximation is
   valid if H3O is substantially (about 10 times)
   greater than 10-7. Check the validity of this
   approximation at the end of the calculation.
3. Write the equilibrium molarities of the species
   of x.
4. Express the acidity constant for the ion in terms
   of x and solve the equations for x assume that x
   is small, but check the validity of this
   approximation at the end of the calculation. If
   Ka is not available, obtain it from the value of
   Kb for the conjugate base by using KaKw/Kb.
   Because x mol/L is the H3O molarity, the pH of
   the solution is logx.

Basic
Follow the same procedure as above, except that
now proton transfer from water to the ion results
in the formation of OH- and the conjugate acid of
the ion. We therefore use Kb ad the equilibrium
table leads to a value of pOH.
An amphiprotic anion can act as both an acid and
a base pH (pKapKb)/2
19
Example

• Estimate the pH of 0.15 M NH4Cl (aq). NH3 Kb
  1.810-5 (Table 15.3)

20
Example

• Estimate the pH of 0.10 M methylammonium
  chloride, CH3NH3Cl (aq). CH3NH3 Kb3.610-4
  (Table 15.3) or Ka 2.810-11 (Table 16.1)

21
Classroom Example

• Estimate the pH of 0.10 M NH4NO3 (aq). NH3 Kb
  1.810-5 (Table 15.3)

22
Example

• Estimate the pH of 0.15 M calcium acetate,
  Ca(CH3CO2)2 (aq). CH3CO2- Ka 1.810-11 (Table
  15.3)

23
Example

• Estimate the pH of 0.1 M potassium heptate,
  KC6H5CO2 (aq). C6H5COOH Ka 6.510-5

24
Classroom Example

• Estimate the pH of 0.02 M potassium fluoride, KF
  (aq). HF Ka 3.510-4

25
Mixed Solution Example

• Estimate the pH of a solution that is 0.5 M HNO2
  and 0.1 M KNO2 (aq). HNO2 Ka 4.310-4 (Table
  15.3)

26
Mixed Solution Example

• Estimate the pH of a solution that is 0.3 M
  CH3NH2 and 0.146 M CH3NH3Cl(aq). CH3NH2 Kb
  3.610-4 (Table 15.3)

27
Classroom Example

• Estimate the pH of a solution that is 0.01 M HClO
  (aq) and 0.2 mM NaClO(aq). HClO Ka 3.010-8
  (Table 15.3)

28
Titrations
Titration A titrant adds H3O (or OH-) to and
analyte so that a neutral solution
(stoichiometric point) is obtained (the initial
pH of the analyte is thus found).

• Strong acid-strong base titrations
• Weak acid-strong base and strong acid-weak base
  titrations
• Titrating a polyprotic acid
• Indicators

29
Figure 16.6 The variation of pH during the
titration of a strong base (the analyte) with a
strong acid (the titrant). This curve is for
25.00 mL of 0.250 M NaOH(aq) titrated with 0.340
M HCl(aq). The stoichiometric point occurs at
pH ? 7 (point S ). The other points are discussed
in the text.
30
Figure 16.7 The variation of pH during a typical
titration of a strong acid (the analyte) with a
strong base (the titrant). The stoichiometric
point occurs at pH ? 7.
31
Figure 16.8 The composition of the solutions
initially (left) and at the stoichiometric point
(right) in the titration of a strong acid (the
analyte) with a strong base (the titrant).
Step 1 calculate the moles of H3O (if the
analyte is a strong acid) or OH- (if the analyte
is a strong base) in the original analyte
solution from its molarity and volume.
Step 2 calculate the moles of H3O (if the
titrant is a strong acid) or OH- (if the titrant
is a strong base) in the volume of titrant added.
Step 3 write the chemical equation for the
neutralization reaction and use reaction
stoichiometry to find the moles of H3O (or OH-
if the analyte is strong base) that remain in the
analyte
solution. Each mole of H3O ions reacts with 1
mol OH- ions therefore subtrsct the number of
of moles of H3O or OH- ions that have reacted
from the initial number of moles of H3O or OH-
ions.
Step 4 Determine the concentration of hydronium
or hydroxide. Step 5 Find the pH (pOH).
32
Calculating points on the pH curve for a strong
acid-strong base titration (before stoichiometric
point)

• Titrant 0.34 M HCl(aq) analyte 25 mL,
  0.25 M NaOH. After addition of 5.0 mL of titrant,
  pH?

• The initial pOH of the analyte
  -log(0.25)0.602?pH13.40.
• The amount of OH- is 0.025 L0.25 M 6.25
  mmol.

• The amount of H3O supplied by the titrant is
• 0.005 L0.34 M 1.70 mmol

3. After the reaction of all the hydronium ions
added, the amount of hydroxide ion remaining
is 6.25 1.70 4.55 mmol.
4. Because the total volume of the solution is
now 30 mL, the molarity of hydroxide ion is
4.55 mmol/30 mL 0.152 M
5. pOH-log(0.152)0.82? pH 13.18.
33
Example titration (before stoichiometric point)

• Titrant 0.34 M HCl(aq) analyte 25 mL,
  0.25 M NaOH. After addition of 10.0 mL of
  titrant, pH?

• The initial pOH of the analyte
  -log(0.25)0.602?pH13.40.
• The amount of OH- is 0.025 L0.25 M 6.25
  mmol.

• The amount of H3O supplied by the titrant is
• 0.01 L0.34 M 3.40 mmol

3. After the reaction of all the hydronium ions
added, the amount of hydroxide ion remaining
is 6.25 3.40 2.85 mmol.
4. Because the total volume of the solution is
now 35 mL, the molarity of hydroxide ion is
2.85 mmol/35 mL 0.0814 M
5. pOH-log(0.0814)1.09? pH 12.91.
34
Classroom Exercise

• Titrant 0.34 M HCl(aq) analyte 25 mL,
  0.25 M NaOH. After addition of 12.0 mL of
  titrant, pH?

• The initial pOH of the analyte
  -log(0.25)0.602?pH13.40.
• The amount of OH- is 0.025 L0.25 M 6.25
  mmol.

• The amount of H3O supplied by the titrant is
• 0.012 L0.34 M 4.08 mmol

3. After the reaction of all the hydronium ions
added, the amount of hydroxide ion remaining
is 6.25 4.08 2.17 mmol.
4. Because the total volume of the solution is
now 37 mL, the molarity of hydroxide ion is
2.17 mmol/37 mL 0.0586 M
5. pOH-log(0.0586)1.232? pH 12.77.
35
Stoichiometric Point

• Titrant 0.34 M HCl(aq) analyte 25 mL,
  0.25 M NaOH. How much titrant is needed to reach
  stoichiometric point?

• The initial pOH of the analyte
  -log(0.25)0.602?pH13.40.
• The amount of OH- is 0.025 L0.25 M 6.25
  mmol.

• The amount of H3O supplied by the titrant is
  assumed to be x at the stoichiometric point, then
• 6.25 mmol x340 mM ? x 18.38 mL

For strong acids-strong base titration, the pH is
7.0, but for other cases, the pH may be not 7.0
at the stoichiometric point!
36
Calculating points on the pH curve for a strong
acid-strong base titration (after stoichiometric
point)

• Titrant 0.34 M HCl(aq) analyte 25 mL,
  0.25 M NaOH. After addition of 19.4 mL of
  titrant, pH?

• The initial pOH of the analyte
  -log(0.25)0.602?pH13.40.
• The amount of OH- is 0.025 L0.25 M 6.25
  mmol.

• The amount of H3O supplied by the titrant is
• 0.0194 L0.34 M 6.6 mmol

3. After the reaction of all the hydroxide ions,
the amount of hydronium ion remaining is 6.60
6.25 0.35 mmol.
4. Because the total volume of the solution is
now 44.4 mL, the molarity of hydronium ion
is 0.35 mmol/44.4 mL 7.9 mM
5. pH-log(0.0079)2.10.
37
Classroom Exercise (pH after stoichiometric point)

• Titrant 0.34 M HCl(aq) analyte 25 mL,
  0.25 M NaOH. After addition of 25 mL of titrant,
  pH?

• The initial pOH of the analyte
  -log(0.25)0.602?pH13.40.
• The amount of OH- is 0.025 L0.25 M 6.25
  mmol.

• The amount of H3O supplied by the titrant is
• 0.025 L0.34 M 8.5 mmol

3. After the reaction of all the hydroxide ions,
the amount of hydronium ion remaining is 8.5
6.25 2.25 mmol.
4. Because the total volume of the solution is
now 50 mL, the molarity of hydronium ion is
2.25 mmol/50 mL 45 mM
5. pH-log(0.045)1.35.
38
Titration of NaOH by HCl
0.34 M HCl
0.25 M NaOH
pH
HCl
25 ml
13.40
0 ml
13.18
25 ml
5 ml
12.91
25 ml
10 ml
12.77
25 ml
12 ml
7.00
NaOH
25 ml
18.38 ml
2.10
25 ml
19.4 ml
1.35
25 ml
25 ml
39
(No Transcript)
40
Animation of Acid-Base Titration
http//science.csustan.edu/chem/titrate/Acid/titra
teZ.swf
Follow the instructions and try a few times with
different initial concentrations/volumes of acid
and base. Click the yellow button on the valve
to see the generation of the titration
curve. (The titration curve might grow rather
slow. Be patient.)
41
Weak acid-strong base and strong acid-weak base
titrations
Examples Chemical Engineering General Industry
or Lab Use (e.g. cleaning agents and surface
treatments with liquid cleaners, degreasers,
strippers, passivators, etchants, solutions and
additives for cleaning and surface preparation. )
Biochemistry (intracellular/extracellular
equilibrium, neuron activation) Medicine (IV,
pharmacology, toxicology) Printed Circuit Board
(PCB) Integrated Circuits Semiconductor
Manufacture Nanomaterials (nanoparticles) Photogra
phy
42
Weak acid-strong base titrations
HCOOHNaOH??NaHCO2H2O
Na HCO2-
HCO2- H2O ??HCOOH OH-
A strong base dominates a weak acid the solution
at the stoichiometric point is basic.
43
Strong acid-weak base titrations
HClO2NH4OH ?? NH4ClH2O
Cl- NH4
NH4 H2O ??NH3 H3O
A strong acid dominates a weak base the solution
at the stoichiometric point is acidic.
44
Figure 16.9 The composition of the solutions in
the course of the titration of a weak acid with a
strong base. From left to right at the start of
the titration, just before the stoichiometric
point, at the stoichiometric point, and well
after the stoichiometric point. At the
stoichiometric point, the OH? ions are mainly
those arising from proton transfer from H2O to A?.
(basic)
45
Calculating the pH during a titration of a weak
acid or a weak base
Step 1 calculate the moles of weak acid (or weak
base) in the original analyte solution from its
molarity and volume.
Step 2 calculate the moles of H3O (if the
analyte is a weak acid) or OH- (if the analyte is
a weak base) in the in the volume of titrant
added.
Step 3 write the chemical equation for the
neutralization reaction and use reaction
stoichiometry to find
Weak acid-strong base titration the moles of
conjugate base formed by the reaction of the
acid with the added base, and the moles of weak
acid remaining. Weak base-strong acid titration
the moles of conjugate acid formed by the
reaction of the base with the added acid, and
the moles of weak base remaining.
Step 4 find the molarity of the conjugate acid
and base in solution by dividing the moles of
each species by the total volume of the solution.
Step 5 for a weak acid, use the equilibrium
table to find the H3O concentration and in each
case, assume that the contribution of the
autoptolysis of water to the pH is insignificant
if the pH is less than 6 or larger than 8.
Step 6 find pH from the hydronium concentration.
For a weak base, use an equilibrium table to
find the OH- concentration and find pOH?pH.
46
Calculating the pH during a titration of a weak
acid
(at stoichiometric point)
1
2
3
HCOOHNaOH??NaHCO2H2O
(No HCOOH left)
4
Na HCO2-
5
HCO2- H2O ??HCOOH OH-
Basic!
6
weak base
pOH?pH
Kb
47
Calculating the pH during a titration of a weak
acid
(at stoichiometric point)
1
2
3
HA MOH?? MA H2O
(No HA left)
4
M A-
MA
MB
MA-
MA-
5
Basic!
A- H2O ??HA OH-
6
weak base
0 0
MA-
pOH?pH
Kb
x x
-x
x x
MA- - x
48
Example Weak Acid-Strong Base Titration

• Estimate the pH at the stoichiometric point
  of the titration of 25.00 mL of 0.1 M HCOOH (aq)
  with 0.15 M NaOH (aq).

Moles of HCOOH 0.0025 L 0.1 mol/L 2.5 mmol.
Because all HCOOH has reacted to form HCO2-, the
moles of HCO2- at the stoichiometric point is
also 2.5 mmol.
From the chemical equation, 1 mol OH- is needed
for 1 mol HCOOH
The moles of NaOH required is then 2.5 mmol/0.15
M 16.7 mL.
The total volume of the solution at the
stoichiometric point is 25 mL 16.7 mL 41.7
mL. The molarity of sodium formate is
2.5 mmol/41.7 mL 0.06 M.
Because the formate ion is a weak base, the
equilibrium to consider is
49
Because the formate ion is a weak base, the
equilibrium to consider is
From Table 15.3, Ka1.8 10-4 for formic acid?Kb
5.6 10-11.
Basic!
50
Figure 16.10 The pH curve for the titration of a
weak acid with a strong base. This curve is for
the titration of 25.00 mL of 0.100 M HCOOH(aq)
with 0.150 M NaOH(aq). The stoichiometric point
(S ) occurs on the basic side of pH ? 7 because
the anion HCO2? is a base.
51
Classroom Exercise Weak Acid-Strong Base
Titration

• Estimate the pH at the stoichiometric point
  of the titration of 25.00 mL of 0.01 M HClO (aq)
  with 0.02 M KOH (aq).

Moles of HClO 0.0025 L 0.01 mol/L 0.25 mmol.
Because all HClO has reacted to form ClO-, the
moles of ClO- at the stoichiometric point is also
0.25 mmol.
From the chemical equation, 1 mol OH- is needed
for 1 mol HClO
The moles of KOH required is then 0.25 mmol/0.02
M 12.5 mL.
The total volume of the solution at the
stoichiometric point is 25 mL 12.5 mL 37.5
mL. The molarity of KClO is 0.25
mmol/37.5 mL 6.67 mM.
Because the ClO- is a weak base, the equilibrium
to consider is
52
Because the ClO- is a weak base, the equilibrium
to consider is
From Table 15.3, Ka3.010-8 for HClO?Kb 3.33
10-7.
Basic!
53
Calculating the pH during a titration of a weak
acid
(before stoichiometric point)
1
2
3
HA MOH?? MA H2O
4
M A-
MA
MB
MA-
MA-
5
Acidic!
HA H2O ?? A- H3O
weak acid
6
MA-
MA-MB
0
pH -log(x)
Ka
x x
-x
MA- x x
MA-MB-x
54
Example Weak Acid-Strong Base Titration(before
stoichiometric point)

• Estimate the pH 5.0 mL of 0.15 M NaOH (aq)
  is added to 25.00 mL of 0.1 M HCOOH (aq).

Moles of HCOOH 0.0025 L 0.1 mol/L 2.5 mmol.
The amount of OH- in 5.0 mL of the titrant is
5.0 mL 0.15 M 0.75 mmol.
From the chemical equation,
0.75 mmol of OH- produces 0.75 mmol of HCO2- and
leaves 2.5 mmol 0.75 mmol 1.75 mmol of
HCOOH.
The total volume of the solution at this stage is
25 mL 5 mL 30 mL, so the molarities of acid
and conjugate base are HCOOH 1.75
mmol/30 mL 0.0583 M. HCO2- 0.75
mmol/30 mL 0.025 M
55
The proton transfer equilibrium in water is
From Table 15.3, Ka1.8 10-4 for formic acid.
Acidic, of cause (before stoichiometric point)
56
Example Weak Acid-Strong Base Titration(before
stoichiometric point)

• Estimate the pH 10.0 mL of 0.15 M NaOH (aq)
  is added to 25.00 mL of 0.1 M HCOOH (aq).

Moles of HCOOH 0.0025 L 0.1 mol/L 2.5 mmol.
The amount of OH- in 10.0 mL of the titrant is
10.0 mL 0.15 M 1.5 mmol.
From the chemical equation,
1.5 mmol of OH- produces 1.5 mmol of HCO2- and
leaves 2.5 mmol 1.5 mmol 1.0 mmol of HCOOH.
The total volume of the solution at this stage is
25 mL 10 mL 35 mL, so the molarities of
acid and conjugate base are HCOOH
1.0 mmol/35 mL 0.0286 M. HCO2- 1.5
mmol/35 mL 0.0428 M
57
The proton transfer equilibrium in water is
From Table 15.3, Ka1.8 10-4 for formic acid.
Less acidic, of cause (more NaOH has been added)
58
Classroom Exercise Weak Acid-Strong Base
Titration(before stoichiometric point))

• Estimate the pH 15.0 mL of 0.15 M NaOH (aq)
  is added to 25.00 mL of 0.1 M HCOOH (aq).

Moles of HCOOH 0.0025 L 0.1 mol/L 2.5 mmol.
The amount of OH- in 15.0 mL of the titrant is
15.0 mL 0.15 M 2.15 mmol.
From the chemical equation,
2.25 mmol of OH- produces 2.25 mmol of HCO2- and
leaves 2.5 mmol 2.25 mmol 0.25 mmol of
HCOOH.
The total volume of the solution at this stage is
25 mL 15 mL 40 mL, so the molarities of
acid and conjugate base are HCOOH
0.25 mmol/40 mL 0.00625 M. HCO2-
2.25 mmol/40 mL 0.0563 M
59
The proton transfer equilibrium in water is
From Table 15.3, Ka1.8 10-4 for formic acid.
Acidic, of cause (before stoichiometric point)
60
Figure 16.12 The pKa of an acid can be
determined by carrying out a titration of the
weak acid with a strong base (or vice versa) and
locating the pH of the solution after the
addition of half the volume of titrant needed to
reach the stoichiometric point.
61
Calculating the pH during a titration of a weak
acid
(after stoichiometric point)
1
2
3
HA MOH?? MA H2O
(No HA left)
4
M A-
MA
MB
MA-
MA-
5
Basic!
A- H2O ??HA OH-
6
Kb
weak base
0 MB - MA-
MA-
pOH?pH
x x
-x
MA- - x
x xMB - MA-
62
Weak-acid-strong-base titration curve
HPr Propionic Acid
63
The Four Major Differences Between a Strong
Acid-Strong Base Titration Curve and a Weak
Acid-Strong Base Titration Curve

1. The initial pH is higher.
2. A gradually rising portion of the curve, called
   the buffer region, appears before the steep rise
   to the equivalence point.
3. The pH at the equivalence point is greater than
   7.00.
4. The steep rise interval is less pronounced.

64
Strong acid-weak base titrations
HClO2NH4OH ?? NH4ClH2O
Cl- NH4
NH4 H2O ??NH3 H3O
A strong acid dominates a weak base the solution
at the stoichiometric point is acidic.
65
Calculating the pH during a titration of a weak
base
(at stoichiometric point)
1
2
3
B HA ?? BHA
4
BH A-
5
weak acid
BH H2O ?? B H3O
Acidic!
6
Ka
pH
66
Example Strong Acid-Weak Base Titration

• Estimate the pH at the stoichiometric point
  of the titration of 25.00 mL of 0.02 M NH3 (aq)
  with 0.015 M HCl (aq). NH4 Ka 5.610-10.

Moles of NH3 0.025 L 0.02 mol/L 0.5 mmol.
Because all NH3 has reacted to form NH4, the
moles of NH4 at the stoichiometric point is also
0.5 mmol.
From the chemical equation, 1 mol H is needed
for 1 mol NH3
The moles of HCl required is then 0.5 mmol/0.015
M 33.33 mL.
The total volume of the solution at the
stoichiometric point is 25 mL 33.33 mL 58.33
mL. The molarity of NH4Cl is 0.5
mmol/58.33 mL 8.57 mM.
Because the NH4 is a weak acid, the equilibrium
to consider is
67
Because the NH4 is a weak acid, the equilibrium
to consider is
Ka5.6 10-10 for NH4.
Acidic!
68
Figure 16.11 A typical pH curve for the titration
of a weak base with a strong acid. The
stoichiometric point (S) occurs on the acidic
side of pH ? 7 because the salt formed by the
neutralization reaction has an acidic cation.
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The four Major Differences Between a Weak
Acid-Strong Base Titration Curve and a Weak
Base-Strong Acid Titration Curve

1. The initial pH is above 7.00.
2. A gradually decreasing portion of the curve,
   called the buffer region, appears before a steep
   fall to the equivalence point.
3. The pH at the equivalence point is less than
   7.00.
4. Thereafter, the pH decreases slowly as excess
   strong acid is added.

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Titrating a Polyprotic Acid
A-
More than one stoichiometric point can be
detected.
72
Titration of H3PO4 (aq)
1 mol
1 mol
1 mol
1 mol
1 mol
1 mol
73
Figure 16.13 The variation of the pH of the
analyte solution during the titration of a
triprotic acid (phosphoric acid) and the major
species present in solution. SP1, SP2, and SP3
indicate the stoichiometric points. Points A, C
and E are the points at which pH? pKa for each
deprotonation. Points B, D, and F are discussed
in the text.
74
Example

• What volume of 0.02 M NaOH (aq) is required
  to reach the (a) first (b) second stoichiometric
  point in a titration of 30 mL of 0.01 M H3PO4
  (aq)?

1 mol
1 mol
30 mL 0.01 M 0.3 mmol
x mL 0.02 M 0.03 mmol
x 15 mL
1 mol
1 mol
0.3 mmol
y mL 0.02 M 0.3 mmol
xy 30 mL
y 15 mL
1 mol
1 mol
z mL 0.02 M 0.3 mmol
xyz 45 mL
0.3 mmol
z 15 mL
75
Classroom Exercise

• What volume of 0.01 M NaOH (aq) is required
  to reach the (a) first (b) second stoichiometric
  point in a titration of 25 mL of 0.01 M H2SO3
  (aq)?

1 mol
1 mol
25 mL 0.01 M 0.025 mmol
x mL 0.01 M 0.025 mmol
x 25 mL
1 mol
1 mol
0.025 mmol
y mL 0.01 M 0.025 mmol
xy 50 mL
y 25 mL
76
Figure 16.14 A commercially available automatic
titrator. The stoichiometric point of the
titration is detected by a sudden change in pH
the pH is monitored electrically, using a
technique described in Investigating Chemistry
18.1. The pH can be plotted as the reaction
proceeds, as shown on the left.
77
Acid-Base Indicators

• An acid-base indictor is a weak acid that has one
  color in its acid form (HIn) and another color in
  its conjugate base form (In-). (The light
  absorption characteristics of HIn are different
  from those of In-.)

In-
In
H
The end point of a titration is defined as the
point at which the concentrations of the acid
and base forms of the indicator are equal HIn
In-. At this point, the color of the
indicator is halfway between the colors of its
acid and base forms.
78
Figure 16.15 The stoichiometric point of an
acid-base titration may be detected by the color
change of an indicator. Here we see the colors of
solutions containing a few drops of
phenolphthalein (?? ) at (from left to right) pHs
of 7.0, 8.5, 9.4 (its end point), 9.8, and 12.0.
At the end point, equal amounts of the conjugate
acid and base forms of the indicator are present.
End point
79
Figure 16.16 The same dye is responsible for the
red of poppies (a) and the blue of
cornflowers(???) (b). The color difference is a
consequence of the more acidic sap of poppies.
Natural Indicators
80
Figure 16.17 Ideally, an indicator should have a
sharp color change close to the stoichiometric
point of the titration, which is at pH ? 7 for a
strong acid-strong base titration. However, the
change in pH is so abrupt that phenolphthalein
can also be used. The color change of methyl
orange, however, would give a less accurate
result.
81
Figure 16.18 Phenolphthalein can be used to
detect the stoichiometric point of a weak
acid-strong base titration, but methyl orange
would give a very inaccurate indication of the
stoichiometric point. The pH curves are
superimposed on approximations to the colors of
the indicators in the neighborhoods of their end
points.
82
Figure 16.19 Methyl orange can be used for a weak
base-strong acid titration. Phenolphthalein would
be inappropriate because its color change occurs
well away from the stoichiometric point. The pH
curves are superimposed on approximations to the
colors of the indicators in the neighborhoods of
their end points.
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A Question

• Q How many methods can you list to evaluate the
  pH of a solution?

A At least three (1) pH meter, (2) automatic
indicator, (3) acid-base indictors.
Other possible methods Absorption
spectroscopy, NMR, electrochemical method etc.
89
Buffer Solutions
Solutions that resist changes in pH when small
amounts of strong acid or base are added.

• The action of buffers
• Selecting a buffer
• Buffer capacity

90
The Action of Buffers

• An acid buffer solution consists of a weak
  acid and its conjugate base HA/A- it has
  pHlt7.0.
• An base buffer solution consists of a weak
  base and its conjugate acid BH/B it has pHgt7.0.

Blood and other cell fluids buffered at pH
7.4. Oceans are buffered at pH8.4.
91
Acid Buffer Action
pH change very small.
HA ?H3O ? CH3COOH
BOH ?OH- ? consumes H3O?CH3COOH ? CH3CO2-
pH change very small.
Acid buffer action the weak acid transfers
protons to the OH- ions from added strong base.
The conjugate base of the weak acid accepts
protons from the H3O ions supplied by a strong
acid.
92
Base Buffer Action
HA ?H3O ? promotes NH3 to protonate and consumed
by OH-
pH change very small.
BOH ?OH- ?stimulates NH3 to deprotonate ? NH4
pH change very small.
Base buffer action the conjugate acid of the
weak base transfers protons to the OH- ions from
added strong base. The weak base accepts protons
from the H3O ions supplied by a strong acid.
93
Figure 16.20 Buffer action depends on the
donation of protons by the weak acid molecules,
HA, when a strong base is added and the
acceptance of protons by the conjugate base ions,
A?, when a strong acid is added. In the inset, a
solid blue color represents water.
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Design a Buffer with Specified pH
This is a problem of mixed solutions (both an
acid and the salt of its conjugate base are
contained).
Acid buffer
96
Example

• Calculate the pH of a buffer solution that
  is 0.04 M NaCH3CO2 (aq) and 0.08 M CH3COOH (aq)
  at 25 oC.
• CH3COOH pKa 4.75 (Table 15.3)

97
Design a Buffer with Specified pH
Base buffer
98
Example

• Calculate the pH of a buffer solution that
  is 0.04 M NH4Cl (aq) and 0.03 M NH3 (aq) at 25
  oC.
• NH3 pKb 4.75 (Table 15.3)

99
Classroom Exercise

• Calculate the pH of a buffer solution that
  is 0.15 M HNO2 (aq) and 0.2 M NaNO2 (aq) at 25
  oC.
• HNO2 pKa 3.37 (Table 15.3)

100
Calculating the pH change when acid or base added
to a buffer

• 1.2 g (0.03 mol) of NaOH is dissolved in 500
  mL of a buffer solution that is 0.04 M NaCH3CO2
  (aq) and 0.08 M CH3COOH (aq) with pH4.45.
  Calculate the pH of the resulting solution and
  the change in pH. Assume volume is not changed.

101
Figure 16.21 A buffer solution contains a sink
for protons (a weak base) that are supplied when
a strong acid is added and a source of protons (a
weak acid) to supply to a strong base that is
added. The joint action of the source and the
sink keeps the pH constant when a small amount of
either strong acid or strong base is added.
102
Figure 16.22 When conjugate acid and base are
present at similar concentrations, the pH changes
very little as more strong base (or strong acid)
is added. As the inset shows, the pH lies between
pKa ? 1 for a wide range of concentrations.
103
Solubility Equilibria

• The solubility product
• The common-ion effect
• Predicting precipitation
• Selective precipitation
• Dissolving precipitates
• Complex ions and solubilities
• Qualitative analysis

104
Figure 16.23 A pollution control officer
collects samples of river water in England. The
water will be tested for the presence of heavy
metal ions by adding a solution containing anions
such as sulfide ions. If a precipitate appears,
further tests will be conducted to identify the
particular ions present.
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Case Study 16 (a) Patients undergoing surgery
often need to be supplied with several kinds of
intravenous solutions. These doctors are
operating on a patients liver, a procedure that
requires the administration of three different
intravenous solutions and a blood transfusion.
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Case Study 16 (b) Three solutions commonly given
intravenously lactated Ringers solution, 0.9
sodium chloride, and 5 dextrose (glucose). The
first two help to control electrolyte levels, the
third maintains the blood sugar level, and all
three help to maintain blood volume.
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Common Ion Effect
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Figure 16.24 If the concentration of one of the
ions in solution is increased, then the
concentration of the other is decreased to
maintain a constant value of Ksp. (a) The cations
(pink) and anions (green) in solution. (b) When
more anions are added (together with their
accompanying spectator ions, which are not
shown), the concentration of cations decreases.
In other words, the solubility of the original
compound is reduced by the presence of a common
ion. In the insets, a solid blue color represents
the water molecules.
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Figure 16.25 (a) A saturated solution of zinc
acetate in water. (b) When acetate ions are added
(as solid sodium acetate in the spatula shown in
(a)), the solubility of the zinc acetate is
significantly reduced, and additional zinc
acetate precipitates.
112
Figure 16.26 The solution in the middle is at
equilibrium. The solution on the left is
supersaturated, so it forms a precipitate until
solid and solute are in equilibrium. In the
solution on the right, solid has been added to
pure water. As the solid begins to dissolve, the
solute concentration is lower than the
equilibrium concentration. In this case, the
solid dissolves until the equilibrium
concentration is reached.
113
Predicting Precipitation
114
Figure 16.27 When a lead(II) nitrate solution is
added to a solution of potassium iodide, yellow
lead(II) iodide immediately precipitates.
115
Selective Precipitation
116
Dissolving Precipitates
117
Complex Ions and Solubilities
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Qualitative Analysis
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Figure 16.28 The sequence for the analysis of
cations by selective precipitation. (a) The
original solution contains Pb2?, Hg22?, Ag?,
Cu2?, and Zn2? ions. Addition of HCl(aq)
precipitates AgCl, Hg2Cl2, and PbCl2, which can
be removed by decanting and filtration. (b)
Addition of H2S to the remaining solution
precipitates CuS, which can also be removed. (c)
Making the resulting solution basic by adding
ammonia precipitates ZnS.
122
Figure 16.29 When an aqueous solution of ammonia
is added to a silver chloride precipitate, the
precipitate dissolves. However, when ammonia is
added to a precipitate of mercury(I) chloride,
mercury metal and mercury(II) ions are formed by
disproportionation and the mass turns gray. Left
to right silver chloride in water, silver
chloride in aqueous ammonia, mercury(I) chloride
in water, mercury(I) chloride in aqueous ammonia.
The different responses of AgCl and Hg2Cl2 to the
addition of ammonia allow them to be
distinguished in a mixture.
123
Assignment for Chapter 16

• 16.5,16.14,16.26,16.31,16.38,
• 16.47,16.52,16.61,16.71,16.76.