Applications of Aqueous Equilibria

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Chapter 8

• Applications of Aqueous Equilibria

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Chapter 8 Applications of Aqueous Equilibria
8.1 Solutions of Acids or Bases Containing a
Common Ion 8.2 Buffered Solutions 8.3 Exact
Treatment of Buffered Solutions 8.4 Buffer
Capacity 8.5 Titrations and pH Curves 8.6
Acid-Base Indicators 8.7 Titration of Polyprotic
Acids 8.8 Solubility Equilibria and The
Solubility Product 8.9 Precipitation and
Qualitative Analysis 8.10 Complex Ion Equilibria
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A base swirling in a solution containing
phenolphthalein
4
Le Châteliers principle for the dissociation
equilibrium for HF
HF(aq) H(aq)
F-(aq)
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Molecular model F-, Na, HF, H2O
6
Like Example 8.1 (P 278-9) - I
Nitrous acid, a very weak acid, is only 2.0
ionized in a 0.12 M solution. Calculate the
H, the pH, and the percent dissociation of
HNO2 in a 1.0 M solution that is also 1.0 M in
NaNO2!
HNO2(aq) H(aq) NO2-(aq)
Initial Concentration (mol/L) Equilibrium
Concentration (mol/L)
HNO20 1.0 M
HNO2 1.0 x (from dissolved
HNO2) NO2-0 1.0 M
NO2- 1.0 x (from
dissolved NaNO2) H0 0
H
x (neglect the contribution from water)
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Like Example 8.1 (P 274-5) - II
H NO2- HNO2
( x ) ( 1.0 x ) (1.0 x )
Ka
4.0 x 10-4
Assume x is small as compared to 1.0
X (1.0) (1.0)
H
4.0 x 10-4 or x 4.0 x 10-4
Therefore pH - log H - log ( 4.0 x 10-4 )
_________
The percent dissociation is
Nitrous acid Nitrous acid
alone NaNO2
H 2.0 x 10-2 4.0 x 10-4 pH
1.70 _______ Diss
2.0 _______
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Example 8.2 (P279-82) - I
A buffered solution contains 0.50 M acetic acid
(HC2H3O2, Ka 1.8 x 10-5) and o.50 M sodium
acetate (NaC2H3O2). Calculate the pH of this
solution, and the pH when 0.010 M of solid NaOH
is added to 1.0 L of this buffer and to pure
water.
HC2H3O2 (aq) H(aq)
C2H3O2 (aq)
H C2H3O2- HC2H3O2
Ka 1.8 x 10-5
Initial Concentration (mol/L) Equilibrium
Concentration (mol/L)
HC2H3O20 0.50
HC2H3O2 0.50 x C2H3O2-0
0.50
C2H3O2- 0.50 x H0 0
H
x
X mol/L of HC2H3O2 dissociates to reach
equilibrium

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Example 8.2 (P279-82) - II
HC2H3O2- HC2H3O2
( x ) ( 0.50 x) 0.50 - x
(x) (0.50) 0.50

Ka 1.8 x 10-5

x 1.8 x 10-5
The approximation by the 5 rule is fine
H x 1.8 x 10-5 M and pH 4.74
To calculate the pH and concentrations after
adding the base
OH- HC2H3O2 H2O C2H3O2-
Before reaction 0.010 mol 0.50 mol
- 0.50 mol After reaction
0.010 0.010 0.50 0.10 -
0.50 0.10 0 mol
0.49 mol 0.51 mol
Note that 0.01 mol of acetic acid has been
converted to acetate ion by the addition of the
base.
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Example 8.2 (P279-82) - III
Initial Concentration (mol/L) Equilibrium
concentration (mol/L)
X mol/L of HC2H3O2 Dissociates to reach
equilibrium
HC2H3O20 0.49
HC2H3O2 0.49 x
C2H3O2-0 0.51
C2H3O2- 0.51 x
H0 0
H x
HC2H3O2- HC2H3O2
(x)(0.51 x) 0.49 - x
(x)(0.51) 0.49
Ka 1.8 x 10-5

x ______________ and pH ___________
If the base is added to pure water without the
buffer being present we get an entirely different
solution
If the 0.01 mol of NaOH is added to 1.0 L of pure
water the Concentration of hydroxide ion is 0.01
M.
Kw OH-
1.0 x 10-14 1.0 x 10-2
H
__________ and the pH ______
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When a strong acid or base is added to a buffered
solution, it is best to deal with the
stoichiometry of the resulting reaction first.
After the stoichiometric calculations are
completed, then consider the equilibrium
calculations. This procedure can be represented
as follows
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OH- ions are not allowed to accumulate but are
replaced by A- ions.
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When the OH- is added, the concentrations of HA
and A- change, but only by small amounts. Under
these conditions the HA/A- ratio and thus the
H stay virtually constant.
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How Does a Buffer Work
Lets add a strong base to a weak acid and see
what happens
OH- HA A- H2O
Final pH of buffer close to original-
Original buffer pH
Added OH- ions Replaced by A- ions
H A- HA
HA A-
Ka
H Ka
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The Effect of Added Acetate Ion on the
Dissociation
of Acetic Acid
CH3COOH CH3COO-added
Dissociation pH
0.10 0.00
1.3 2.89 0.10
0.050
0.036 4.44 0.10
0.10
0.018 4.74 0.10
0.15 0.012
4.92
CH3COOHdissoc
Dissociation x
100
CH3COOHinit
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Human blood is a buffered solution
CO2 (g), H2CO3 (aq) and HCO3-(aq) are the
buffering components in Blood that hold the pH
to a range that will allow Hemoglobin to
transport oxygen from the lungs to the cells of
the body for metabolism.
Source Visuals Unlimited
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Pure water at pH 7.00
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Molecular model HC2H3O2, C2H3O2-
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How a Buffer WorksI
A buffer consists of a solution that contains
high concentrations of the acidic and basic
components. This is normally a weak acid and the
anion of that weak acid, or a weak base and the
corresponding cation of the weak base. When
small quantities of H3O or OH- are added to the
buffer, they cause a small amount of one buffer
component to convert into the other. As long as
the amounts of H3O and OH- are small as compared
to the concentrations of the acid and base in the
buffer, the added ions will have little effect
on the pH since they are consumed by the buffer
components.
Consider a buffer made from acetic acid and
sodium acetate
CH3COOH(aq) H2O(l)
CH3COO-(aq) H3O(aq)
CH3COO- H3O
Ka
or H3O Ka x
CH3COOH
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How a Buffer WorksII
Lets consider a buffer made by placing 0.25 mol
of acetic acid and 0.25 mol of sodium acetate
per liter of solution. What is the pH of the
buffer? And what will be the pH of 100.00 mL of
the buffer before and after 1.00 mL of
concentrated HCl (12.0 M) is added to the buffer?
What will be the pH of 300.00 mL of pure water if
the same acid is added?
(0.25)
H3O Ka x 1.8
x 10-5 x 1.8 x 10-5
(0.25)
pH -logH3O -log(1.8 x 10-5) pH
_____ Before acid added!
1.00 mL conc. HCl 1.00 mL x 12.0 mol/L
0.012 mol H3O
Added to 300.00 mL of water
pH -log(0.0399 M) pH _____ Without buffer!

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How a Buffer WorksIII
After acid is added
Conc. (M) CH3COOH(aq) H2O(aq)
CH3COO- H3O
Initial 0.250
---- 0.250
0 Change 0.012
---- -0.012
0.012 Equilibrium 0.262
---- 0.238 0.012
Solving for the quantity ionized
Conc. (M) CH3COOH(aq) H2O(aq)
CH3COO- H3O
Initial 0.262
---- 0.238
0 Change -x
---- x
x Equilibrium 0.262 - x
---- 0.238 x x
Assuming 0.262 - x 0.262 0.238 x
0.238
(0.262)
H3O Ka x 1.8 x
10-5 x 1.982 x 10-5
(0.238)
pH -log(1.982 x 10-5) 5.000 - 0.297
_____After the acid is added!
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How a Buffer WorksIV
Suppose we add 1.0 mL of a concentrated base
instead of an acid. Add 1.0 mL of 12.0 M NaOH to
pure water and our buffer, and lets see what
the impact is 1.00 mL x 12.0 mol OH-/1000mL
0.012 mol OH- This will reduce the quantity
of acid present and force the equilibrium to
produce more hydronium ion to replace that
neutralized by the addition of the base!
Conc. (M) CH3COOH(aq) H2O(aq)
CH3COO- H3O
Initial 0.250
---- 0.250
0 Change - 0.012
---- 0.012
0.012 Equilibrium 0.238
---- 0.262
0.012
Assuming Again, using x as the quantity of acid
dissociated we get our normal
assumptions 0.262 x 0.262 0.238 - x
0.238
0.238
H3O 1.8 x 10-5 x 1.635
x 10-5
0.262
pH -log(1.635 x 10-5) 5.000 - 0.214 _____
After base is added!
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How a Buffer WorksV
By adding the 1.00mL base to 300.00 mL of pure
water we would get a hydroxide ion concentration
of
The hydrogen ion concentration is
This calculates out to give a pH of
pH -log(2.5 6 x 10-10) 10.000 - 0.408 9.59
With 1.0 mL of the

base in pure water!
In summary Buffer
alone pH 4.74 Buffer plus 1.0 mL base
pH 4.79 Base alone pH 9.59
Buffer plus 1.0 mL acid pH 4.70 Acid
alone pH 1.40
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The Relation Between Buffer Capacity and pH
Change
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A digital pH meter shows the pH of the buffered
solution to be 4.74
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When 0.01 mol NaOH is added to 1.0 L of pure
water, the pH jumps to 12.00
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Molecular model Cl-, NH4
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Preparing a Buffer
Problem The ammonia-ammonium ion buffer has a pH
of about 9.2 and can be used to keep solutions
in the basic pH range. What mass of ammonium
chloride must be added to 400.00 mL of a 3.00 M
ammonia solution to prepare a buffer ? Plan The
conjugate pair is the ammonia-ammonium ion pair
which has an equilibrium constant Kb 1.8 x 10
-5. The reaction equation with water can be
written along with the Kb expression, since we
want to add sufficient ammonium ion to equal the
aqueous ammonia concentration. Solution The
reaction for the ammonia-ammonium ion buffer is
NH4Cl 53.49 g/mol
NH4 OH-
Therefore mass NH4Cl 1.20 mol x
53.49g/mol mass ________ g NH4Cl
Kb 1.8 x 10-5
NH3
NH4 3.00 mol x 0.400 L 1.20 mol
L
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The Henderson-Hasselbalch Equation
Take the equilibrium ionization of a weak acid
HA(aq) H2O(aq) H3O(aq) A-(aq)
H3O A-
Ka
HA
Solving for the hydronium ion concentration gives
Taking the negative logarithm of both sides
Generalizing for any conjugate acid-base pair
( )
Henderson-Hasselbalch equation
pH log Ka log
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Like Example 8.3 (P 285-7) -I
Problem Instructions for making a buffer say to
mix 60.0 ml of 0.100 M NH3 with 40.0 ml of 0.100
M NH4Cl. What is the pH of this buffer?
The combined volume is 60.0 ml 40.0 ml 100.0
ml
Moles of Ammonia VolNH3 x MNH3 0.060 L x
0.100 M 0.0060 mol Moles of Ammonium ion
VolNH4Cl x MNH4Cl 0.040 L x 0.100 M

0.0040 mol NH3
0.060 M NH4
0.040 M
0.0060 mol 0.100 L
0.0040 mol 0.100 L
Concentration (M) NH3 (aq) H2O(l)
NH4(aq) OH-(aq)
Starting 0.060
0.040 0 Change
-x
x x Equilibrium 0.060
x 0.040 x x
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Like Example 8.3 (P 285-7) - II
Substituting into the equation for Kb
NH4 OH- NH3
(0.040 x) (x) (0.060 x)
Kb 1.8 x 10-5


Assume 0.060 x 0.060 0.040 x 0.040
0.040 (x) 0.060
Kb 1.8 x 10-5 x
2.7 x 10-5
Check assumptions 0.040 0.000027 0.040
or 0.068
0.060 0.000027 0.060 or 0.045
OH- 2.7 x 10-5 pOH - logOH- - log
(2.7 x 10-5) 5 0.43

pOH 4.57
pH 14.00 pOH 14.00 4.57 ___________
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pH Box
H3O 10-pH
pH H3O
pH -logH3O
Kw 1 x 10-14 _at_ 25oC H3OOH- 1 x
10-14
pH pOH 14 _at_ 25oC
OH- 10-pOH
pOH OH-
pOH -logOH-
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Summary Characteristics of Buffered Solutions
Buffered solutions contain relatively large
concentrations of a weak acid and its
corresponding weak base. They can involve a weak
acid HA and the conjugate base A- or a weak base
B and the conjugate acid BH. When H is added
to a buffered solution, it reacts essentially to
completion with the weak base present
H A- HA or
H B BH When OH- is added to a
buffered solution, it reacts essentially to
completion with the weak acid present.
OH- HA A- H2O or OH- BH
B H2O The pH of the buffered
solution is determined by the ratio of the
concentrations of the weak base and weak acid.
As long as this ratio remains virtually constant,
the pH will remain virtually constant. This will
be the case as long as the concentrations of the
buffering materials (HA and A- or G and BH) are
large compared with the amounts of H or OH-
added.
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Exact Treatment of Buffer Solutions
We can use several relationships to calculate the
exact solution to buffered solution problems
Charge balance equation Na
H A- OH- Material Balance equation
A-0 HA0 HA A-
Since A-0 Na and Kw OH-H , we can
rewrite the charge balance equation, and solve
for A-
H2 Kw H
From the mass balance equation solved for HA we
get
HA A-0 HA0 A-
Substituting the expression for A-, and
substituting into the Ka expression for HA we
obtain
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Example 8.4 (P 289-90) - I
Calculate the pH of a buffered solution
containing 3.0 x 10-4 M HOCl (Ka 3.5 x 10-8)
and 1.0 x 10-4 M NaOCl.
Let x H then OCl- 1.0 x 10-4 x HOCl
3.0 x 10-4 - x
H OCl- HOCl
Ka 3.5 x 10-8
H OCl- HOCl
(x)(1.0 x 10-4 x) (3.0 x 10-4 x )
3.5 x 10-8
Assuming x is small compared to 1.0 x 10-4 and
solving for x we have
1.05 x 10-11 1.0 x 10-4
Since this is close to that of water we must use
the equation that uses water, and takes its
ionization into account.
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Example 8.4 (P 289-90) - II
H2 1.0 x 10-14
HOCl-0

H
Ka 3.5 x 10-8
Where
OCl-0 1.0 x 10-4 M HOCl0 3.0 x 10-4 M
We expect H to be close to 1.0 x 10-7, so
H2 to be about 1.0 x 10-14
H2 1.0 x 10-14
HOCl0 1.0 x 10-4 M gtgtgt
H
The expression becomes
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Example 8.4 (P 289-90) - III
H 1.05 x 10-7 M 1.1 x 10-7 M
Using this result, we can check the magnitude of
the neglected term
(1.05 x 10-7)2 1.0 x 10-14
H2 1.0 x 10-14

9.8 x 10-9
1.05 x 10-7
H
This result suggests that the approximation was
fine!
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pH and Capacity of Buffered Solutions
The pH of a buffered solution is determined by
the ratio A-/HA. The capacity of a
buffered solution is determined by the
magnitudes of HA and A-
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Example 8.5 (P 290-2) - I
Calculate the change in pH that occurs when 0.010
mol of gaseous HCl is added to 1.0 L of each of
the following solutions Solution A 5.00 M
HC2H3O2 and 5.00 M NaC2H3O2 Solution B 0.050
M HC2H3O2 and 0.0500 M NaC2H3O2 For Acetic acid,
Ka 1.8 x 10-5
Use the Henderson-Hasselbalch equation for
initial pH
C2H3O2- H C2H3O2
Since C2H3O2- H C2H3O2 The equation
becomes
pH pKa log
pH pKa log (1) pKa -log(1.8 x 10-5) 4.74
Adding 0.010 mol of HCl will cause a shift in the
equilibrium due to
H(aq) C2H3O2-(aq) H C2H3O2
(aq)
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Original solution and new solution
44
Original solution and new solution
45
Example 8.5 (P 290-2) - II
For Solution A H
C2H3O2- H C2H3O2
Before reaction 0.010 M 5.00
M 5.00 M After reaction
0 4.99 M
5.01 M
Calculate the new pH using the Henderson-Hasselbal
ch equation
For Solution B H
C2H3O2- H C2H3O2
Before reaction 0.010 M 0.050
M 0.050 M After reaction
0 0.040 M
0.060 M
0.040 0.060
The new pH is pH 4.74 log( )
4.74 0.18 4.56
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Solution A and Solution B
47
A buret valve
Source American Color
48
Figure 8.1 The pH curve for the titration of
50.0 ml of Nitric acid with 0.10M NaOH
49
Vol NaOH added (mL)
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Figure 8.2 The pH curve for the titration of
100.0 ml of 0.50 M NaOH with 0.10 M HCl.
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Weak acid
53
Figure 8.3 The pH curve for the titration of
50.0 ml of Acetic acid with 0.10 M NaOH
54
Treat the stoichiometry
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Figure 8.4 The pH curves for the titrations of
50.0
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Calculating the pH During a Weak Acid-Strong Base
TitrationI
Problem Calculate the pH during the titration of
20.00 mL of 0.250 M nitrous acid (HNO2 Ka 4.5
x 10-4) after adding different volumes of 0.150 M
NaOH (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d)
35.00 mL. Plan (a) We just calculate the pH of a
weak acid. (b)-(d) We calculate the amounts of
acid remaining after the reaction with the base,
and the anion concentration, and plug these into
the Henderson-Hasselbalch eq. Solution
HNO2 (aq) NaOH(aq)
H2O(l) NaNO2 (aq)
(a)
H3O NO2-
x (x)
Ka
4.5 x 10-4
x2 1.125 x 10-4 x 1.061 x 10-2
HNO2
0.250 M
pH -log(1.061 x 10-2) 2.000 - 0.0257
_______ no base added
58
Calculating the pH During a Weak Acid-Strong Base
TitrationII
(b) 15.00 mL of 0.150 M NaOH is added to the
20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250
mmol/mL 5.00 mmol HNO2) which will neutralize
(15.00 mL x 0.150 mmol/mL 2.25 mmol of HNO2),
leaving 2.75 mmol HNO2, and generating 2.25 mmol
of nitrite anion.
Concentration (M)
Initial 0.00275 ----
0
0.00225 Change -x
---- x
x Equilibrium 0.00275 - x ----
x 0.00225 x
pH 3.35 -0.0872 ________ with 15.0 mL of
NaOH added
59
Calculating the pH During a Weak Acid-Strong Base
TitrationIII
(c) 20.00 mL of 0.150 M NaOH is added to the
20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250
mmol/mL 5.00 mmol HNO2) which will neutralize
(20.00 mL x 0.150 mmol/mL 3.00 mmol of HNO2),
leaving 2.00 mmol HNO2, and generating 3.00 mmol
of nitrite anion.
Concentration (M)
Initial 0.00200
---- x 0.00300
pH 3.35 0.176 ____________ with 20.00 mL
of base added
(d) 35.00 mL of 0.150 M NaOH is added to the
20.00 mL of 0.250 M HNO2 (20.00 mL x 0.250
mmol/mL 5.00 mmol HNO2) which will neutralize
(35.00 mL x 0.150 mmol/mL 5.25 mmol of HNO2),
leaving no HNO2, and generating 5.00 mmol of
nitrite anion. There will be an excess of 0.25
mmol of NaOH which will control the pH.
60
Calculating the pH During a Weak Acid-Strong Base
TitrationIV
(d) continued Since all of the HNO2 has
been neutralized, we only have to look at the
concentration of hydroxide ion in the total
volume of the solution to calculate the pH of
the resultant solution.
combined volume 20.00 mL 35.00 mL 55.00 mL
0.000250 mol OH-
OH- 0.004545
M
0.05500 L
Kw
1 x 10-14
H3O
2.200 x 10-12
OH-
0.004545
pH -log (2.200 x 10-12) 12.000 - 0.342
______ when all of the acid

neutralized, and there

is an excess of NaOH
61
Summary Titration Curve Calculations
A Stoichiometry problem. The reaction of
hydroxide ion with the weak acid is assumed to
run to completion, and the concentrations of the
acid remaining and the conjugate base formed are
determined. An equilibrium problem. The
position of the weak acid equilibrium is
determined, and the pH is calculated.
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Figure 8.5 The pH curve for the titration of
100.0 ml of 0.050 M NH3 with ).10 M HCl
64
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Figure 8.6 The indicator phenolphthalein is
pink in basic solution and colorless in acidic
solution.
66
Figure 8.7 (a) Yellow acid form of bromthymol
blue (b) a greenish tint is seen when the
solution contains 1 part blue and 10 parts
yellow (c) blue basic form.
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Figure 8.8 The useful pH ranges for several
common indicators
69
Colors and Approximate pH Range of Some Common
Acid-Base Indicators
70
Figure 8.9 pH curve of 0.10 M HCI being
titratedwith 0.10 M NaOH
71
Figure 8.10 pH of 0.10 M HC2H3O2 being titrated
with NaOH
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Figure 8.11 A summary of the important
equilibria at various points in the titration of
a triprotic acid
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An X ray of the upper gastrointestinal
Source Photo Researchers, Inc.
76
Precipitation of bismuth
77
Equilibria of Slightly Soluble Ionic Compounds
When a solution becomes saturated and a
precipitate forms we move into the area of
insoluble material in solution, and we begin to
calculate the quantity of material that remains
in solution. We are working with what we call
the
Solubility Product The equilibrium constant
that is used for these calculations is called the
Solubility-product constant Ksp
Example Lead chromate
PbCrO4 (s) Pb2(aq) CrO42-(aq)
Since the concentration of a solid is
constant, we can move it to the other side of the
equals sign and combine it with the constant
yielding the solubility product constant Ksp
Pb2CrO42-
Qc
PbCrO4
PbCrO4 x Qc Pb2CrO42- Ksp
78
Like Example 8.12 (P 320)
The Ksp value for the mineral fluorite, CaF2 is
3.4 x 10-11 . Calculate The solubility of
fluorite in units of grams per liter.
Concentration (M) CaF2 (s)
Ca2(aq) 2 F-(aq)
Starting
0 0 Change

x 2x Equilibrium
x
2x
Substituting into Ksp

• Ca2F-2 Ksp
• (2x)2 3.4 x 10-11
• 4x3 3.4 x 10-11

3.4 x 10-11 4
3
x x 2.0 x 10-4
The solubility is 2.0 x 10-4 moles CaF2 per liter
of water. To get mass we must multiply by the
molar mass of CaF2 (78.1 g/mol).
78.1 g CaF2 1 mol CaF2
2.0 x 10-4 mol CaF2 x
_________ g CaF2 per L
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Writing Ion-Product Expressions for Slightly
Soluble Ionic
Compounds
Problem Write the ion-product expression for (a)
silver bromide (b) strontium phosphate (c)
aluminum carbonate (d) nickel(III)
sulfide. Plan Write the equation for a saturated
solution, then write the expression for the
solubility product. Solution
(a) Silver bromide AgBr(s)
Ag(aq) Br -(aq)
Ksp Ag Br -
(b) Strontium phosphate Sr3(PO4)(s)
3 Sr2(aq) 2 PO43-(aq)
Ksp Sr23PO43-2
(c) Aluminum carbonate Al2(CO3)3 (s)
2 Al3(aq) 3 CO32-(aq)
Ksp Al32CO32-3
(d) Nickel(III) sulfide Ni2S3 (s) 3 H2O(l)
2 Ni3(aq) 3 HS -(aq) 3 OH-(aq)
Ksp Ni32HS-3OH-3
81
Table 8.5 Ksp Values at 25 C for Common Ionic
Solids
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Determining Ksp from Solubility
Problem Lead chromate is an insoluble compound
that at one time was used as the pigment in the
yellow stripes on highways. Its solubility
is 4.33 x 10 -6g/100mL water. What is the
Ksp? Plan We write an equation for the
dissolution of the compound to see the number of
ions formed, then write the ion-product
expression. Solution
PbCrO4 (s) Pb2(aq) CrO42-(aq)
4.33 x 10 -6g
1000 ml 1 L
1mol PbCrO4
Molar solubility of PbCrO4
x x
100 mL
323.2 g
1.34 x 10 -8 M PbCrO4
1 Mole PbCrO4 1 mole Pb2 and 1 mole CrO42-
Therefore Pb2 CrO42- 1.34 x 10-8 M
Ksp Pb2 CrO42- (1.34 x 10 -8 M)2
________________
85
Determining Solubility from Ksp
Problem Lead chromate used to be used as the
pigment for the yellow lines on roads, and is a
very insoluble compound. Calculate the solubility
of PbCrO4 in water if the Ksp is equal to 2.00 x
10-16. Plan We write the dissolution equation,
and the ion-product expression. Solution Writing
the dissolution equation, and the ion-product
expression
Ksp 2.00 x 10-16 Pb2CrO42
Concentration (M) PbCrO4
Pb2 CrO42-
Initial
---------- 0
0 Change
---------- x
x Equilibrium
---------- x
x
Ksp Pb2 CrO42- (x)(x ) 2.00 x 10-16
x 1.41 x 10-8
Therefore the solubility of PbCrO4 in water is
_______________ M
86
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87
Relationship Between Ksp and Solubility at 25oC
No. of Ions Formula CationAnion
Ksp Solubility (M)
2 MgCO3 11
3.5 x 10-8 1.9 x 10-4 2
PbSO4 11
1.6 x 10-8 1.3 x 10-4 2
BaCrO4 11
2.1 x 10-10 1.4 x 10-5 3
Ca(OH)2 12 6.5 x
10-6 1.2 x 10-2 3
BaF2 12 1.5 x
10-6 7.2 x 10-3 3
CaF2 12 3.2 x
10-11 2.0 x 10-4 3
Ag2CrO4 21 2.6 x
10-12 8.7 x 10-5
88
The Effect of a Common Ion on Solubility
PbCrO4(s) Pb2(aq) CrO42-(aq)
PbCrO4(s) Pb2(aq) CrO42-(aq

added)
89
Calculating the Effect of a Common Ion on
Solubility
Problem What is the solubility of silver
chromate in 0.0600 M silver nitrate solution?
Ksp 2.6 x 10-12 . Plan From the equation and
the ion-product expression for Ag2CrO4, we
predict that the addition of silver ion will
decrease the solubility. Solution Writing the
equation and ion-product expression
Ag2CrO4 (s) 2 Ag(aq)
CrO42-(aq) Ksp Ag2CrO42-
Concentration (M) Ag2CrO4 (s)
2 Ag(aq) CrO42-(aq)
Initial ---------
0.0600
0 Change ---------
2x x
Equilibrium ---------
0.0600 2x x
Assuming that Ksp is small, 0.0600 M 2x 0.600
M
Ksp 2.6 x 10-12 (0.0600)2(x) x
7.22 x 10-10 M
Therefore, the solubility of silver chromate is
7.22 x 10-10 M
90
Test for the Presence of a Carbonate
91
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 (a) Iron(II)
cyanide (b) Potassium bromide (c) Aluminum
hydroxide Plan Write the balanced dissolution
equation and note the anion. Anions of weak
acids react with H3O and shift the equilibrium
position toward more dissolution. Strong acid
anions do not react, so added acid has
no effect. Solution (a) Fe(CN)2 (s)
Fe2(aq) 2 CN-(a) Increases solubility We
noted earlier that CN- ion reacts with water to
form the weak acid HCN, so it would be removed
from the solubility expression. (b) KBr(s)
K(aq) Br -(aq) No effect This occurs
since Br- is the anion of a strong acid, and K
is the cation of a strong base. (c) Al(OH)3 (s)
Al3(aq) 3 OH-(aq)
Increases solubility The OH- is the anion of
water, a very weak acid, so it reacts with the
added acid to produce water in a simple
acid-base reaction.
92
The Chemistry of Limestone Formation
Gaseous CO2 is in equilibrium with aqueous CO2 in
natural waters
H2O(l)
CO2 (g) CO2 (aq)
The concentration of CO2 is proportional to the
partial pressure of CO2 (g) in contact with the
water (Henrys law section 13.3)
CO2 (aq) (proportional to) PCO2
The reaction of CO2 with water produces H3O
CO2 (aq) 2 H2O(l) H3O(aq)
HCO3-(aq)
Thus, the presence of CO2 (aq) forms H3O, which
increases the solubility of CaCO3
CaCO3 (s) CO2 (aq) H2O(l)
Ca2(aq) 2 HCO3-(aq)
93
Predicting the Formation of a Precipitate Qsp
vs. Ksp
The solubility produce constant, Ksp, can be
compared to the ion-product constant, Qsp to
understand the characteristics of a solution with
respect to forming a precipitate.
Qsp Ksp When a solution becomes saturated,
no more solute will dissolve,
and the solution is called saturated. There
will be no changes that will
occur.
Qsp gt Ksp Precipitates will form until the
solution becomes saturated.
Qsplt Ksp Solution is unsaturated, and no
precipitate will form.
94
Predicting Whether a Precipitate Will FormI
Problem Will a precipitate form when 0.100 L
of a solution containing 0.055 M barium nitrate
is added to 200.00 mL of a 0.100 M solution of
sodium chromate? Plan We first see if the
solutions will yield soluble ions, then we
calculate the concentrations, adding the two
volumes together to get the total volume of the
solution, then we calculate the product constant
(Qsp), and compare it to the solubility product
constant to see if a precipitate will
form. Solution Both Na2CrO4 and Ba(NO3) are
soluble, so we will have Na, CrO42-, Ba2 and
NO3- ions present in 0.300 L of solution. We
change partners, look up solubilities, and we
find that BaCrO4 would be insoluble, so we
calculate its ion-product constant and compare
it to the solubility product constant of 2.1 x
10-10
For Ba2 0.100 L Ba(NO3)2 0.55 M 0.055mol
Ba2
0.055 mol Ba2
Ba2 ___________ M
in Ba2
0.300 L
95
Predicting Whether a Precipitate Will FormII
Solution cont.
For CrO42- 0.100 M Na2CrO4 0.200 L
0.0200 mol CrO42-
Qsp Ba2 CrO42- (0.183 M Ba2)(0.667 M
CrO42-) 0.121
Since Ksp 2.1 x 10-10 and Qsp 0.121, Qsp
gtgt Ksp and a precipitate

will form.
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97
Figure 8.12 Separation of Cu2 and Hg2
98
Figure 8.13 Separation of common cations by
precipitation
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100
The Stepwise Exchange of NH3 for H2O in M(H2O)42
101
Formation Constants (Kf) of Some Complex Ions
at 25oCI
Complex Ion Kf
Ag(CN)2- 3.0 x
1020 Ag(NH3)2 1.7 x
107 Ag(S2O3)23- 4.7 x
1013 AlF63-
4 x 1019 Al(OH)4-
3 x 1033 Be(OH)42-
4 x 1018 CdI42-
1 x 106 Co(OH)42-
5 x 109 Cr(OH)4-
8.0 x 1029 Cu(NH3)42
5.6 x 1011 Fe(CN)64-
3 x 1035 Fe(CN)63-
4.0 x 1043
102
Formation Constants (Kf) of
Some Complex Ions at 25oCII
Complex Ion Kf
Hg(CN)42- 9.3 x
1038 Ni(OH)42- 2 x
1028 Pb(OH)3 - 8 x
1013 Sn(OH)3 - 3 x
1025 Zn(CN)42- 4.2 x
1019 Zn(NH3)42 7.8 x
108 Zn(OH)42- 3 x
1015
103
Calculating the Concentrations of Complex IonsI
Problem A chemist converts Ag(H2O)2 to the more
stable form Ag(NH3)2 by mixing 50.0 L of
0.0020 M Ag(H2O)2 and 25.0 L of 0.15 M NH3. What
is the final Ag(H2O)2? Kf Ag(NH3)2 1.7 x
107. Plan We write the equation and the Kf
expression, set up the table for the
calculation, then substitute into Kf and
solve. Solution Writing the equation and Kf
expression
Ag(NH3)2
Kf 1.7
x 107
Ag(H2O)2NH32
Finding the initial concentrations
50.0 L (0.0020 M)
Ag(H2O)2init
1.3 x 10-3 M
50.0 L 25.0 L
25.0 L (0.15 M)
NH3init
__________ M
50.0 L 25.0 L
104
Calculating the Concentration of Complex IonsII
We assume that all of the Ag(H2O)2 is converted
Ag(NH3)2, so we set up the table with x
Ag(H2O)2 at equilibrium. Ammonia reacted
NH3reacted 2(1.3 x 10-3 M) 2.6 x 10-3 M
Concentration (M) Ag(H2O)2(aq) 2NH3 (aq)
Ag(NH3)2 2 H2O(aq)
Initial 1.3 x 10-3
5.0 x 10-2 0
---- Change (-1.3 x 10-3)
(-2.6 x 10-3) (1.3 x 10-3) ---- Equilibrium
x 4.7 x 10-2
1.3 x 10-3 ----
Ag(NH3)2
1.3 x 10-3
Kf
1.7 x 107
Ag(H2O)2NH32
x(4.7 x 10-2)2
x _________________ M Ag(H2O)2
105
The Amphoteric Behavior of Aluminum Hydroxide
106
Separating Ions by Selective PrecipitationI
Problem A solution consists of 0.10 M AgNO3 and
0.15 M Cu(NO3)2. Calculate the I - that would
separate the metal ions as their iodides. Kspof
AgI 8.3 x 10-17 Kspof CuI 1.0 x
10-12. Plan Since the two iodides have the same
formula type (11), we compare their Ksp values
and we see that CuI is about 100,000 times
more soluble than AgI. Therefore, AgI
precipitates first, and we solve for I - that
will give a saturated solution of AgI. Solution
Writing chemical equations and ion-product
expressions
H2O
AgI(s) Ag(aq) I -(aq)
Ksp AgI -
H2O
CuI(s) Cu(aq) I -(aq)
Ksp CuI -
Calculating the quantity of iodide needed to give
a saturated solution of CuI
Ksp
1.0 x 10-12
I -
_________________ M
Cu
0.15 M
107
Separating Ions by Selective PrecipitationII
Thus, the concentration of iodide ion that will
give a saturated solution of copper(I) iodide is
1.0 x 10-11 M, and this will not precipitate the
copper(I) ion, but should remove most of the
silver ion. Calculating the quantity of silver
ion remaining in solution we get
Ksp
8.3 x 10-17
Ag
1.2 x 10-6 M
I -
6.7 x 10-11
Since the initial silver ion was 0.10 M, most of
it has been removed, and essentially none of the
copper(I) was removed, so the separation was
quite complete. If the iodide was added as sodium
iodide, you would have to add only a few
nanograms of NaI to remove nearly all of the
silver from solution
1 molNaI
149.9 g NaI
6.7 x 10-11 mol I - x x
_____ ng NaI
mol NaI
mol I -
108
The General Procedure for Separating Ions in
Qualitative Analysis
109
Separation into Ion Groups
Ion Group 1 Insoluble chlorides

Ag, Hg22, Pb2 Ion Group 2 Acid-insoluble
sulfides Cu2,
Cd2, Hg2, As3, Sb3, Bi3, Sn2, Sn4,
Pb2 Ion Group 3 Base-insoluble sulfides and
hydroxides
Zn2, Mn2, Ni2, Fe2, Co2 as
sulfides,
and Al3, Cr3 as
hydroxides Ion Group 4 Insoluble phosphates

Mg2, Ca2, Ba2 Ion Group 5
Alkali metal and ammonium ions

Na, K, NH4
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111
Tests to Determine the Presence of Cations in Ion
Group 5
OH - NH4 NH3 H2O
Na ions
K ions
plus litmus paper
112
A Qualitative Analysis Scheme for Ag, Al3, Cu2
and Fe3
113
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114
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115
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116
Figure 8.14 Separation of Group I ions
117
The Acid Rain ProblemI
Normally the pH of precipitation is controlled by
the reaction of carbon dioxide with water to
form carbonic acid, which keeps the pH of clean
rain in the slightly acid range, of about 5.6
CO2 (g) H2O(l) H2CO3 (aq)
H3O(aq) HCO3-(aq)
Sulfurous acid is produced by the reaction of
sulfur dioxide with water, and even though it is
a weak acid, it does add to the acidity of rain.
A greater worry is the oxidation of the SO2 to
form SO3 which reacts with water to form the
strong acid, sulfuric acid.
SO2 (g) H2O(l) H2SO3 (aq)
2 SO2 (g) O2 (g) 2 SO3 (g)
SO3 (g) H2O(l) H2SO4 (aq)
118
The Acid Rain ProblemII
Another strong acid is formed from the nitrogen
oxides that are produced in all internal
combustion engines, that is nitric acid, which is
not only a strong acid, but also a strongly
oxidizing acid.
N2 (g) O2 (g) 2 NO(g)
2 NO(g) O2 (g) 2 NO2 (g)
3 NO2 (g) H2O(l) 2 HNO3 (aq)
NO(g)
(Final step in the Ostwald process)
These acids together make the problem of acid
rain so severe in the eastern United States, and
far worse in other parts of the world. The
average pH of rain in the eastern U.S. back in
1984 was 4.2. Sweden and Pennsylvania share
second place with a pH of 2.7, but the record was
in Wheeling, West Virginia where the pH was 1.8.
Some areas of California also reach a pH of
1.6. The problems are global in nature.
119
Formation of Acidic Precipitation