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Chapter 19
Ionic Equilibria in Aqueous Systems
19-1
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Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffer Systems
19.2 Acid-Base Titration Curves
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions
19.5 Application of Ionic Equilibria to Chemical Analysis
19-2
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Buffered solutions
 A solution
that resists a change in its pH
when either OH- or H+ ions are added.
 Ex: Blood, absorbs acid and bases without
change in pH.
 The components of a buffer are the
conjugate acid-base pair of a weak acid.
 WA and its salt-HF and NaF
 WB and its salt- NH3 and NH4Cl.
19-3
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Figure 19.1
The effect of addition of acid or base to …
acid added
Figure 19.2
an unbuffered solution
acid added
or a buffered solution
19-4
base added
base added
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Common ion effect
 The
shift in equilibrium position that occurs
because of the addition of an ion already
involved in equilibrium reaction is called
common ion effect.
 Consider a weak acid HF and its salt NaF.
 NaF (s) _______ Na+ (aq)+ F- (aq) , Major
species: Na+ F- HF H2O
 HF (aq)↔ H+ (aq) + F- (aq)
19-5
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Common ion effect
 F-
is the common ion.
 F- from the added NaF moves the equ
position to the left according to LC
principle.
 If we dissolve HF in a NaF solution, the Fion and H+ ion enter the solution. The Fion already present combines with the H+,
which lowers the H+.
19-6
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Buffered Solution
Characteristics
 Buffers
contain relatively large amounts of
weak acid and corresponding base.
 Added H+ is consumed by A Added OH is consumed by HA
 The pH is determined by the ratio of the
concentrations of the weak acid and its
conjugate base.
19-7
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Table 19.1
The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]initial
[CH3COO-]added
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
* % Dissociation =
[CH3COOH]dissoc
[CH3COOH]initial
19-8
% Dissociation*
x 100
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Figure 19.3
How a buffer works.
Buffer after addition of H3O+
Buffer with equal
concentrations of
conjugate base and acid
H3O+
H2O + CH3COOH
19-9
H3O+ + CH3COO-
Buffer after addition of OH-
OH-
CH3COOH + OH-
H2O + CH3COO-
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Problem
P.1. A buffered solution contains 0.50 M acetic acid
(Ka= 1.8 x 10-5) and 0.50 M sodium acetate.
a) Calculate pH of this solution.
b) Calculate the change in pH that occurs when
0.010 mol solid NaOH is added to 1.0 L of
buffered solution described in the above
problem.Compare this pH change with the one
that occurs when 0.010 mol solid NaOH is
added to 1.0 L of water.
c) Calculate the change in pH that occurs when
0.010 mol solid HCl is added to 1.0 L of buffered
solution.
19-10
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The Henderson-Hasselbalch Equation
H3O+ + A-
HA + H2O
Ka =
[H3O+] [A-]
[HA]
[H3O+] =
Ka [HA]
[A-]
[A-]
- log[H3O+] = - log Ka + log
pH = pKa + log
[HA]
[base]
[acid]
19-11
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Problems
P. 2. Calculate the pH of a solution 0.75 M
lactic acid (Ka= 1.4 x10-4 ) and 0.25 M
sodium lactate.
P.3. A buffer solution contains 0.25 M NH3
(Kb= 1.8 x 10-5) and 0.40 M NH4Cl.
Calculate pH of this solution.
19-12
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Buffering Capacity
It represents the amount of H+ or OH the buffer
can absorb without a significant change in pH.
 More concentrated the components of a buffer ,
the greater the buffer capacity.
 pH of a buffered solution is determined by the
ratio [A-]/[HA] .
 If an acid or a base is added , the concentration
ratio changes less when the buffer component
concentrations are similar than they are
different.
19-13
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Buffering Capacity:
 Buffer
has highest capacity when
component concentrations are equal.
 pH=pKa highest buffer capacity.
 The Buffer Range is the pH range over
which the buffer works effectively.
 The further the buffering component
concentration ratio is from 1 less effective
is the buffering action.
19-14
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Buffer Capacity and Buffer Range
Buffer capacity is the ability to resist pH change.
The more concentrated the components of a buffer, the greater
the buffer capacity.
The pH of a buffer is distinct from its buffer capacity.
A buffer has the highest capacity when the component
concentrations are equal.
Buffer range is the pH range over which the buffer acts effectively.
Buffers have a usable range within ± 1 pH unit of the pKa of
its acid component.
19-15
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Figure 19.4
19-16
The relation between buffer capacity and pH change.
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Preparing a Buffer




19-17
Choose the conjugate acid-base pair. Ratio
of concentrations should be close to1.
pH=pKa
Calculate ratio of buffer component
concentrations. Use the Henderson
Hasselbalch equation.
Determine the buffer concentration. Higher
the buffer concentrations greater the buffer
capacity.
Mix the solutions and adjust the pH by
adding strong acid or strong base with the
help of a pH probe.
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Preparing a Buffer
1. Choose the conjugate acid-base pair.
2. Calculate the ratio of buffer component concentrations.
3. Determine the buffer concentration.
4. Mix the solution and adjust the pH.
19-18
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Problem

P.4. A chemist needs a solution buffered at pH=
4.30 and can choose from the following acids
and their sodium salts:
1. chloroacetic acid Ka= 1.35 x 10-3
2. 2. propanoic acid Ka= 1.3 x 10-5
3. benzoic acid Ka= 6.4 x 10-5
4. hypochlorous acid. Ka= 3.5 x 10-8
Calculate the ratio [HA]/[A-] required for each
system to yield a pH of 4.30, which system will
work the best?
19-19
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Acid base Titration Curves
A plot of pH of the solution being analyzed
as a function of the amount of titrant
added.
19-20
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Acid-Base Indicators:
 It
marks the end point of a titration by
changing color.
 Ex: phenolphthalein is colorless in its HIn
form and pink in In- form, or basic form.
 Indicator changes color over a range of
about 2 pH units.
19-21
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Figure 19.5
Colors and approximate pH range of some
common acid-base indicators.
pH
19-22
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Figure 19.6 The color change of the indicator bromthymol blue.
basic
acidic
19-23
change occurs
over ~2pH units
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Strong Acid- Strong Base Titration
curves:
 pH
is low initially, as base is added the pH
increases slowly.
 The pH rises steeply when the moles of
OH- nearly equals the moles of H3O+
 The additional drop of base neutralizes the
tiny excess acid and introduces a tiny
excess of base.
 Then pH increases smoothly as more
base is added.
19-24
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Equivalence (stoichiometric)
point
Enough titrant has been added to react
exactly with the solution being analyzed.
 Equivalence point is nearly the vertical
portion of the curve.
 The point at which the number of moles of
added OH- =the number of moles of
H3O+
 For a SA vs SB titration pH =7(neutral
ions)

19-25
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End point
It occurs when indicator changes color. The
indicator chosen should be close to the
equivalence point.
19-26
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Figure 19.7
19-27
Curve for a strong acid-strong
base titration
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Calculating pH:




19-28
Initial pH is the pH of the acid.
Before the equ. Point find initial moles
present, moles reacted, change in
volume and then the pH.
At the equ point, pH =7
After the equ point find the excess of
OH- present and then calculate pH.
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Problem
P.5. 50.0 mL of 0.200 m HNO3 is titrated
with 0.100M NaOH. Calculate the pH of
the solution at selected points during the
course of titration, where 0 mL, 20.0 mL,
100.0 mL, 150.0 mL of 0.100 M NaOH has
been added.
19-29
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Weak acid-Strong Base
Titrations:
 Initial
pH is high as weak acid dissociates
slightly.
 Buffer Region: A gradual rising portion of
the curve appearing before the equ.point.
 At midpoint of the buffer region pH=pKa.
 pH at equ point is > 7.
 Beyond equ point pH increases slowly as
excess OH- is added.
19-30
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Figure 19.8
Titration of 40.00mL of 0.1000M HPr with
0.1000M NaOH
Curve for a
weak acidstrong base
titration
pKa of HPr
= 4.89
pH = 8.80 at
equivalence point
[HPr] = [Pr-]
19-31
methyl red
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Calculating pH
 1.
When only HA present, Use ICE table
calculate H+ and pH like a weak acid type
calculation.

2. As we add base, find change in
moles and solve for H+

3. At equ. Point pH >7, Find Kb from
Ka and then calculate OH- and pH.

4. Beyond the equ. point , calculate
excess of OH- present and calculate pH.
19-32
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Problem
P.6. 50.0 mL of 0.1 M acetic acid
(Ka=1.8 x 10-5) with 0.10 M NaOH.
Calculate the pH at various points
representing volumes of 0mL, 10.0 mL ,
25.0 mL, 50.0 mL , 60.0 mL of added
NaOH.
19-33
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Weak base –strong acid Titration
curves:







Same shape curve as WA vs SB , but inverted.
Initially pH above 7 as it is a weak base.
pH decreases in buffer region.
At midpoint pH=pKa
After buffer region curve drops vertically to
equ.point .
pH at equ.point is below 7
Beyond equ point , pH decreases slowly as
more acid is added.
19-34
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Figure 19.9
Titration of 40.00mL of 0.1000M NH3
with 0.1000M HCl
pKa of NH4+ =
9.25
Curve for a
weak basestrong acid
titration
19-35
pH = 5.27 at
equivalence
point
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Problems
P.7. 100.0 mL of 0.050 M NH3 is titrated with
0.10 M HCl. Calculate the pH at various
points: 1) before adding HCl 2) Before
equ.point 3) At equ point (setup)
P.8 If 2.00 mmol of solid acid in 100.0 mL
water is titrated with 0.0500 M NaOH. After
20.0 mL NaOH has been added the pH is
6.00. What is the Ka value for acid?
19-36
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Titration curves of polyprotic
acids:
Number of curves =Number of H+ ions.
19-37
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Figure 19.10
Curve for the titration of a weak polyprotic acid.
pKa = 7.19
pKa = 1.85
Titration of 40.00mL of 0.1000M
H2SO3 with 0.1000M NaOH
19-38
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Sample Problem 19.3
Calculating the pH During a Weak AcidStrong Base Titration
PROBLEM:
Calculate the pH during the titration of 40.00 mL of 0.1000M
propanoic acid (HPr; Ka = 1.3x10-5) after adding the following
volumes of 0.1000M NaOH:
(a) 0.00mL
(b) 30.00mL (c) 40.00mL (d) 50.00mL
PLAN: The amounts of HPr and Pr- will be changing during the titration.
Remember to adjust the total volume of solution after each addition.
SOLUTION: (a) Find the starting pH using the methods of Chapter 18.
Ka = [Pr-][H3O+]/[HPr]
[Pr-] = x = [H3O+]
x  (1.3x105 )(0.10)
(b)
x = 1.1x10-3 ; pH = 2.96
Amount (mol)
HPr(aq) + OH-(aq)
Before addition
0.04000
Addition
After addition
19-39
[Pr-] = x = [H3O+]
0.01000
Pr-(aq) + H2O (l)
-
0
-
0.03000
-
-
0
0.03000
-
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Sample Problem 19.3
Calculating the pH During a Weak AcidStrong Base Titration
continued
[H3O+] = 1.3x10-5
0.001000 mol
= 4.3x10-6M
pH = 5.37
0.003000 mol
(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -]
will be
(0.004000 mol)
= 0.05000M
(0.004000L) + (0.004000L)
K a x K b = Kw
Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10
-9
[H3O+] = Kw / K b x[Pr  =
] 1.6x10 M
pH = 8.80
(d) 50.00mL of NaOH will produce an excess of OH-.
mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol
[H3O+] = 1.0x10-14/0.01111 = 9.0x10-11M
pH = 12.05
19-40
M = (0.00100)
(0.0900L)
M = 0.01111
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Figure 19.11
19-41
Sickle shape of red blood cells in sickle cell anemia.
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Equilibria of slightly soluble
ionic compounds:






An ionic substance dissociates completely in
water to form hydrate cations and anions.
Both the forward and reverse reactions take
place.
A stage is reached when no more solid
dissolves. This is called as dynamic equilibrium.
CaF2 (s) ↔Ca2+ (aq) + 2F- (aq)
Ksp= [Ca2+ ] [F-]2
Ksp is called as the solubility product constant or
solubility product for the equilibrium expression.
19-42
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Equilibria of slightly soluble
ionic compounds:





For the hypothetical compound, MpXq
At equilibrium Qsp = [Mn+]p [Xz-]q= Ksp
Higher the Ksp greater the solubility for formulas
containing same total number of ions.
If salts being compared produce different
number of ions.
Compare their solubilities, which is opposite of
Ksp
19-43
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Ion-Product Expression (Qsp)
and Solubility Product Constant (Ksp)
For the hypothetical compound, MpXq
At equilibrium
19-44
Qsp = [Mn+]p [Xz-]q = Ksp
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Sample Problem 19.4
Writing Ion-Product Expressions for Slightly
Soluble Ionic Compounds
PROBLEM: Write the ion-product expression for each of the following:
(a) Magnesium carbonate
(c) Calcium phosphate
(b) Iron (II) hydroxide
(d) Silver sulfide
PLAN: Write an equation which describes a saturated solution. Take
note of the sulfide ion produced in part (d).
SOLUTION:
Mg2+(aq) + CO32-(aq)
(a) MgCO3(s)
Fe2+(aq) + 2OH- (aq)
(b) Fe(OH)2(s)
2Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l)
Ag2S(s) + H2O(l)
19-45
Ksp = [Fe2+][OH-] 2
3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO43-]2
(c) Ca3(PO4)2(s)
(d) Ag2S(s)
Ksp = [Mg2+][CO32-]
HS-(aq) + OH-(aq)
Ksp = [Ag+]2[HS-][OH-]
2Ag+(aq) + HS-(aq) + OH-(aq)
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Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 250C
Name, Formula
19-46
Ksp
Aluminum hydroxide, Al(OH)3
3 x 10-34
Cobalt (II) carbonate, CoCO3
1.0 x 10-10
Iron (II) hydroxide, Fe(OH)2
4.1 x 10-15
Lead (II) fluoride, PbF2
3.6 x 10-8
Lead (II) sulfate, PbSO4
1.6 x 10-8
Mercury (I) iodide, Hg2I2
4.7 x 10-29
Silver sulfide, Ag2S
8 x 10-48
Zinc iodate, Zn(IO3)2
3.9 x 10-6
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Problems
 P.10.
CuBr has a measured solubility of
2.0 x 10-4 mol/L at 25 C. Calculate Ksp.
 P.11. Calculate Ksp value for Bi2S3 which
has a solubility of 1.0 x 10-15 mol/L at 25
C.
 P.12. Calculate the solubility of copper (II)
iodate at 25 C, if Ksp = 1.4 x 10-7
19-47
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Table 19.3 Relationship Between Ksp and Solubility at 250C
No. of Ions
19-48
Formula
Cation:Anion
Ksp
Solubility (M)
2
MgCO3
1:1
3.5 x 10-8
1.9 x 10-4
2
PbSO4
1:1
1.6 x 10-8
1.3 x 10-4
2
BaCrO4
1:1
2.1 x 10-10
1.4 x 10-5
3
Ca(OH)2
1:2
5.5 x 10-6
1.2 x 10-2
3
BaF2
1:2
1.5 x 10-6
7.2 x 10-3
3
CaF2
1:2
3.2 x 10-11
2.0 x 10-4
3
Ag2CrO4
2:1
2.6 x 10-12
8.7 x 10-5
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Common ion Effect:

Presence of a common ion decreases
solubility of a slightly soluble ionic
compound.
19-49
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Figure 19.12
The effect of a common ion on solubility
CrO42- added
PbCrO4(s)
19-50
Pb2+(aq) + CrO42-(aq)
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
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Problem
 .13.
Calculate the solubility of CaF2
(Ksp=4.0 x 10-11) in a 0.025 M NaF
solution.
19-51
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Figure 19.13
19-52
Test for the presence of a carbonate.
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pH and solubility:
 If
the compound contains the anion of a
weak acid, addition of H3O+ increases its
solubility.
19-53
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Sample Problem 19.8
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 these ionic compounds:
(a) Lead (II) bromide
(b) Copper (II) hydroxide
(c) Iron (II) sulfide
PLAN: Write dissolution equations and consider how strong acid would affect
the anion component.
SOLUTION: (a) PbBr2(s)
Pb2+(aq) + 2Br-(aq) Br- is the anion of a strong acid.
No effect.
(b) Cu(OH)2(s)
Cu2+(aq) + 2OH-(aq)
OH- is the anion of water, which is a weak acid. Therefore it will shift the
solubility equation to the right and increase solubility.
Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will
react with water to produce OH-.
FeS(s) + H2O(l)
Fe2+(aq) + HS-(aq) + OH-(aq)
(c) FeS(s)
Both weak acids serve to increase the solubility of FeS.
19-54
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Predicting formation of precipitate:
 If
Qsp > Ksp ppt occurs
 If Qsp < Ksp solution is unsaturated, no
ppt occurs.
 If Qsp = Ksp solution is saturated and no
change occurs.
19-55
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Sample Problem 19.9
PROBLEM:
PLAN:
Predicting Whether a Precipitate Will Form
A common laboratory method for preparing a precipitate is to mix
solutions of the component ions. Does a precipitate form when
0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?
Write out a reaction equation to see which salt would be formed. Look
up the Ksp valus in a table. Treat this as a reaction quotient, Q,
problem and calculate whether the concentrations of ions are > or <
Ksp. Remember to consider the final diluted solution when calculating
concentrations.
SOLUTION:
CaF2(s)
Ca2+(aq) + 2F-(aq)
mol Ca2+ = 0.100L(0.30mol/L) = 0.030mol
mol F- = 0.200L(0.060mol/L) = 0.012mol
Q = [Ca2+][F-]2 =
[Ca2+] = 0.030mol/0.300L = 0.10M
[F-] = 0.012mol/0.300L = 0.040M
(0.10)(0.040)2 = 1.6x10-4
Q is >> Ksp and the CaF2 WILL precipitate.
19-56
Ksp = 3.2x10-11
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Equilibria Involving Complex
Ions:
 Complex
ion: A central metal ion
covalently bonded to two or more anions,
or molecules called ligands.
 Ligand: Lewis base that contains a lone
pair of electron that can be donated to an
empty orbital on the metal ion to form a
covalent bond.

Ex: H2O, NH3, CN-,Cl19-57
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Equilibria Involving Complex
Ions:
 Coordination
number: Number of ligands
attached to a metal ion.ex; 6,4,2
 Formation constants/Stability Constants :
Metal ions add ligands one at a time in
steps.
 A ligand increases the solubility of a
slightly soluble ionic compound if it forms a
complex ion with the cation.
19-58
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Figure 19.14
19-59
Cr(NH3)63+, a typical complex ion.
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Figure 19.15
The stepwise exchange of NH3 for H2O in M(H2O)42+.
NH3
M(H2O)42+
3NH3
M(H2O)3(NH3)2+
M(NH3)42+
19-60
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19-61
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Sample Problem 19.10
Calculating the Concentration of a Complex Ion
PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more stable
Zn(NH3)42+ by mixing 50.0L of 0.0020M Zn (H2O)42+ and 25.0L of
0.15M NH3. What is the final [Zn (H2O)42+]?
Kf of Zn(NH3)42+ is 7.8x108.
PLAN:
Write the reaction equation and Kf expression. Use a reaction table
to list various concentrations. Remember that components will be
diluted when mixed as you calculate final concentrations. It is
obvious that there is a huge excess of NH3 and therefore it will drive
the reaction to completion.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq)
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
Zn(NH3)42+(aq) + 4H2O(l)
[Zn(H2O)42+]initial = (50.0L)(0.0020M) = 1.3x10-3 M
75.0L
[NH3]initial = (25.0L)(0.15M) = 5.0x10-2 M
75.0L
19-62
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Sample Problem 19.10
Calculating the Concentration of a Complex Ion
continued
Since we assume that all of the Zn(H2O)42+ has reacted, it would use 4
times its amount in NH3.
[NH3]used = 4(1.3x10-3M) = 5.2x10-3M
[Zn(H2O)42+]remaining = x(a very small amount)
Concentration(M)
Zn(H2O)42+(aq) + 4NH3(aq)
1.3x10-3
Initial
Change
~(-1.3x10-3)
Equilibrium
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
19-63
5.0x10-2
~(-5.2x10-3)
4.5x10-2
x
=
7.8x108
=
(1.3x10-3)
x(4.5x10-2)
Zn(NH3)42+(aq) + 4H2O(l)
0
-
~(+1.3x10-3)
-
1.3x10-3
-
x = 4.1x10-7M
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Sample Problem 19.11
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), through formation of the complex ion
Ag(S2O3)23-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0M
hypo. Kf of Ag(S2O3)23- is 4.7x1013 and Ksp AgBr is 5.0x10-13.
PLAN: Write equations for the reactions involved. Use Ksp to find S, the molar
solubility. Consider the shifts in equilibria upon the addition of the
complexing agent.
SOLUTION:
AgBr(s)
Ag+(aq) + Br-(aq)
(a) S = [AgBr]dissolved = [Ag+] = [Br-]
(b)
19-64
AgBr(s)
Ksp = [Ag+][Br-] = 5.0x10-13
Ksp = S2 = 5.0x10-13 ; S = 7.1x10-7M
Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq)
Ag(S2O3)23-(aq)
AgBr(s) + 2S2O32-(aq)
Br -(aq) + Ag(S2O3)23-(aq)
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Sample Problem 19.11
Calculating the Effect of Complex-Ion Formation
on Solubility
continued
Koverall = Ksp x Kf =
Concentration(M)
[Br-][Ag(S2O3]23[AgBr][S2O32-]2
AgBr(s) + 2S2O32-(aq)
Br-(aq) + Ag(S2O3)23-(aq)
Initial
-
1.0
0
0
Change
-
-2S
+S
+S
Equilibrium
-
1.0-2S
S
S
Koverall =
S2
(1.0-2S)2
S
= 24
1.0-2S
S = [Ag(S2O3)23-] = 0.45M
19-65
= (5.0x10-13)(4.7x1013) = 24
= (24)1/2
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Figure 19.16
The amphoteric behavior of aluminum hydroxide.
3H2O(l) + Al(H2O)3(OH)3(s)
19-66
Al(H2O)3(OH)3(s)
Al(H2O)3(OH)4-(s) + H2O(l)
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Sample Problem 19.12
Separating Ions by Selective Precipitation
PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M 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 precipitates are of the same ion ratio, 1:2, so we can compare
their Ksp values to determine which has the greater solubility.
It is obvious that Cu(OH)2 will precipitate first so we calculate the
[OH-] needed for a saturated solution of Mg(OH)2. This should
ensure that we do not precipitate Mg(OH)2. Then we can check
how much Cu2+ remains in solution.
SOLUTION:
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10
Cu(OH)2(s)
Cu2+(aq) + 2OH-(aq)
[OH-] needed for a saturated Mg(OH)2 solution =
Ksp = 2.2x10-20
K sp
[Mg 2 ]
= 5.6x10-5M
19-67
6.3x1010

0.20
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Sample Problem 19.12
Separating Ions by Selective Precipitation
continued
Use the Ksp for Cu(OH)2 to find the amount of Cu remaining.
[Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 = 7.0x10-12M
Since the solution was 0.10M CuCl2, virtually none of the Cu2+
remains in solution.
19-68
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 19.17
The general procedure for separating ions in qualitative analysis.
19-69
Add
precipitating
ion
Centrifuge
Centrifuge
Add
precipitating
ion
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A qualitative analysis scheme for separating cations
into five ion groups.
19-70
Add
(NH4)2HPO4
Centrifuge
Add
NH3/NH4+
buffer(pH 8)
Centrifuge
Centrifuge
Add
6M HCl
Acidify to
pH 0.5;
add H2S
Centrifuge
Figure 19.18
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+
Centrifuge
Step 1
Add
NH3(aq)
Step 2
Add HCl
Centrifuge
Extra:
Step 3 Add
NaOH
Centrifuge
Step 4
Add HCl,
Na2HPO4
19-71
Step 5
Dissolve in
HCl and
add KSCN
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Figure B19.1
19-72
A view inside Carlsbad Caverns, New Mexico
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure B19.3
19-73
Formation of acidic precipitation.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure B19.4
A forest damaged by acid rain
19-74
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure B19.5
The effect of acid rain on marble statuary.
1944
1994
Location: New York City
19-75