Chapter 19
Ionic Equilibria in Aqueous Systems
19-1
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
The effect of addition of acid or base to …
acid added
base added
an unbuffered solution.
acid added
Figure 19.1
19-3
base added
a buffered solution.
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Acid-Base Buffer Systems
Buffers function by reducing changes in [H3O+]
that result from additions of acid or base to the solution.
Buffers are composed of the conjugate
acid-base pair of a weak acid.
Buffers function via the common ion effect.
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
The common ion effect occurs when a reactant containing
a given ion is added to an equilibrium mixture that already
contains that ion and the position of the equilibrium shifts
away from forming more of it.
19-4
Table 19.1
The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]initial
[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
* % dissociation =
[CH3COOH]dissoc
x 100
[CH3COOH]initial
19-5
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How an acetic acid/acetate buffer works
buffer after addition of H3O+
buffer with equal
concentrations of weak
acid and its conjugate
base
H3O+
H2O + CH3COOH
H3O+ + CH3COO-
buffer after addition of OH-
OH-
CH3COOH + OH-
H2O + CH3COO-
HA scavenges OH-, A- scavenges H+
Figure 19.2
19-6
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Some Details
Ka = [CH3COO-][H3O+]/[CH3COOH]
[H3O+] = Ka x [CH3COOH]/[CH3COO-]
Since Ka is constant, [H3O+] depends directly on the ratio
of the concentrations of HA and its conjugate base, A-.
19-7
Sample Problem 19.1
PROBLEM:
Calculating the effect of added H3O+ and OHon buffer pH
Calculate the pH of the following solutions.
(a) A buffer solution consisting of 0.50 M CH3COOH and 0.50 M CH3COONa
(b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution in part (a)
(c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in part (a)
Ka of CH3COOH = 1.8 x 10-5 (assume the additions cause negligible volume
changes)
PLAN: We know Ka and can find initial concentrations of conjugate acid and
base. Make assumptions about the amount of acid dissociating relative to its
initial concentration. Proceed stepwise through changes in the system.
SOLUTION:
concentration (M)
(a)
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
initial
0.50
0.50
0
-x
+x
+x
change
equilibrium
0.50 - x
0.50 + x
x
___________________________________________________________
19-8
Sample Problem 19.1
(continued)
[H3O+] = x
[H3O+][CH3COO-]
Ka =
[CH3COO-]eq ≈ 0.50 M
[CH3COOH]eq ≈ 0.50 M
[H3O+] = x = Ka
[CH3COOH]
[CH3COOH]
[CH3COO-]
Check the assumption: 1.8 x 10-5/0.50
(b)
[OH-]added =
concentration (M)
0.020 mol
X
= 1.8 x 10-5 M
pH = 4.74
100 = 3.6 x 10-3 %
= 0.020 M NaOH
1.0 L soln
CH3COOH(aq) + OH-(aq)
CH3COO-(aq) + H2O (l)
before addition
0.50
-
0.50
-
addition
after addition
0.48
0.020
0
0.52
-
___________________________________________________________
19-9
Sample Problem 19.1 (continued)
Set up a reaction table with the new values.
concentration (M)
CH3COOH(aq) + H2O(l)
initial
change
0.48
-x
-
CH3COO-(aq) + H3O+(aq)
0.52
+x
0
+x
equilibrium
0.48 - x
0.52 + x
x
___________________________________________________________
[H3O+] = 1.8 x 10-5 x
(c)
[H3O+]added =
concentration (M)
0.48
0.52
0.020 mol
1.0 L soln
= 1.7 x 10-5
pH = 4.77
= 0.020 M H3O+
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O (l)
0.50
0.50
before addition
0.020
addition
0.48
0
0.52
after addition
___________________________________________________________
19-10
Sample Problem 19.1 (continued)
Set up a reaction table with the new values.
concentration (M)
CH3COOH(aq) + H2O(l)
initial
change
equilibrium
CH3COO-(aq) + H3O+(aq)
0.52
-x
-
0.48
+x
0
+x
0.52 - x
-
0.48 + x
x
___________________________________________________________
[H3O+] = 1.8 x 10-5 x
0.52
0.48
= 2.0 x 10-5
pH = 4.70
Note that this buffer resists changes in pH from
additions of either strong acid or strong base!
(What is the solution pH of 0.020 M HCl or 0.020 M NaOH?)
19-11
The Henderson-Hasselbalch Equation
An equation that relates pH, pKa and the ratio of
weak acid to its conjugate base for a buffer solution.
H3O+ + A-
HA + H2O
Ka = [H3O+][A-]/[HA]
[H3O+] = Ka x [HA]/[A-]
-log [H3O+] = -log Ka - log ([HA]/[A-])
pH = pKa + log ([A-]/[HA])
or
pH = pKa + log ([base]/[acid])
Special case: when [base] = [acid], the pH of the buffer
solution equals the pKa of the weak acid.
19-12
The relationship between buffer capacity and pH change
Buffer capacity refers to the
ability of a buffer to resist
pH change; buffer capacity
increases as the
concentrations
of its components (i.e., the
weak acid and its conjugate
base) increase.
Buffer pH and buffer capacity
are different concepts.
Figure 19.3
19-13
For an acetic acid/
acetate buffer
Key Concepts
For a given addition of acid or base, the concentration ratio
([A-]/[HA]) changes less for similar buffer component
concentrations than it does for different concentrations.
A buffer has the highest capacity when the
concentrations of HA and A- are equal.
A buffer whose pH is equal to or near the pKa of its acid
component has the highest buffer capacity.
Buffer range: the pH range over which the buffer acts effectively;
defined as pKa +/- 1 pH unit
19-14
Sample Problem 19.2
PROBLEM:
PLAN:
Preparing a buffer
An environmental chemist needs a carbonate buffer of pH 10.00 to
study the effects of acid rain on limestone-rich soils. How many
grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M
NaHCO3 to prepare the buffer? Ka of HCO3- is 4.7 x 10-11.
We know the Ka and the conjugate acid-base pair. Convert pH to
[H3O+], find the number of moles of carbonate and convert to mass.
SOLUTION:
[CO32-][H3O+]
HCO3-(aq) + H2O(l)
CO32-(aq) + H3O+(aq)
pH = 10.00; [H3O+] = 1.0 x 1010
4.7 x 10-11 =
[CO32-][10-10]
[0.20]
moles of Na2CO3 = (1.5 L)(0.094 mol/L) = 0.14 moles
105.99 g
0.14 moles x
19-15
mol
Ka =
= 15 g Na2CO3
[HCO3-]
[CO32-] = 0.094 M
Acid-Base Titration Curves
Acid-Base Indicator: a weak organic acid (HIn) that has a
different color than its conjugate base (In-); small amounts
are used in acid-base titrations; change color over different
pH ranges
19-16
Colors and approximate pH ranges of some
common acid-base indicators
Figure 19.4
19-17
The color change of the indicator bromthymol blue
pH 7.5
pH 6
Figure 19.5
acidic
19-18
basic
change occurs
over ~2 pH
units
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Curve for a strong acid-strong base titration
40 mL of 0.1000 M HCl
19-19
Figure 19.6
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Titration of 40.00 mL of 0.1000 M HPr with 0.1000 M NaOH
HPr = CH3CH2COOH
Curve for a
weak acidstrong base
titration
pH = 8.80 at
equivalence point
pKa of HPr
= 4.89
[HPr] = [Pr-]
Figure 19.7
19-20
methyl red
Sample Problem 19.3
PROBLEM:
Calculating the pH during a weak acid-strong
base titration
Calculate the pH during the titration of 40.00 mL of 0.1000 M
propanoic acid (HPr; Ka = 1.3 x 10-5) after adding the following
volumes of 0.1000 M NaOH:
(a) 0.00 mL
(b) 30.00 mL (c) 40.00 mL
(d) 50.00 mL
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.
PLAN:
Ka = [Pr-][H3O+]/[HPr]
x  (1.3x105 )(0.10)
(b)
[Pr-] = x = [H3O+]
x = 1.1 x 10-3 ; pH = 2.96
amount (mol)
HPr(aq) + OH-(aq)
before addition
0.004000
addition
after addition
-
Pr-(aq) + H2O (l)
-
0
-
0.003000
-
-
0.001000
0
0.003000
-
_________________________________________________
19-21
Sample Problem 19.3 (continued)
[H3O+] = 1.3 x 10-5 x
0.001000 mol
= 4.3 x 10-6 M
pH = 5.37
0.003000 mol
(c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr-]
will be:
0.004000 mol
= 0.05000 M
0.04000 L + 0.04000 L
Ka x Kb = Kw
Kb = Kw/Ka = 1.0 x 10-14/1.3 x 10-5 = 7.7 x 10-10
[H3O+] = Kw / K b x[Pr  ] = 1.6 x 10-9 M
pH = 8.80
(d) 50.00 mL of NaOH will produce an excess of OH-.
mol excess OH- = (0.1000 M)(0.05000 L - 0.04000 L) = 0.00100 mol
[H3O+] = 1.0 x 10-14/0.01111 = 9.0 x 10-13 M
pH = 12.05
19-22
M = 0.00100 mol
0.0900 L
M = 0.01111
Titration of 40.00 mL of 0.1000 M NH3 with 0.1000 M HCl
pKa of NH4+
= 9.25
Curve for a
weak basestrong acid
titration
pH = 5.27 at
equivalence
point
Figure 19.8
19-23
Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH
sulfurous acid: a diprotic weak acid
Curve for
the titration
of a weak
polyprotic
acid.
pKa2 = 7.19
pKa1 = 1.85
Figure 19.9
19-24
Sickle shape of red blood cells in sickle cell anemia
Single-site mutations
in the hemoglobin
molecule can change the
net charge on the
protein, which causes
protein aggregation
and a consequent
change in cell
morphology
Figure 19.10
19-25
Dealing with Solubility Equilibria
Slightly soluble ionic compounds
Ion-Product Expressions (Qsp); Solubility-Product Constants (Ksp)
PbSO4(s)
Pb2+(aq) + SO42-(aq)
Qc = [Pb2+][SO4 2-]/[PbSO4]
Qsp = [Pb2+][SO42-]
At saturation, Qsp attains a constant value (equilibrium
has been established); thus, Qsp = Ksp
Generally, for a saturated solution of a slightly soluble ionic compound
with formula MpXq: Qsp = [Mn+]p[Xz-]q = Ksp
19-26
A slightly soluble
ionic compound
PbCl2
19-27
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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
PLAN:
(b) iron(II) hydroxide
(d) silver sulfide
Write an equation which describes a saturated solution. Take note
of the unusual behavior of the sulfide ion produced in (d).
SOLUTION:
(a) MgCO3(s)
(aq)
(b) Fe(OH) (s)
2
(c) Ca3(PO4)2(s)
Mg2+(aq) + CO3 2Fe2+(aq) + 2OH- (aq)
Ksp = [Mg2+][CO32-]
Ksp = [Fe2+][OH-]2
3Ca2+(aq) + 2PO43-(aq)
Ksp = [Ca2+]3[PO43-]2
(d) Ag2S(s)
2Ag+(aq) + S 2(aq) 2S (aq) + H2O(l)
HS-(aq) + OH-(aq)
Ag2S(s) + H2O(l)
19-28
2Ag+(aq) + HS-(aq) + OH-(aq) Ksp = [Ag+]2[HS-][OH-]
Table 19.2
Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 25 oC
Name and Formula
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
The magnitude of Ksp is a measure of how far to the right the dissolution
proceeds at equilibrium (i.e., at saturation).
19-29
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Sample Problem 19.5
Determining Ksp from solubility data
PROBLEM: (a) Lead(II) sulfate is a key component in car batteries. Its solubility
in water at 25 oC is 4.25 x 10-3 g/100 mL solution. What is the Ksp
of PbSO4?
(b) When lead(II) fluoride (PbF2) is shaken with pure water at 25 oC,
the solubility is found to be 0.64 g/L. Calculate the Ksp of PbF2.
PLAN:
Write the dissolution equation; find moles of dissociated ions;
convert solubility to M and substitute values into the solubility
product constant expression.
SOLUTION: (a) PbSO4(s)
4.25 x 10-3 g
100 mL soln
x
1000 mL
L
Pb2+(aq) + SO42-(aq)
x
mol PbSO4
Ksp = [Pb2+][SO42-]
= 1.40 x 10-4 M PbSO4
303.3 g PbSO4
Ksp = [Pb2+][SO42-] = (1.40 x 10-4)2 = 1.96 x 10-8
19-30
Sample Problem 19.5 (continued)
Pb2+(aq) + 2F-(aq)
(b) PbF2(s)
0.64 g
L soln
x
Ksp = [Pb2+][F-]2
mol PbF2
245.2 g PbF2
= 2.6 x 10-3 M PbF2
Ksp = (2.6 x 10-3)(5.2 x 10-3)2 = 7.0 x 10-8
[Pb2+]
19-31
[F-]
Sample Problem 19.6
Determining solubility from Ksp
PROBLEM: Calcium hydroxide is a major component of mortar, plaster and
cement, and solutions of Ca(OH)2 are used in industry as a
cheap, strong base. Calculate the solubility of Ca(OH)2 in water
at 25 oC if its Ksp is 6.5 x 10-6.
PLAN:
Write a dissociation equation and Ksp expression; find the molar
solubility (S) using a table.
SOLUTION:
Ca2+(aq) + 2OH-(aq)
Ca(OH)2(s)
concentration (M)
Ca(OH)2(s)
Ksp = [Ca2+][OH-]2
Ca2+(aq) + 2OH-(aq)
initial
-
0
0
change
-
+S
+ 2S
equilibrium
-
S
2S
____________________________________________________
Ksp = (S)(2S)2 = 4S3
19-32
S=
3
6.5x106
4
= 1.2 x 10-2 M
Ksp and Solubilities
Ksp values are used to determine relative solubilities provided that comparisons
are made between compounds whose formulas contain the same total
number of ions. The compound having the higher Ksp is more soluble.
19-33
Relationship Between Ksp and Solubility at 25 oC
Table 19.3
no. of ions
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
6.5 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
Ag2CrO
2:1
2.6 x 10-12
8.7 x 10-5
4
19-34
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Common Ion Effects on Solubility
The presence of a common ion decreases the solubility of a
slightly soluble ionic compound.
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
Add Na2CrO4 (a very soluble salt; a strong electrolyte)
Result: equilibrium shifts to the left (LeChâtelier’s principle)
19-35
The effect of a common ion on solubility
CrO42- added
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
Figure 19.11
19-36
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Calculating the effect of a common ion on
solubility
Sample Problem 19.7
PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2
in water. What is its solubility in 0.10 M Ca(NO3)2? Ksp of
Ca(OH)2 is 6.5 x 10-6.
PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2.
The Ca(NO3)2 will supply extra [Ca2+], which will influence the
solubility of the Ca(OH)2 through the common ion effect.
SOLUTION:
concentration (M)
Ca(OH)2(s)
Ca2+(aq) + 2OH-(aq)
initial
-
0.10
0
change
-
+S
+2S
equilibrium
-
0.10 + S
2S
_____________________________________________
Ksp = 6.5 x 10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2
S ≈ (6.5 x 10-5/4)0.5
19-37
= 4.0 x 10 -3 M
= 1.2 x 10-2 M
(S << 0.10)
Check the assumption:
4.0 x 10 -3 M
x 100 = 4.0 %
0.10 M
The Effect of pH on Solubility
If the compound contains the anion of a weak acid, addition
of H3O+ (from a strong acid) increases its solubility
(LeChâtelier’s principle)
CaCO3(s)
Ca2+(aq) + CO32-(aq)
CO32-(aq) + H3O+(aq)
HCO3-(aq) + H3O+(aq)
H2CO3(aq)
HCO3- (aq) + H2O(l)
H2CO3(aq) + H2O(l)
2H2O(l) + CO2(g)
Adding H3O+ shifts the equilibrium to the right.
19-38
Test for the
presence of a
carbonate
Effect of the addition
of a strong acid
(release of carbon
dioxide)
Figure 19.12
19-39
Predicting the effect of adding a strong acid
on solubility
Sample Problem 19.8
PROBLEM: Write balanced equations to explain whether addition of H3O+ from a
strong acid will affect the solubility of the following 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.
(b) Cu(OH)2(s)
No effect.
Cu2+(aq) + 2OH-(aq)
The OH- reacts with the added hydronium ion to form water. The
equilibrium will shift to the right and thus solubility will increase.
2will
Fe2+(aq) + S2-(aq) S is the anion of a weak acid and
react with water to produce OH .
FeS(s) + H2O(l)
Fe2+(aq) + HS-(aq) + OH-(aq)
(c) FeS(s)
19-40
Both weak acids serve to increase the solubility of FeS.
Predicting Whether a Precipitate Will Form
Compare Qsp with Ksp.
When Qsp = Ksp, the solution is saturated and no change occurs.
When Qsp > Ksp, a precipitate forms until the solution is saturated.
When Qsp < Ksp, the solution is unsaturated and no precipitate forms.
19-41
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
0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?
Write out a reaction equation to see which salt could form. Look up
the Ksp values 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)
Ksp = 3.2 x 10-11
mol Ca2+ = 0.100 L(0.30 mol/L) = 0.030 mol [Ca2+] = 0.030 mol/0.300 L = 0.10 M
mol F- = 0.200 L(0.060 mol/L) = 0.012 mol
[F-] = 0.012 mol/0.300 L = 0.040 M
Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6 x 10-4
Q is >> Ksp, thus CaF2 will precipitate.
19-42
Equilibria Involving Complex Ions
A complex ion consists of a central metal ion covalently bonded
to two or more anions (or molecules) called ligands. Ionic ligands include
hydroxide (OH-), chloride (Cl-) and cyanide (CN-) anions. Water, CO
and NH3 are examples of molecular ligands.
All complex ions are Lewis adducts. The metal acts as a Lewis acid
and the ligand acts as a Lewis base.
We will consider equilibria of hydrated ions with ligands
other than water.
19-43
Cr(NH3)63+ is a
typical complex
ion.
Cr3+ is the central metal,
and is surrounded by six
NH3 ligands
Figure 19.13
19-44
Formation Constants, Kf, of Complex Ions
M(H2O)42+(aq) + 4NH3(aq)
M(NH3)42+(aq) + 4H2O(l)
Kc = [M(NH3)42+][H2O]4/[M(H2O)42+][NH3]4
Kf = Kc/[H2O]4 = [M(NH3)42+]/[M(H2O)42+][NH3]4
In fact, four sequential reactions occur, defined by four Kf values,
for the systematic substitution of H2O ligands by NH3 ligands.
Thus, Kf = Kf1 x Kf2 x Kf3 x Kf4
19-45
The stepwise exchange of
NH3 for H2O in M(H2O)42+
NH3
M(H2O)42+
3NH3
M(H2O)3(NH3)2+
Figure 19.14
19-46
M(NH3)42+
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-47
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 50.0 L of 0.0020 M Zn(H2O)42+ and 25.0 L
of 0.15 M NH3. What is the final [Zn(H2O)42+]? Kf of Zn(NH3)42+
is 7.8 x 108.
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. The large
excess of NH3 will drive the reaction to completion.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq)
Kf =
[Zn(NH3)42+]
Zn(NH3)42+(aq) + 4H2O(l)
[Zn(H2O)42+]initial = (50.0 L)(0.0020 M)
[Zn(H2O)42+][NH3]4
[NH3]initial =
75.0 L
(25.0 L)(0.15 M)
75.0 L
19-48
= 1.3 x 10-3 M
= 5.0 x 10-2 M
Sample Problem 19.10 (continued)
Since we assume that all of the Zn(H2O)42+ has reacted, it would consume
four times its amount in NH3..
[NH3]used = 4 (1.3 x 10-3 M) = 5.2 x 10-3 M
[Zn(H2O)42+]remaining = x (a very small amount)
concentration (M)
initial
change
Zn(H2O)42+(aq) + 4NH3(aq)
1.3 x 10-3
~(-1.3 x 10-3)
equilibrium
Zn(NH3)42+(aq) + 4H2O(l)
5.0 x 10-2
~(-5.2 x 10-3)
4.5 x 10-2
x
0
-
~(+1.3 x 10-3)
1.3 x 10-3
-
_____________________________________________________________
[Zn(NH3)42+
Kf =
]
[Zn(H2O)42+][NH3]4
7.8 x
108
=
(1.3 x 10-3)
x (4.5 x 10-2)4
x = 4.1 x 10-7 M
A result of the
very large Kf!
19-49
Solubility Issues
A ligand increases the solubility of a slightly soluble ionic compound
if it forms a complex ion with the cation.
ZnS(s) + H2O(l)
Zn2+(aq) + HS-(aq) + OH-(aq)
Ksp = 2.0 x 10-22
Upon addition of some 1.0 M NaCN:
Zn2+(aq) + 4CN-(aq)
Zn(CN)42+(aq)
Kf = 4.2 x 1019
The overall equation is:
ZnS(s) + 4CN-(aq) + H2O(l)
Zn(CN)42+(aq) + HS-(aq) + OH-(aq)
Koverall = Ksp x Kf = 8.4 x 10-3
19-50
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) 1.0 M
hypo. Kf of Ag(S2O3)23- is 4.7 x 1013 and Ksp AgBr is 5.0 x 10-13.
PLAN: Write equations for the reactions involved. Use Ksp to find S, the molar
solubility. Consider the shift in the equilibrium upon the addition of the
complexing agent.
SOLUTION:
(a)
(b)
19-51
AgBr(s)
Ag+(aq) + Br-(aq)
Ksp = [Ag+][Br-] = 5.0 x 10-13
S = [AgBr]dissolved = [Ag+] = [Br-] Ksp = S2 = 5.0 x 10-13; S = 7.1 x 10-7 M
AgBr(s)
Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq)
Ag(S2O3)23-(aq)
AgBr(s) + 2S2O32-(aq)
Br-(aq) + Ag(S2O3)23-(aq)
Sample Problem 19.11 (continued)
[Br-][Ag(S2O3)23- ]
Koverall = Ksp x Kf =
= (5.0 x 10-13)(4.7 x 1013)= 24
[S2O32-]2
concentration (M)
AgBr(s) + 2S2O32-(aq)
Br-(aq) + Ag(S2O3)23-(aq)
initial
-
1.0
0
0
change
-
-2S
+S
+S
equilibrium
-
S
S
1.0 - 2S
__________________________________________________________
Koverall =
S2
= 24
(1.0 - 2S)2
S
1.0 - 2S
S = [Ag(S2O3)23-] = 0.45 M
19-52
= (24)1/2
Complex Ions of Amphoteric Hydroxides
Amphoteric hydroxides: compounds that dissolve very little in water
but to a much greater extent in acidic and basic solutions
Al(OH)3(s)
Al3+(aq) + 3OH-(aq)
Ksp = 3 x 10-34
Al(OH)3 dissolves in acid: 3H3O+(aq) + 3OH- (aq)
Al(OH)3(s) + 3H3O+(aq)
6H2O(l)
Al3+(aq) + 6H2O(l)
Al(OH)3 dissolves in base through the formation of a complex ion:
Al(OH)3(s) + OH-(aq)
19-53
Al(OH)4-(aq)
The amphoteric behavior of aluminum hydroxide
19-54
Figure 19.15
Selective Precipitation
Selection of an ion in solution by precipitation; achieved by
adding a precipitating agent (ion) to the solution until the
Qsp of the more soluble compound is almost equal to its
Ksp (but Qsp < Ksp). This ion remains in solution, whereas
the other ion(s) having Qsp > Ksp precipitate.
19-55
Sample Problem 19.12
Separating ions by selective precipitation
PROBLEM: A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2. Calculate
the [OH-] that would separate the metal ions as their hydroxides.
Ksp of Mg(OH)2 = is 6.3 x 10-10; Ksp of Cu(OH)2 = 2.2 x 10-20.
PLAN: Both precipitates have the same ion ratio, 1:2, so we can compare
their Ksp values to determine which has the greater solubility.
Cu(OH)2 will precipitate first (it has the smaller Ksp) so we calculate
the [OH-] needed for a saturated solution of Mg(OH)2. This will
ensure that we do not precipitate Mg(OH)2. We can then check how
much Cu2+ remains in solution.
SOLUTION:
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq) Ksp = 6.3 x 10-10
Cu(OH)2(s)
Cu2+(aq) + 2OH-(aq)
[OH-] needed for a saturated Mg(OH)2 solution =
Ksp = 2.2 x 10-20
K sp
[Mg 2 ]
6.3x1010

0.20
= 5.6 x 10-5 M
19-56
Sample Problem 19.12 (continued)
Use the Ksp for Cu(OH)2 to find the amount of Cu2+ remaining in
solution.
[Cu2+] = Ksp/[OH-]2 = 2.2 x 10-20/(5.6 x 10-5)2 =7.0 x 10-12 M
Since the solution was 0.10 M CuCl2, virtually none of the
Cu2+ remains in solution.
19-57
Qualitative Analysis
Self-Study: pp. 841-845 in textbook
Only an overview of this material
will be provided in class. You will be responsible,
however, for this material for the Final Exam.
19-58
General procedure for separating ions in qualitative analysis
19-59
add
precipitating
ion
centrifuge
centrifuge
add
precipitating
ion
Figure 19.16
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A qualitative analysis scheme for separating cations into five ion
groups
Figure 19.17
19-60
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
add
(NH4)2HPO4
centrifuge
add
NH3/NH4+
buffer(pH 8)
centrifuge
centrifuge
centrifuge
add
6 M HCl
acidify to
pH 0.5;
add H2S
19-61
Tests to determine the
presence of cations in
ion group 5
(a) flame test for Na+: yellow-orange
(b) flame test for K+: violet
(c) NH4+ + OH- : pH test for NH3 gas
Figure 19.18
19-62
QuickTime™ and a
Photo - JPEG decompressor
are needed to see this picture.
A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+
Figure 19.18
Step 3: Add
NaOH
centrifuge
Step 4:
Add HCl,
Na2HPO4
19-63
Step 2:
Add HCl
centrifuge
centrifuge
Step 1:
Add
NH3(aq)
Step 5:
Dissolve in
HCl and
add KSCN
End of Assigned Material
19-64
A view inside Carlsbad Caverns, New Mexico
Figure B19.1
19-65
Formation of acidic precipitation
Figure B19.2
19-66
The effect of acid rain on statuary
Location: New York City
Figure B19.4
19-67
A forest
damaged by
acid rain
A forest damaged by acid rain
Figure B19.3
19-68