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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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Figure 19.1
The effect of addition of acid or base to …
acid added
Figure 19.2
base added
an unbuffered solution
acid added
base added
or a buffered solution
19-3
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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
19-4
% 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
Buffer after addition of OH-
OH-
H3O+ + CH3COO-
CH3COOH + OH-
H2O + CH3COO-
19-5
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Calculating the Effect of Added H3O+ or OHon Buffer pH
Calculate the pH:
Sample Problem 19.1
PROBLEM:
(a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa
(b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a)
(c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a)
Ka of CH3COOH = 1.8x10-5. (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 step-wise through changes in the system.
SOLUTION:
(a)
Concentration (M)
Initial
Change
Equilibrium
19-6
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
0.50
-x
-
0.50
+x
0
+x
0.50-x
-
0.50 +x
x
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Calculating the Effect of Added H3O+ and OHon Buffer pH
Sample Problem 19.1
continued (2 of 4)
[CH3COOH]equil ≈ 0.50M
[H3O+] = x
Ka =
[H3O+][CH3COO-]
[CH3COOH]
Check the assumption:
(b)
[H3O+] = x = Ka
[CH3COOH]
[CH3COO-]
1.0L soln
= 0.020M NaOH
CH3COOH(aq) + OH-(aq)
Concentration (M)
Before addition
Addition
After addition
= 1.8x10-5M
1.8x10-5/0.50 X 100 = 3.6x10-3 %
0.020 mol
[OH-]added =
[CH3COO-]initial ≈ 0.50M
CH3COO-(aq) + H2O (l)
0.50
-
0.020
0.50
-
-
0.48
0
0.52
-
19-7
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Calculating the Effect of Added H3O+ and OHon Buffer pH
Sample Problem 19.1
continued (3 of 4)
Concentration (M)
Set up a reaction table with the new values.
Initial
Change
Equilibrium
[H3O+] = 1.8x10-5
(c)
[H3O+]added =
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
0.48
0.48
-x
-
0.52
+x
0
+x
0.48 -x
-
0.52 +x
x
= 1.7x10-5
0.52
0.020 mol
1.0L soln
pH = 4.77
= 0.020M H3O+
Concentration (M) CH3COO-(aq) + H3O+(aq)
Before addition
Addition
After addition
19-8
0.50
0.48
0.020
0
CH3COOH(aq) + H2O (l)
0.50
0.52
-
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Calculating the Effect of Added H3O+ and OHon Buffer pH
Sample Problem 19.1
continued (4 of 4)
Concentration (M)
Set up a reaction table with the new values.
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
Initial
Change
Equilibrium
0.52
[H3O+] = 1.8x10-5
0.48
0.52
-x
-
0.48
+x
0
+x
0.52 -x
-
0.48 +x
x
= 2.0x10-5
pH = 4.70
19-9
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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]
19-10
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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.
19-11
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Figure 19.4
19-12
The relation between buffer capacity and pH change.
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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.
19-13
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Sample Problem 19.2
PROBLEM:
Preparing a Buffer
An environmental chemist needs a carbonate buffer of pH 10.00
to study the effects of the acid rain on limsetone-rich soils. How
many grams of Na2CO3 must she add to 1.5L of freshly prepared
0.20M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.
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+]
CO32-(aq) + H3O+(aq) Ka =
HCO3-(aq) + H2O(l)
[HCO3-]
PLAN:
pH = 10.00; [H3O+] = 1.0x10-10
4.7x10-11 =
[CO32-](0.20)
1.0x10-10
moles of Na2CO3 = (1.5L)(0.094mols/L) = 0.14
0.14 moles
19-14
105.99g
mol
= 15 g Na2CO3
[CO32-] = 0.094M
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Figure 19.5
Colors and approximate pH range of some
common acid-base indicators.
pH
19-15
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Figure 19.6 The color change of the indicator bromthymol blue.
basic
acidic
19-16
change occurs
over ~2pH units
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Figure 19.7
Curve for a strong acid-strong
base titration
19-17
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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-]
19-18
methyl red
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Sample Problem 19.3
Calculating the pH During a Weak AcidStrong Base Titration
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.
PROBLEM:
Ka = [Pr-][H3O+]/[HPr]
[Pr-] = x = [H3O+]
x = (1.3x10−5 )(0.10)
(b)
[Pr-] = x = [H3O+]
x = 1.1x10-3 ; pH = 2.96
Amount (mol)
HPr(aq) + OH-(aq)
Before addition
Addition
0.04000
0.03000
After addition
0.01000
Pr-(aq) + H2O (l)
0
-
0
-
0.03000
-
19-19
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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
0.003000 mol
= 4.3x10-6M
pH = 5.37
(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)
Ka x Kb = Kw
Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10
[H3O+] = Kw / K b x[Pr − =
] 1.6x10-9M
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
19-20
M = (0.00100)
(0.0900L)
M = 0.01111
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Titration of 40.00mL of 0.1000M NH3
with 0.1000M HCl
Figure 19.9
pKa of NH4+ =
9.25
Curve for a
weak basestrong acid
titration
pH = 5.27 at
equivalence
point
19-21
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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
19-22
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Figure 19.11
Sickle shape of red blood cells in sickle cell anemia.
19-23
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Ion-Product Expression (Qsp)
and Solubility Product Constant (Ksp)
For the hypothetical compound, MpXq
At equilibrium
19-24
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:
(a) MgCO3(s)
Mg2+(aq) + CO32-(aq)
(b) Fe(OH)2(s)
Fe2+(aq) + 2OH- (aq)
(c) Ca3(PO4)2(s)
(d) Ag2S(s)
Ksp = [Mg2+][CO32-]
Ksp = [Fe2+][OH-] 2
3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO43-]2
2Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l)
Ag2S(s) + H2O(l)
HS-(aq) + OH-(aq)
Ksp = [Ag+]2[HS-][OH-]
2Ag+(aq) + HS-(aq) + OH-(aq)
19-25
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Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 250C
Name, Formula
19-26
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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Sample Problem 19.5
Determining Ksp from Solubility
PROBLEM: (a) Lead (II) sulfate is a key component in lead-acid car batteries.
Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is
the Ksp of PbSO4?
(b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C,
the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2.
PLAN:
Write the dissolution equation; find moles of dissociated ions;
convert solubility to M and substitute values into solubility product
constant expression.
SOLUTION: (a) PbSO4(s)
Pb2+(aq) + SO42-(aq)
4.25x10-3g
1000mL
mol PbSO4
100mL soln
L
303.3g PbSO4
Ksp = [Pb2+][SO42-] = (1.40x10-4)2 =
Ksp = [Pb2+][SO42-]
= 1.40x10-4M PbSO4
1.96x10-8
19-27
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Sample Problem 19.5
Determining Ksp from Solubility
continued
(b) PbF2(s)
0.64g
L soln
Pb2+(aq) + 2F-(aq)
mol PbF2
245.2g PbF2
= 2.6x10-3 M
Ksp = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8
19-28
Ksp = [Pb2+][F-]2
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Sample Problem 19.6
Determining Solubility from Ksp
PROBLEM: Calcium hydroxide (slaked lime) 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 if the Ksp is 6.5x10-6.
PLAN:
Write out a dissociation equation and Ksp expression; Find the molar
solubility (S) using a table.
SOLUTION:
Ca(OH)2(s)
Concentration (M)
Ca2+(aq) + 2OH-(aq)
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
S=
3
6.5x10−6
4
= 1.2x10x-2M
19-29
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Table 19.3 Relationship Between Ksp and Solubility at 250C
No. of Ions
19-30
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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Figure 19.12
The effect of a common ion on solubility
CrO42- added
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
19-31
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Sample Problem 19.7
Calculating the Effect of a Common Ion on
Solubility
PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2
in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of
Ca(OH)2 is 6.5x10-6.
PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2.
The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar
solubility of the ions involved.
SOLUTION:
Concentration(M)
Ca(OH)2(s)
Initial
-
0.10
0
Change
-
+S
+2S
Equilibrium
-
0.10 + S
2S
Ksp = 6.5x10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2
S=
6.5x10−6
4
= 4.0x10-3
S << 0.10
Check the assumption:
4.0x10-3
0.10M
19-32
Ca2+(aq) + 2OH-(aq)
x 100 = 4.0%
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Figure 19.13
Test for the presence of a carbonate.
19-33
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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.
(c) FeS(s)
Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will
react with water to produce OH-.
2+
Fe (aq) + HS (aq) + OH-(aq)
FeS(s) + H2O(l)
Both weak acids serve to increase the solubility of FeS.
19-34
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Sample Problem 19.9
Predicting Whether a Precipitate Will Form
PROBLEM: 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?
PLAN:
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 =
Ksp = 3.2x10-11
[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.
19-35
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Figure 19.14
19-36
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+
19-37
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19-38
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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.
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.
PLAN:
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq)
[Zn(NH3)42+]
Kf =
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
[Zn(H2O)42+][NH3]4
75.0L
19-39
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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
19-40
5.0x10-2
~(-5.2x10-3)
x
= 7.8x108 =
4.5x10-2
(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.
AgBr(s)
SOLUTION:
Ag+(aq) + Br-(aq)
(a) S = [AgBr]dissolved = [Ag+] = [Br-]
AgBr(s)
(b)
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)
19-41
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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
19-42
= (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)
Al(H2O)3(OH)3(s)
Al(H2O)3(OH)4-(s) + H2O(l)
19-43
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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
[Mg2+ ]
= 5.6x10-5M
19-44
=
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.
19-45
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Figure 19.17
The general procedure for separating ions in qualitative analysis.
19-46
Add
precipitating
ion
Centrifuge
Centrifuge
Add
precipitating
ion
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A qualitative analysis scheme for separating cations
into five ion groups.
Add
NH3/NH4+
buffer(pH 8)
Centrifuge
Add
(NH4)2HPO4
Centrifuge
Centrifuge
Add
6M HCl
Acidify to
pH 0.5;
add H2S
Centrifuge
Figure 19.18
19-47
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
19-48
Step 5
Dissolve in
HCl and
add KSCN
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Figure B19.1
A view inside Carlsbad Caverns, New Mexico
19-49
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Figure B19.3
19-50
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
19-51
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Figure B19.5
The effect of acid rain on marble statuary.
1944
1994
Location: New York City
19-52