19-2
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
19-3
Figure 19.1
The effect of addition of acid or base to …
an unbuffered solution.
a buffered solution.
acid added base added
acid added base added
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19-4
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 conjugateacid-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 containinga given ion is added to an equilibrium mixture that alreadycontains that ion and the position of the equilibrium shiftsaway from forming more of it.
19-5
Table 19.1
The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]initial [CH3COO-]added% dissociation* pH
* % dissociation = [CH3COOH]dissoc
[CH3COOH]initial
x 100
0.10 0.00
0.10 0.050
0.10
0.10 0.10
0.15
1.3
0.036
0.018
0.012
2.89
4.44
4.74
4.92
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-6
Figure 19.2
How an acetic acid/acetate buffer works
buffer with equal concentrations of weak acid and its conjugate
base
OH-H3O+
buffer after addition of H3O+
H2O + CH3COOH H3O+ + CH3COO-
buffer after addition of OH-
CH3COOH + OH- H2O + CH3COO-
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
HA scavenges OH-, A- scavenges H+
19-7
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-.
19-8
Sample Problem 19.1 Calculating the effect of added H3O+ and OH-
on buffer pH
PROBLEM: 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: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.
initial
change
equilibrium
0.50+ x
0.50 - x
--
-
0.50 0+ x
0.50 + x x
- x
SOLUTION:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration (M)
(a)
___________________________________________________________
19-9
Sample Problem 19.1 (continued)
[CH3COOH]eq ≈ 0.50 M [CH3COO-]eq ≈ 0.50 M[H3O+] = x
Ka =[H3O
+][CH3COO-]
[CH3COOH][H3O
+] = x = Ka
[CH3COO-]
[CH3COOH]= 1.8 x 10-5 M
Check the assumption: 1.8 x 10-5/0.50 X 100 = 3.6 x 10-3 %
CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l)concentration (M)
before addition
addition
after addition
(b)[OH-]added =
0.020 mol
1.0 L soln= 0.020 M NaOH
0.50 - 0.50 -
-
-
- 0.020 -
0.48 0 0.52___________________________________________________________
pH = 4.74
19-10
Sample Problem 19.1 (continued)
Set up a reaction table with the new values.
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration (M)
initialchange
equilibrium
0.48 -- x
0.48 - x
-
-
0.52 0
x
+ x + x
0.52 + x
[H3O+] = 1.8 x 10-5 x
0.48
0.52= 1.7 x 10-5 pH = 4.77
CH3COO-(aq) + H3O+(aq) CH3COOH(aq) + H2O
(l)concentration (M)
before addition
addition
after addition
(c) [H3O+]added =
0.020 mol
1.0 L soln= 0.020 M H3O
+
0.50 - 0.50 -
-
-
- 0.020 -
0.48 0 0.52
___________________________________________________________
___________________________________________________________
19-11
Sample Problem 19.1 (continued)
Set up a reaction table with the new values.
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration (M)
initialchange
equilibrium
0.52 -- x
0.52 - x
-
-
0.48 0
x
+ x + x
0.48 + x
[H3O+] = 1.8 x 10-5 x
0.48
0.52= 2.0 x 10-5 pH = 4.70
___________________________________________________________
Note that this buffer resists changes in pH fromadditions of either strong acid or strong base!
(What is the solution pH of 0.020 M HCl or 0.020 M NaOH?)
19-12
The Henderson-Hasselbalch Equation
An equation that relates pH, pKa and the ratio ofweak acid to its conjugate base for a buffer solution.
HA + H2O H3O+ + A-
Ka = [H3O+][A-]/[HA]
[H3O+] = Ka x [HA]/[A-]
-log [H3O+] = -log Ka - log ([HA]/[A-])
pH = pKa + log ([A-]/[HA])
pH = pKa + log ([base]/[acid])
Special case: when [base] = [acid], the pH of the buffersolution equals the pKa of the weak acid.
or
19-13
Figure 19.3
The relationship between buffer capacity and pH change
Buffer capacity refers to theability of a buffer to resist
pH change; buffer capacity increases as the concentrations
of its components (i.e., theweak acid and its conjugate
base) increase.
Buffer pH and buffer capacityare different concepts.
For an acetic acid/acetate buffer
19-14
A buffer has the highest capacity when the
concentrations of HA and A- are equal.
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.
Key Concepts
Buffer range: the pH range over which the buffer acts effectively;defined as pKa +/- 1 pH unit
A buffer whose pH is equal to or near the pKa of its acid component has the highest buffer capacity.
19-15
Sample Problem 19.2 Preparing a buffer
SOLUTION:
PROBLEM: 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.
PLAN: 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.
HCO3-(aq) + H2O(l) CO3
2-(aq) + H3O+(aq)
Ka =[CO3
2-][H3O+]
[HCO3-]
pH = 10.00; [H3O+] = 1.0 x 10-10 4.7 x 10-11 =
[CO32-][10-10]
[0.20][CO3
2-] = 0.094 M
moles of Na2CO3 = (1.5 L)(0.094 mol/L) = 0.14 moles
0.14 moles 105.99 g
mol= 15 g Na2CO3
x
19-16
Acid-Base Titration Curves
Acid-Base Indicator: a weak organic acid (HIn) that has a
different color than its conjugate base (In-); small amountsare used in acid-base titrations; change color over differentpH ranges
19-18
Figure 19.5
The color change of the indicator bromthymol blue
acidic
basic
change occurs over ~2 pH
units
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pH 6
pH 7.5
19-19 Figure 19.6
Curve for a strong acid-strong base titration
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
40 mL of 0.1000 M HCl
19-20
Figure 19.7
Curve for a weak acid-
strong base titration
Titration of 40.00 mL of 0.1000 M HPr with 0.1000 M NaOH
[HPr] = [Pr-]
pH = 8.80 at equivalence pointpKa of HPr =
4.89
methyl red
HPr = CH3CH2COOH
19-21
Sample Problem 19.3 Calculating the pH during a weak acid-strong base titration
PROBLEM: 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
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) x = 1.1 x 10-3 ; pH = 2.96
(b)
before addition
addition
after addition
0.004000
0.003000
0.0030000.001000
0 -
-
-0
-
- -
HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)amount (mol)
_________________________________________________
19-22
Sample Problem 19.3 (continued)
[H3O+] = 1.3 x 10-5 x
0.001000 mol
0.003000 mol = 4.3 x 10-6 M pH = 5.37
(c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr-]will be: 0.004000 mol
0.04000 L + 0.04000 L= 0.05000 M
Ka x Kb = Kw Kb = Kw/Ka = 1.0 x 10-14/1.3 x 10-5 = 7.7 x 10-10
[H3O+] = Kw / = 1.6 x 10-9 M
€
Kbx[Pr−] 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
M = 0.00100 mol0.0900 L
M = 0.01111
[H3O+] = 1.0 x 10-14/0.01111 = 9.0 x 10-13 M
pH = 12.05
19-23
Figure 19.8
Curve for a weak base-strong acid
titration
Titration of 40.00 mL of 0.1000 M NH3 with 0.1000 M HCl
pH = 5.27 at equivalence
point
pKa of NH4+
= 9.25
19-24
pKa2 = 7.19
pKa1 = 1.85
Figure 19.9
Curve for the titration of a weak polyprotic
acid.
Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH
sulfurous acid: a diprotic weak acid
19-25
Figure 19.10
Sickle shape of red blood cells in sickle cell anemia
Single-site mutationsin the hemoglobin
molecule can change thenet charge on the
protein, which causesprotein aggregationand a consequent
change in cellmorphology
19-26
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
19-27
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture.
A slightly solubleionic compound
PbCl2
19-28
Sample Problem 19.4 Writing ion-product expressions for slightly soluble ionic compounds
SOLUTION:
PROBLEM: Write the ion-product expression for each of the following:
(a) magnesium carbonate (b) iron(II) hydroxide(c) calcium phosphate (d) silver sulfide
PLAN: Write an equation which describes a saturated solution. Take note of the unusual behavior of the sulfide ion produced in (d).
Ksp = [Mg2+][CO32-]
(a) MgCO3(s) Mg2+(aq) + CO32-(aq)
Ksp = [Fe2+][OH-]2
(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)
Ksp = [Ca2+]3[PO4
3-]2
(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)
(d) Ag2S(s) 2Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l) HS-(aq) + OH-(aq)
Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq) Ksp = [Ag+]2[HS-][OH-]
19-29
Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 25 oC
Name and Formula Ksp
aluminum hydroxide, Al(OH)3
cobalt(II) carbonate, CoCO3
iron(II) hydroxide, Fe(OH)2
lead(II) fluoride, PbF2
lead(II) sulfate, PbSO4
silver sulfide, Ag2S
zinc iodate, Zn(IO3)2
3 x 10-34
1.0 x 10-10
4.1 x 10-15
3.6 x 10-8
1.6 x 10-8
4.7 x 10-29
8 x 10-48
mercury(I) iodide, Hg2I2
3.9 x 10-6
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The magnitude of Ksp is a measure of how far to the right the dissolutionproceeds at equilibrium (i.e., at saturation).
19-30
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.
Ksp = [Pb2+][SO42-]
4.25 x 10-3 g100 mL soln
1000 mL
L 303.3 g PbSO4
mol PbSO4= 1.40 x 10-4 M PbSO4
Ksp = [Pb2+][SO42-] = (1.40 x 10-4)2 = 1.96 x 10-8
SOLUTION: PbSO4(s) Pb2+(aq) + SO42-(aq)(a)
x x
19-31
Sample Problem 19.5 (continued)
(b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2
0.64 g
L soln 245.2 g PbF2
mol 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
x
[Pb2+] [F-]
19-32
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: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2
-initial
change
equilibrium
-
-
0 0
+S + 2S
S 2S
Ksp = (S)(2S)2 = 4S3 S =
€
6.5x10−6
43
= 1.2 x 10-2 M
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)concentration (M)
____________________________________________________
19-33
Ksp and Solubilities
Ksp values are used to determine relative solubilities provided that comparisonsare made between compounds whose formulas contain the same totalnumber of ions. The compound having the higher Ksp is more soluble.
19-34
Table 19.3 Relationship Between Ksp and Solubility at 25 oC
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)21:2 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 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5
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19-35
Common Ion Effects on Solubility
The presence of a common ion decreases the solubility of aslightly 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)
19-36
Figure 19.11
The effect of a common ion on solubility
PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4
2-(aq)
CrO42- added
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-37
Sample Problem 19.7Calculating 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 0.10 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: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)concentration (M)
initial
change
equilibrium
-
-
-
0.10 0
+S +2S
0.10 + S 2S
Ksp = 6.5 x 10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2 (S << 0.10)
S ≈ = 4.0 x 10-3 M Check the assumption:
4.0 x 10-3 M 0.10 M
x 100 = 4.0 %
_____________________________________________
(6.5 x 10-5/4)0.5
= 1.2 x 10-2 M
19-38
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) + H2O(l)
HCO3-(aq) + H3O
+(aq) H2CO3(aq) + H2O(l)
H2CO3(aq) 2H2O(l) + CO2(g)
Adding H3O+ shifts the equilibrium to the right.
19-39
Figure 19.12
Test for the presence of a
carbonate
Effect of the additionof a strong acid
(release of carbondioxide)
19-40
Sample Problem 19.8Predicting the effect of adding a strong acid
on solubility
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.
Br- is the anion of a strong acid. No effect.
SOLUTION: (a) PbBr2(s) Pb2+(aq) + 2Br-(aq)
(b) Cu(OH)2(s) 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.
(c) FeS(s) Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will
react with water to produce OH-.
Both weak acids serve to increase the solubility of FeS.
FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)
19-41
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.
19-42
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 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?
PLAN: 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.
19-43
Equilibria Involving Complex Ions
A complex ion consists of a central metal ion covalently bondedto two or more anions (or molecules) called ligands. Ionic ligands include
hydroxide (OH-), chloride (Cl-) and cyanide (CN-) anions. Water, COand NH3 are examples of molecular ligands.
All complex ions are Lewis adducts. The metal acts as a Lewis acidand the ligand acts as a Lewis base.
We will consider equilibria of hydrated ions with ligands other than water.
19-44
Figure 19.13
Cr(NH3)63+
is a typical complex ion.
Cr3+ is the central metal,and is surrounded by six
NH3 ligands
19-45
Formation Constants, Kf, of Complex Ions
M(H2O)42+(aq) + 4NH3(aq) M(NH3)4
2+(aq) + 4H2O(l)
Kc = [M(NH3)42+][H2O]4/[M(H2O)4
2+][NH3]4
Kf = Kc/[H2O]4 = [M(NH3)42+]/[M(H2O)4
2+][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
19-46
Figure 19.14
The stepwise exchange of NH3 for H2O in
M(H2O)42+
M(H2O)42+
M(H2O)3(NH3)2+
M(NH3)42+
NH3
3NH3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-48
Sample Problem 19.10 Calculating the concentration of a complex ion
SOLUTION:
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)4
2+ and 25.0 L
of 0.15 M NH3. What is the final [Zn(H2O)42+]? Kf of Zn(NH3)4
2+ 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.
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4
2+(aq) + 4H2O(l)
Kf =[Zn(NH3)4
2+]
[Zn(H2O)42+][NH3]
4
[Zn(H2O)42+]initial = (50.0 L)(0.0020 M)
75.0 L= 1.3 x 10-3 M
[NH3]initial = (25.0 L)(0.15 M)
75.0 L= 5.0 x 10-2 M
19-49
Sample Problem 19.10 (continued)
Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4
2+(aq) + 4H2O(l)concentration (M)
initial
change
equilibrium
1.3 x 10-3 5.0 x 10-2 0 -
~(-1.3 x 10-3) ~(-5.2 x 10-3)
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
~(+1.3 x 10-3) -
x 4.5 x 10-2 1.3 x 10-3
[Zn(H2O)42+]remaining = x (a very small amount)
-
Kf =[Zn(NH3)4
2+]
[Zn(H2O)42+][NH3]
4 7.8 x 108 = (1.3 x 10-3)
x (4.5 x 10-2)4x = 4.1 x 10-7 M
_____________________________________________________________
A result of thevery large Kf!
19-50
Solubility Issues
A ligand increases the solubility of a slightly soluble ionic compoundif 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
19-51
Sample Problem 19.11 Calculating the effect of complex ion formation on solubility
SOLUTION:
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.
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(a)
(b) AgBr(s) Ag+(aq) + Br-(aq)
Ag+(aq) + 2S2O32-(aq) Ag(S2O3)2
3-(aq)
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2
3-(aq)
19-52
Sample Problem 19.11 (continued)
Koverall = Ksp x Kf = [Br-][Ag(S2O3)2
3- ]
[S2O32-]2
= (5.0 x 10-13)(4.7 x 1013) = 24
AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2
3-(aq)concentration (M)
initial
change
equilibrium
-
-
-
1.0
-2S
1.0 - 2S
0 0
+S +S
S S
Koverall =S2
(1.0 - 2S)2
= 24S
1.0 - 2S= (24)1/2
S = [Ag(S2O3)23-] = 0.45 M
__________________________________________________________
19-53
Complex Ions of Amphoteric Hydroxides
Amphoteric hydroxides: compounds that dissolve very little in waterbut 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) 6H2O(l)
Al(OH)3 dissolves in base through the formation of a complex ion:
Al(OH)3(s) + OH-(aq) Al(OH)4-(aq)
Al(OH)3(s) + 3H3O+(aq) Al3+(aq) + 6H2O(l)
19-55
Selective Precipitation
Selection of an ion in solution by precipitation; achieved byadding a precipitating agent (ion) to the solution until theQsp 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.
19-56
Sample Problem 19.12 Separating ions by selective precipitation
SOLUTION:
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.Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3 x 10-10
Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2 x 10-20
[OH-] needed for a saturated Mg(OH)2 solution =
€
Ksp
[Mg2+]
€
=6.3x10−10
0.20
= 5.6 x 10-5 M
19-57
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.
19-58
Qualitative Analysis
Self-Study: pp. 841-845 in textbook
Only an overview of this materialwill be provided in class. You will be responsible,
however, for this material for the Final Exam.
19-59 Figure 19.16
General procedure for separating ions in qualitative analysis
add precipitating
ion
cent
rifug
e
add precipitating
ion
cent
rifug
e
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-60
Figure 19.17
A qualitative analysis scheme for separating cations into five ion groups
add 6 M HCl
cent
rifug
eacidify to pH 0.5;
add H2S
cent
rifug
e
add
NH3/NH4+
buffer(pH 8)
cent
rifug
e
add (NH4)2HPO4
cent
rifug
e
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
19-62
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture.
Tests to determine thepresence 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
19-63
Figure 19.18
Step 1: Add
NH3(aq)
cent
rifug
e
cent
rifug
e
Step 2: Add HCl
Step 3: Add NaOH
cent
rifug
eStep 4:
Add HCl, Na2HPO4
Step 5: Dissolve in
HCl and add KSCN
A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+