John C. Kotz
Paul M. Treichel
John Townsend
http://academic.cengage.com/kotz
Chapter 18
Principles of Reactivity:
Other Aspects of Aqueous Equilibria
John C. Kotz • State University of New York, College at Oneonta
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3
More About
Chemical Equilibria
Acid-Base & Precipitation Reactions
Chapter 18
PLAY MOVIE
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
4
Stomach Acidity &
Acid-Base Reactions
PLAY MOVIE
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
Acid-Base Reactions
• Strong acid + strong base
HCl + NaOH f
• Strong acid + weak base
HCl + NH3 f
• Weak acid + strong base
HOAc + NaOH f
• Weak acid + weak base
HOAc + NH3 f
© 2009 Brooks/Cole - Cengage
5
What is relative pH
before, during, &
after reaction?
Need to study:
a) Common ion
effect and buffers
b) Titrations
The Common Ion Effect
Section 18.1
QUESTION: What is the effect on the pH of adding
NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O e NH4+(aq) + OH-(aq)
Here we are adding NH4+, an ion COMMON to the
equilibrium.
Le Chatelier predicts that the equilibrium will shift to the
left (1), right (2), no change (3).
The pH will go up (1), down (2), no change (3).
NH4+ is an acid!
© 2009 Brooks/Cole - Cengage
6
pH of Aqueous NH3
QUESTION: What is the effect on the pH of adding NH4Cl to
0.25 M NH3(aq)?
NH3(aq) + H2O e NH4+(aq) + OH-(aq)
Let us first calculate the pH of a 0.25 M NH3
solution.
[NH3]
[NH4+]
[OH-]
initial
0.25
0
0
change
-x
+x
+x
equilib
0.25 - x
x
x
© 2009 Brooks/Cole - Cengage
7
pH of Aqueous NH3
QUESTION: What is the effect on the pH of adding NH4Cl
to 0.25 M NH3(aq)?
NH3(aq) + H2O e NH4+(aq) + OH-(aq)
-5
Kb = 1.8 x 10
=
[NH4+ ][OH- ]
[NH3 ]
x2
=
0.25 - x
Assuming x is << 0.25, we have
[OH-] = x = [Kb(0.25)]1/2 = 0.0021 M
This gives pOH = 2.67
and so pH = 14.00 - 2.67
= 11.33 for 0.25 M NH3
© 2009 Brooks/Cole - Cengage
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9
pH of NH3/NH4+ Mixture
Problem: What is the pH of a solution with 0.10
M NH4Cl and 0.25 M NH3(aq)?
NH3(aq) + H2O e NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding
NH4Cl. Let’s test that!
[NH3]
[NH4+]
[OH-]
initial
change
equilib
© 2009 Brooks/Cole - Cengage
10
pH of NH3/NH4+ Mixture
Problem: What is the pH of a solution with 0.10
M NH4Cl and 0.25 M NH3(aq)?
NH3(aq) + H2O e NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding
NH4Cl. Let’s test that!
[NH3]
[NH4+]
[OH-]
initial
0.25
0.10
0
change
-x
+x
+x
equilib
0.25 - x
0.10 + x
x
© 2009 Brooks/Cole - Cengage
11
pH of NH3/NH4+ Mixture
Problem: What is the pH of a solution with 0.10 M NH4Cl
and 0.25 M NH3(aq)?
NH3(aq) + H2O e NH4+(aq) + OH-(aq)
Kb = 1.8 x 10-5 =
[NH4+ ][OH- ]
[NH3 ]
=
x(0.10 + x)
0.25 - x
Assuming x is very small,
[OH-] = x = (0.25 / 0.10)(Kb) = 4.5 x 10-5 M
This gives pOH = 4.35 and pH = 9.65
pH drops from 11.33 to 9.65
on adding a common ion
© 2009 Brooks/Cole - Cengage
Buffer Solutions
Section 18.2
HCl is added to
pure water.
PLAY MOVIE
HCl is added to a
solution of a weak
acid H2PO4- and its
conjugate base
HPO42-.
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
12
Buffer Solutions
A buffer solution is a special case of the
common ion effect.
The function of a buffer is to resist changes
in the pH of a solution.
Buffer Composition
Weak Acid
+
Conj. Base
HOAc
+
OAcH2PO4+
HPO42NH4+
+
NH3
© 2009 Brooks/Cole - Cengage
13
Buffer Solutions
Consider HOAc/OAc- to see how buffers work
ACID USES UP ADDED OHWe know that
OAc- + H2O e HOAc + OHhas Kb = 5.6 x 10-10
Therefore, the reverse reaction of the WEAK
ACID with added OHhas Kreverse = 1/ Kb = 1.8 x 109
Kreverse is VERY LARGE, so HOAc completely
snarfs up OH- !!!!
© 2009 Brooks/Cole - Cengage
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Buffer Solutions
Consider HOAc/OAc- to see how buffers
work
CONJ. BASE USES UP ADDED H+
HOAc + H2O e OAc- + H3O+
has Ka = 1.8 x 10-5
Therefore, the reverse reaction of the WEAK
BASE with added H+
has Kreverse = 1/ Ka = 5.6 x 104
Kreverse is VERY LARGE, so OAc- completely
snarfs up H+ !
© 2009 Brooks/Cole - Cengage
15
Buffer Solutions
Problem: What is the pH of a buffer that has
[HOAc] = 0.700 M and [OAc-] = 0.600 M?
HOAc + H2O e OAc- + H3O+
Ka = 1.8 x 10-5
0.700 M HOAc has pH = 2.45
The pH of the buffer will have
© 2009 Brooks/Cole - Cengage
1.
pH < 2.45
2.
pH > 2.45
3.
pH = 2.45
16
Buffer Solutions
Problem: What is the pH of a buffer that has
[HOAc] = 0.700 M and [OAc-] = 0.600 M?
HOAc + H2O e OAc- + H3O+
Ka = 1.8 x 10-5
initial
change
equilib
© 2009 Brooks/Cole - Cengage
[HOAc]
[OAc-]
0.700
-x
0.600
+x
0.700 - x
[H3O+]
0
+x
x
0.600 + x
17
Buffer Solutions
Problem: What is the pH of a buffer that has [HOAc] = 0.700 M
and [OAc-] = 0.600 M?
HOAc + H2O e OAc- + H3O+
Ka = 1.8 x 10-5
[HOAc]
[OAc-]
[H3O+]
equilib
0.700 - x 0.600 + x x
Assuming that x << 0.700 and 0.600, we have
K a = 1.8 x 10-5 =
[H3 O+ ](0.600)
0.700
[H3O+] = 2.1 x 10-5 and pH = 4.68
© 2009 Brooks/Cole - Cengage
18
Buffer Solutions
Notice that the expression for calculating the
H+ conc. of the buffer is
[H3O+ ] =
[H3O+ ] =
Orig. conc. of HOAc
Orig. conc. of OAc
[Acid]
x Ka
[Conj. base]
-
x Ka
[OH- ] =
[Base]
x Kb
[Conj. acid]
Notice that the H3O+ or OH- concs.
depend on (1) K and (2) the ratio of
acid and base concs.
© 2009 Brooks/Cole - Cengage
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20
Henderson-Hasselbalch Equation
[Acid]
[H3O ] =
x Ka
[Conj. base]
+
Take the negative log of both sides of this
equation
[Acid]
pH  pK a - log
[Conj. base]
pH = pKa + log
[Conj. base]
[Acid]
The pH is determined largely by the pKa of the acid and
then adjusted by the ratio of acid and conjugate base.
© 2009 Brooks/Cole - Cengage
Adding an Acid to a Buffer
Problem: What is the pH when 1.00 mL of 1.00 M HCl
is added to
a)1.00 L of pure water (before HCl, pH = 7.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (a)
Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of
water
C1•V1 = C2 • V2
C2 = 1.00 x 10-3 M = [H3O+]
pH = 3.00
© 2009 Brooks/Cole - Cengage
21
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (after HCl, pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH before = 4.68)
Solution to Part (b)
Step 1 — do the stoichiometry
H3O+ (from HCl) + OAc- (from buffer) f
HOAc (from buffer)
The reaction occurs completely because K is
very large.
© 2009 Brooks/Cole - Cengage
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Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 1—Stoichiometry
[H3O+] + [OAc-]
[HOAc]
Before rxn
0.00100 mol 0.600 mol
0.700 mol
Change
-0.00100
-0.00100
+0.00100
After rxn
0.599 mol
0
0.701 mol
© 2009 Brooks/Cole - Cengage
23
Adding an Acid to a Buffer
24
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O e H3O+ + OAc[HOAc]
[H3O+]
Before rxn (M) 0.701 mol/L 0
-x
Change (M)
+x
0.701-x
x
After rxn (M)
© 2009 Brooks/Cole - Cengage
[OAc-]
0.599 mol/L
+x
0.599 + x
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O e H3O+ + OAc[HOAc] [H3O+]
[OAc-]
After rxn
0.701-x
0.599+x
x
Because [H3O+] = 2.1 x 10-5 M BEFORE adding
HCl, we again neglect x relative to 0.701 and
0.599.
© 2009 Brooks/Cole - Cengage
25
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
a)
1.00 L of pure water (pH = 3.00)
b)
1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O e H3O+ + OAc0.701
[H3O ] =
x Ka =
x (1.8 x 10-5 )
0.599
[OAc- ]
+
[HOAc]
[H3O+] = 2.1 x 10-5 M f pH = 4.68
The pH has not changed
on adding HCl to the buffer!
© 2009 Brooks/Cole - Cengage
26
Preparing a Buffer
You want to buffer a solution at pH = 4.30.
This means [H3O+] = 10-pH = 5.0 x 10-5 M
It is best to choose an acid such that [H3O+]
is about equal to Ka (or pH ≈ pKa).
—then you get the exact [H3O+] by adjusting
the ratio of acid to conjugate base.
[Acid]
[H3O ] =
x Ka
[Conj. base]
+
© 2009 Brooks/Cole - Cengage
27
Preparing a Buffer
You want to buffer a solution at pH = 4.30 or
[H3O+] = 5.0 x 10-5 M
POSSIBLE ACIDS
Ka
HSO4- / SO421.2 x 10-2
HOAc / OAc1.8 x 10-5
HCN / CN4.0 x 10-10
Best choice is acetic acid / acetate.
© 2009 Brooks/Cole - Cengage
28
Preparing a Buffer
You want to buffer a solution at pH = 4.30 or
[H3O+] = 5.0 x 10-5 M
+
-5
[H3O ] = 5.0 x 10
Solve for
=
[HOAc]
[OAc - ]
[HOAc]/[OAc-]
(1.8 x 10-5 )
2.78
ratio = 1
Therefore, if you use 0.100 mol of NaOAc
and 0.278 mol of HOAc, you will have pH =
4.30.
© 2009 Brooks/Cole - Cengage
29
Preparing a Buffer
A final point —
CONCENTRATION of the acid and conjugate
base are not important.
It is the RATIO OF THE NUMBER OF MOLES
of each.
Result: diluting a buffer solution does
not change its pH
© 2009 Brooks/Cole - Cengage
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31
Commercial Buffers
• The solid acid and
conjugate base in
the packet are mixed
with water to give
the specified pH.
• Note that the
quantity of water
does not affect the
pH of the buffer.
© 2009 Brooks/Cole - Cengage
Preparing a Buffer
Buffer prepared from
8.4 g NaHCO3
weak acid
16.0 g Na2CO3
conjugate base
HCO3- + H2O
e H3O+ + CO32What is the pH?
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
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33
Titrations
pH
Titrant volume, mL
© 2009 Brooks/Cole - Cengage
34
Acid-Base Titrations
PLAY MOVIE
Adding NaOH from the buret to acetic acid in the flask, a weak acid.
In the beginning the pH increases very slowly.
© 2009 Brooks/Cole - Cengage
35
Acid-Base Titrations
PLAY MOVIE
Additional NaOH is added. pH rises as equivalence point is
approached.
© 2009 Brooks/Cole - Cengage
36
Acid-Base Titrations
PLAY MOVIE
Additional NaOH is added. pH increases and then levels off as
NaOH is added beyond the equivalence point.
© 2009 Brooks/Cole - Cengage
37
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
pH at
half-way
point?
Benzoic acid
+ NaOH
© 2009 Brooks/Cole - Cengage
pH at
equivalence
point?
pH of solution of
benzoic acid, a
weak acid
Acid-Base Titration
Section 18.3
QUESTION: You titrate 100. mL of a 0.025 M solution
of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
HBz + NaOH f Na+ + Bz- + H2O
C6H5CO2H = HBz
© 2009 Brooks/Cole - Cengage
Benzoate ion = Bz-
38
Acid-Base Titration
Section 18.3
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to
the equivalence point. What is the pH of the
final solution?
HBz + NaOH f Na+ + Bz- + H2O
The pH of the final solution will be
1. Less than 7
2. Equal to 7
3. Greater than 7
© 2009 Brooks/Cole - Cengage
39
40
Acid-Base Titrations
The product of the titration of benzoic acid is the
benzoate ion, Bz- .
Bz- is the conjugate base of a weak acid.
Therefore, final solution is basic.
Bz- + H2O e HBz + OH-
Kb = 1.6 x 10-10
e
+
© 2009 Brooks/Cole - Cengage
+
41
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
pH at
equivalence
point is
basic
Benzoic acid
+ NaOH
© 2009 Brooks/Cole - Cengage
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Strategy — find the conc. of the conjugate
base Bz- in the solution AFTER the
titration, then calculate pH.
This is a two-step problem
1. stoichiometry of acid-base reaction
2. equilibrium calculation
© 2009 Brooks/Cole - Cengage
42
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
STOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d
(0.100 L HBz)(0.025 M) = 0.0025 mol HBz
This requires 0.0025 mol NaOH
2. Calc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH req’d
© 2009 Brooks/Cole - Cengage
43
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
STOICHIOMETRY PORTION
25 mL of NaOH req’d
3. Moles of Bz- produced = moles HBz =
0.0025 mol
4. Calc. conc. of BzThere are 0.0025 mol of Bz- in a TOTAL
SOLUTION VOLUME of
125 mL
[Bz-] = 0.0025 mol / 0.125 L = 0.020 M
© 2009 Brooks/Cole - Cengage
44
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH at equivalence point?
Equivalence Point
Most important species in solution is benzoate
ion, Bz-, the weak conjugate base of benzoic
acid, HBz.
Bz- + H2O e HBz + OHKb = 1.6 x 10-10
[Bz-]
[HBz]
[OH-]
initial
0.020
0
0
change
-x
+x
+x
equilib
0.020 - x x
x
© 2009 Brooks/Cole - Cengage
45
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH at equivalence point?
Equivalence Point
Most important species in solution is benzoate ion, Bz-,
the weak conjugate base of benzoic acid, HBz.
Bz- + H2O e HBz + OHKb = 1.6 x 10-10
Kb = 1.6 x 10-10
x2
=
0.020 - x
x = [OH-] = 1.8 x 10-6
pOH = 5.75 and pH = 8.25
© 2009 Brooks/Cole - Cengage
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47
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH at half-way point?
pH at half-way point?
1.
<7
2.
=7
3.
>7
© 2009 Brooks/Cole - Cengage
Equivalence point
pH = 8.25
48
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH at half-way point?
pH at halfway point
© 2009 Brooks/Cole - Cengage
Equivalence point
pH = 8.25
Acid-Base Reactions
49
You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH.
What is the pH at the half-way point?
HBz + H2O e H3O+ + Bz-
Both HBz and Bzare present.
This is a BUFFER!
Ka = 6.3 x 10-5
+
[H3O ] =
[HBz]
-
[Bz ]
x Ka
At the half-way point, [HBz] = [Bz-]
Therefore, [H3O+] = Ka = 6.3 x 10-5
pH = 4.20 = pKa of the acid
© 2009 Brooks/Cole - Cengage
Acetic acid titrated with NaOH
See Fig 18.5: Weak acid
titrated with a strong base
© 2009 Brooks/Cole - Cengage
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51
Strong acid titrated with a strong base
See Figure 18.4
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52
Weak diprotic
acid (H2C2O4)
titrated with a
strong base
(NaOH)
See Figure 18.6
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53
Titration of a
1. Strong acid with strong base?
2. Weak acid with strong base?
3. Strong base with weak acid?
4. Weak base with strong acid?
5. Weak base with weak acid
6. Weak acid with weak base?
pH
Volume of titrating reagent added -->
© 2009 Brooks/Cole - Cengage
54
Weak base (NH3)
titrated with a strong
acid (HCl)
See Figure 18.7
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55
Acid-Base
Indicators
See Figure
18.8
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56
Indicators for Acid-Base Titrations
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57
Natural Indicators
Red rose extract at different pH’s and with Al3+ ions
Rose extract
In CH3OH
Add Al3+
Add HCl
© 2009 Brooks/Cole - Cengage
Add NH3
Add NH3/NH4+
58
PRECIPITATION REACTIONS
Solubility of Salts
Section 18.4
PLAY MOVIE
Lead(II) iodide
© 2009 Brooks/Cole - Cengage
Types of Chemical Reactions
• EXCHANGE REACTIONS:
AB + CD f AD + CB
– Acid-base:
CH3CO2H + NaOH f NaCH3CO2 + H2O
– Gas forming:
CaCO3 + 2 HCl f CaCl2 + CO2(g) + H2O
– Precipitation:
Pb(NO3) 2 + 2 KI f PbI2(s) + 2 KNO3
• OXIDATION REDUCTION
– 4 Fe + 3 O2 f 2 Fe2O3
• Apply equilibrium principles to acid-base and
precipitation reactions.
© 2009 Brooks/Cole - Cengage
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60
Analysis of Silver Group
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
All salts formed in
this experiment are
said to be
INSOLUBLE and
form when mixing
moderately
concentrated
solutions of the
metal ion with
chloride ions.
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of
Silver Group
Although all salts formed in this experiment are
said to be insoluble, they do dissolve to some
SLIGHT extent.
AgCl(s) e Ag+(aq) + Cl-(aq)
When equilibrium has been established, no more
AgCl dissolves and the solution is
SATURATED.
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61
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of
Silver Group
62
AgCl(s) e Ag+(aq) + Cl-(aq)
When solution is SATURATED, expt. shows that
[Ag+] = 1.67 x 10-5 M.
This is equivalent to the SOLUBILITY of AgCl.
What is [Cl-]?
[Cl-] = [Ag+] = 1.67 x 10-5 M
© 2009 Brooks/Cole - Cengage
+
Ag
Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of
Silver Group
AgCl(s) e Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 1.67 x 10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (1.67 x 10-5)(1.67 x 10-5)
= 2.79 x 10-10
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63
+
Ag
Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of
Silver Group
AgCl(s) e Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 2.79 x 10-10
Because this is the product of “solubilities”, we call it
Ksp = solubility product constant
• See Table 18.2 and Appendix J
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65
Some Values of Ksp
Table 18.2 and Appendix J
© 2009 Brooks/Cole - Cengage
Lead(II) Chloride
PbCl2(s) e Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5 = [Pb2+][Cl–]2
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
66
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) e Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 0.00130 M
Solution
1. Solubility = [Pb2+] = 1.30 x 10-3 M
[I-] = ?
[I-] = 2 x [Pb2+] = 2.60 x 10-3 M
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Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) e Pb2+(aq) + 2 I-(aq)
Calculate Ksp
if solubility = 0.00130 M
Solution
2. Ksp = [Pb2+] [I-]2
= [Pb2+] {2 • [Pb2+]}2
Ksp = 4 [Pb2+]3 = 4 (solubility)3
Ksp = 4 (1.30 x 10-3)3 = 8.79 x 10-9
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Precipitating an Insoluble Salt
Hg2Cl2(s) e Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
If [Hg22+] = 0.010 M, what [Cl-] is req’d to just
begin the precipitation of Hg2Cl2?
That is, what is the maximum [Cl-] that can be
in solution with 0.010 M Hg22+ without
forming Hg2Cl2?
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Precipitating an Insoluble Salt
Hg2Cl2(s) e Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
Recognize that
Ksp = product of
maximum ion concs.
Precip. begins when product of
ion concs. EXCEEDS the Ksp.
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Precipitating an Insoluble Salt
Hg2Cl2(s) e Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = 0.010 M,
-
[Cl ] =
K sp
0.010
= 1.1 x 10-8 M
If this conc. of Cl- is just exceeded, Hg2Cl2
begins to precipitate.
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Precipitating an Insoluble Salt
Hg2Cl2(s) e Hg22+(aq) + 2 Cl-(aq)
Ksp = 1.1 x 10-18
Now raise [Cl-] to 1.0 M. What is the value of
[Hg22+] at this point?
Solution
[Hg22+] = Ksp / [Cl-]2
= Ksp / (1.0)2 = 1.1 x 10-18 M
The concentration of Hg22+ has been reduced
by 1016 !
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73
The Common Ion Effect
Adding an ion “common” to an
equilibrium causes the equilibrium to
shift back to reactant.
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Common Ion Effect
PbCl2(s) e Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5
PLAY MOVIE
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75
Barium Sulfate
Ksp = 1.1 x 10-10
(a) BaSO4 is a
common mineral,
appearing a white
powder or
colorless crystals.
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(b) BaSO4 is opaque
to x-rays. Drinking
a BaSO4 cocktail
enables a physician
to exam the
intestines.
The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure
water and (b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) e Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)1/2 = 1.1 x 10-5 M
Solubility in pure water = 1.0 x 10-5 mol/L
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The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) e Ba2+(aq) + SO42-(aq)
Solution
Solubility in pure water = 1.1 x 10-5 mol/L.
Now dissolve BaSO4 in water already
containing 0.010 M Ba2+.
Which way will the “common ion” shift the
equilibrium? ___ Will solubility of BaSO4 be
less than or greater than in pure water?___
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The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) e Ba2+(aq) + SO42-(aq)
Solution
initial
change
equilib.
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[Ba2+]
0.010
+y
0.010 + y
[SO42-]
0
+y
y
78
The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) e Ba2+(aq) + SO42-(aq)
Solution
Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)
Because y < 1.1 x 10-5 M (= x, the solubility in
pure water), this means 0.010 + y is about
equal to 0.010. Therefore,
Ksp = 1.1 x 10-10 = (0.010)(y)
y = 1.1 x 10-8 M = solubility in presence of
added Ba2+ ion.
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The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and
(b) in 0.010 M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) e Ba2+(aq) + SO42-(aq)
SUMMARY
Solubility in pure water = x = 1.1 x 10-5 M
Solubility in presence of added Ba2+
= 1.1 x 10-8 M
Le Chatelier’s Principle is followed!
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81
Separating Metal Ions
Cu2+, Ag+, Pb2+
Ksp Values
AgCl
PbCl2
PbCrO4
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
1.8 x 10-10
1.7 x 10-5
1.8 x 10-14
Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add
CrO42- to precipitate red Ag2CrO4 and yellow
PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
The substance whose Ksp is first
exceeded precipitates first.
The ion requiring the lesser amount
of CrO42- ppts. first.
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Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate
red Ag2CrO4 and yellow PbCrO4. Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
Calculate [CrO42-] required by each ion.
[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M
[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2
= 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M
PbCrO4 precipitates first
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Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add
CrO42- to precipitate red Ag2CrO4 and yellow
PbCrO4. PbCrO4 ppts. first.
Ksp (Ag2CrO4)= 9.0 x 10-12
Ksp (PbCrO4) = 1.8 x 10-14
How much Pb2+ remains in solution when Ag+
begins to precipitate?
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Separating Salts by Differences in Ksp
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate
red Ag2CrO4 and yellow PbCrO4.
How much Pb2+ remains in solution when Ag+ begins to precipitate?
Solution
We know that [CrO42-] = 2.3 x 10-8 M to begin to
ppt. Ag2CrO4.
What is the Pb2+ conc. at this point?
[Pb2+] = Ksp / [CrO42-] = 1.8 x 10-14 / 2.3 x 10-8 M
= 7.8 x 10-7 M
Lead ion has dropped from 0.020 M to < 10-6 M
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Separating Salts by
Differences in Ksp
• Add CrO42- to solid PbCl2. The less soluble salt,
PbCrO4, precipitates
• PbCl2(s) + CrO42- e PbCrO4 + 2 Cl• Salt
Ksp
PbCl2
1.7 x 10-5
PbCrO4
1.8 x 10-14
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87
Separating Salts by
Differences in Ksp
• PbCl2(s) + CrO42- e PbCrO4 + 2 ClSalt
Ksp
PbCl2
1.7 x 10-5
PbCrO4
1.8 x 10-14
PbCl2(s) e Pb2+ + 2 ClPb2+ + CrO42- e PbCrO4
K1 = Ksp
K2 = 1/Ksp
Knet = (K1)(K2) = 9.4 x 108
Net reaction is product-favored
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PbCl2
PbI2
Lead
Chemistry
• From Chemistry &
Chemical
Reactivity, 5th
Pb(CO3)2
PbCrO4
edition
• Illustrates the
transformation of
one insoluble
compound into an
even less soluble
compound.
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Separations by Difference in Ksp
See Figure 18.18
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Solubility and Complex Ions
Sections 18.6 & 18.7
The combination of metal ions (Lewis acids) with
Lewis bases such as H2O and NH3 leads to
COMPLEX IONS
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91
Reaction of NH3 with Cu2+(aq)
PLAY MOVIE
PLAY MOVIE
© 2009 Brooks/Cole - Cengage
Dissolving Precipitates
by forming Complex Ions
Formation of complex ions explains why you
can dissolve a ppt. by forming a complex
ion.
PLAY MOVIE
PLAY MOVIE
AgCl(s) + 2 NH3 e Ag(NH3)2+ + Cl© 2009 Brooks/Cole - Cengage
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Dissolving Precipitates
by forming Complex Ions
Examine the solubility of AgCl in ammonia.
AgCl(s) e Ag+ + ClAg+ + 2 NH3 e Ag(NH3)2+
Ksp = 1.8 x 10-10
Kform = 1.6 x 107
-------------------------------------
AgCl(s) + 2 NH3 e Ag(NH3)2+ + ClKnet = (Ksp)(Kform) = 2.9 x 10-3
By adding excess NH3, the equilibrium shifts to the
right.
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