1
More About
Chemical Equilibria
Acid-Base & Precipitation Reactions
Chapter 15 & 16
© 2009 Brooks/Cole - Cengage
Objectives – Chapter 15
• Define the Common Ion Effect (15.1)
• Define buffer and show how a buffer
controls pH of a solution (15.2 15.3)
• Identify and Evaluate titration curves
(15.4)
• Use of Indicators (15.5)
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Stomach Acidity &
Acid-Base Reactions
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Acid-Base Reactions
• Strong acid + strong base
HCl + NaOH 
• Strong acid + weak base
HCl + NH3 
• Weak acid + strong base
HOAc + NaOH 
• Weak acid + weak base
HOAc + NH3 
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What is relative pH
before, during, &
after reaction?
Need to study:
a) Common ion
effect and buffers
b) Titrations
The Common Ion Effect
Section 15.1
QUESTION: What is the effect on the pH of adding
NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O  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!
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pH of Aqueous NH3
QUESTION: What is the effect on the pH of adding NH4Cl to
0.25 M NH3(aq)?
NH3(aq) + H2O  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
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pH of Aqueous NH3
QUESTION: What is the effect on the pH of adding NH4Cl
to 0.25 M NH3(aq)?
NH3(aq) + H2O  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
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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  NH4+(aq) + OH-(aq)
We expect that the pH will decline on adding
NH4Cl. Let’s test that!
[NH3]
[NH4+]
[OH-]
initial
change
equilib
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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  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
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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  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
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Common Ion Effect
11
• The addition of a common ion in a base such as ammonia,
limits the ionization of the base and the production of [OH-]
(Le Châtelier’s Principle).
B(aq) + H2O(l) ⇌ HB+(aq) + OH-(aq)
• For acid, addition of conjugate base from salt limit
ionization of acid, which limits [H3O+]
HA(aq) + H2O(l) ⇌A-(aq) + H3O+(aq)
• Doesn’t work if the conjugate base is from a strong acid
because the conjugate base is too weak.
• In other words, adding Cl- to above does not change the pH
because the Cl- comes from strong acid HCl, which has a
very large Ka.
HCl + H2O  H3O+ + Cl© 2009 Brooks/Cole - Cengage
Common Ion Effect Practice
12
Calculate the pH of 0.30 M formic acid (HCO2H) in water and
with enough sodium formate, NaHCO2, to make the solution
0.10 M in the salt. The Ka of formic acid is 1.8x10-4.
HCO2H(aq) + H2O(l) ⇌ HCO2-(aq) + H3O+
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Common Ion Effect Practice
13
Calculate the pH of 0.30 M formic acid (HCO2H) in water and
with enough sodium formate, NaHCO2, to make the solution
0.10 M in the salt. The Ka of formic acid is 1.8x10-4.
HCO2H(aq) + H2O(l) ⇌ HCO2-(aq) + H3O+
With salt:
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Buffered Solutions
Section 15.2
HCl is added to
pure water.
HCl is added to a
solution of a weak
acid H2PO4- and its
conjugate base
HPO42-.
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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,
either by H+ or by OH-.
Buffer Composition
Weak Acid
+
Conj. Base
HOAc
+
OAcH2PO4+
HPO42NH4+
+
NH3
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Buffer Solutions
Consider HOAc/OAc- to see how buffers work
ACID USES UP ADDED OHWe know that
OAc- + H2O  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
consumes OH-!!!!
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Buffer Solutions
Consider HOAc/OAc- to see how buffers
work
CONJ. BASE USES UP ADDED H+
HOAc + H2O  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
consumes H3O+!
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Buffer Solutions
Problem: What is the pH of a buffer that has
[HOAc] = 0.700 M and [OAc-] = 0.600 M?
HOAc + H2O  OAc- + H3O+
Ka = 1.8 x 10-5
0.700 M HOAc has pH = 2.45
The pH of the buffer will have
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1.
pH < 2.45
2.
pH > 2.45
3.
pH = 2.45
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Buffer Solutions
Problem: What is the pH of a buffer that has
[HOAc] = 0.700 M and [OAc-] = 0.600 M?
HOAc + H2O  OAc- + H3O+
Ka = 1.8 x 10-5
initial
change
equilib
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[HOAc]
[OAc-]
0.700
-x
0.600
+x
0.700 - x
[H3O+]
0
+x
x
0.600 + x
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Buffer Solutions
Problem: What is the pH of a buffer that has [HOAc] = 0.700 M
and [OAc-] = 0.600 M?
HOAc + H2O  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
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Buffer Solutions
Notice that the expression for calculating the
H+ conc. of the buffer is
[H3O+ ] =
Orig. conc. of HOAc
Orig. conc. of OAc
[H3O+ ] =
-
x Ka
[Acid]
x Ka
[Conj. base]
Notice that the [H3O+] depends on:
1. the Ka of the acid
2. the ratio of the [acid] and
[conjugate base].
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Henderson-Hasselbalch Equation
or
Take the negative log of both sides of this
equation
Note: HA and A- are
generic terms for acid
and conjugate base,
respectively.
The pH is determined largely by the pKa of the acid and
then adjusted by the ratio of conjugate base and acid.
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Henderson-Hasselbalch Equation
Works the same for base!
B(aq) + H2O(l) ⇌ HB+(aq) + OH-(aq)
𝐾𝑏 =
Solve for
𝐻𝐵+ [𝑂𝐻 − ]
𝐵
[OH-]
= 𝐾𝑏 ∗
𝐵
+
𝐻𝐵
Take the negative log of both sides of this
equation to get:
𝑯𝑩+
𝑝𝑂𝐻 = 𝑝𝐾b + 𝒍𝒐𝒈
𝐵
The pOH (and pH) is determined largely by the pKb of
the base and then adjusted by the ratio of the
conjugate acid and base.
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Buffer Calculation Example
(Acid/Conj. Base)
What is the pH of a buffer that is 0.12 M in
lactic acid, HC3H5O3, and 0.10 M in sodium
lactate (NaC3H5O3)?
Ka for lactic acid is 1.4  10-4.
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Buffer Calculation Example
HC3H5O3 + H2O ⇌ C3H5O3- + H3O+
I
0.12 M
0.10
E 0.12-x
0.10 + x
x
x(0.10 + x) = 1.4x10-4
0.12 - x
We’ll make usual assumption that x is small.
x = 0.12(1.40 x 10-4) = 1.68 x 10-4 = [H3O+]
0.10
So the pH = 3.77
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Buffer Calculation using the
Henderson–Hasselbalch Equation
HC3H5O3 + H2O  C3H5O3- + H3O+
acid
conj base
0.12
0.10
pH = –log (1.4 
10-4)
(0.10)
+ log(0.12)
pH = 3.85 + (–0.08)
pH = 3.77 (same as before!!)
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Buffer Calculation using the
Henderson–Hasselbalch Equation
Calculate the pH of a buffer that has is 0.080 M
ammonia and 0.045 M ammonium chloride.
The Kb for ammonia is 1.8 x 10-5.
NH3(aq) + H2O(l)  NH4+(aq)
+ OH-(aq)
[base]
[conj. acid]
0.080
0.045
(0.045)
pOH = –log (1.8 x 10-5) + log (0.080)
pOH = 4.74 + (–0.25)
pOH = 4.49
pH = 14.00 – 4.49 = 9.51
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Buffer Practice 1
Calculate the pH of a solution that is 0.50 M
HC2H3O2 (acetic acid) and 0.25 M NaC2H3O2
(Ka = 1.8 x 10-5). Use H-H Eqn.
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Buffer Practice 2
Calculate the concentration of sodium benzoate
(NaC7H5O2) that must be present in a 0.20 M solution of
benzoic acid (HC7H5O2) to produce a pH of 4.00.
(Benzoic acid Ka = 6.3 x 10-5).
Use Henderson-Hasselbalch equation.
HC7H5O2 + H2O  H3O+ + C7H5O2-
© 2009 Brooks/Cole - Cengage
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Adding an Acid to a Buffer
(How a Buffer Maintains pH)
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) (See slide 20)
Solution to Part (a)
Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of
water
M1•V1 = M2 • V2
(1.00 M)(1.0 mL) = M2(1001 mL)
M2 = 9.99 x 10-4 M = [H3O+]
pH = 3.00
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Adding an Acid to a Buffer
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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 (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) →
HOAc (from buffer)
The reaction occurs completely because K is
very large.
The stronger acid (H3O+) will react with the conj.
base of the weak acid (OAc-) to make the weak
acid (HOAc).
© 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
0.599 mol
0.701 mol
Equil []’s
0
1.001 L
0.598 mol/L
© 2009 Brooks/Cole - Cengage
1.00l L
0.700 mol/L
33
Adding an Acid to a Buffer
34
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  H3O+ + OAc[HOAc]
[H3O+]
0.700 mol/L 0
Initial (M)
-x
Change (M)
+x
0.700-x
x
Equil. (M)
(re-established)
© 2009 Brooks/Cole - Cengage
[OAc-]
0.598 mol/L
+x
0.598 + 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  H3O+ + OAc[HOAc] [H3O+]
[OAc-]
Equilibrium
0.700-x
x
0.598+x
Because [H3O+] = 2.1 x 10-5 M BEFORE adding
HCl, we again neglect x relative to 0.700 and
0.598.
© 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 2—Equilibrium
HOAc + H2O  H3O+ + OAc-
pH = 4.74 -0.07 = 4.67
The pH has not changed much
on adding HCl to the buffer!
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Adding a Base to a Buffer
Problem: What is the pH when 1.00 mL of 1.00 M
NaOH is added to
a) 1.00 L of pure water (before NaOH, pH = 7.00)
b) 1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M (pH = 4.68) (See slide 20)
Solution to Part (a)
Calc. [OH-] after adding 1.00 mL of NaOH to 1.00 L of
water
M1•V1 = M2 • V2
(1.00 M)(1.0 mL) = M2(1001 mL)
M2 = 1.00 x 10-3 M = [OH-]
pOH = 3.00
pH = 11.00
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Adding an Base to a Buffer
38
What is the pH when 1.00 mL of 1.00 M NaOH is added to
a) 1.00 L of pure water (after NaOH, pH = 11.00)
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
OH- (from NaOH) + HOAc (from buffer) →
OAc- (from buffer and rxn)
The reaction occurs completely because K is
very large.
The strong base (OH-) will react with weak acid
(HOAc) to make the conj. weak base (OAc-).
© 2009 Brooks/Cole - Cengage
Adding a Base to a Buffer
What is the pH when 1.00 mL of 1.00 M NaOH is added to
a) 1.00 L of pure water (pH = 11.00)
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
[OH-] + [HOAc] → [OAc-]
Before rxn
0.00100 mol 0.700 mol
0.600 mol
Change
-0.00100
-0.00100
+0.00100
After rxn
0.699 mol
0
0.601 mol
0.699 mol
0.601 mol
0
Equil. []’s
1.001 L
0.698 mol/L
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1.001 L
0.600 mol/L
39
Adding a Base to a Buffer
What is the pH when 1.00 mL of 1.00 M NaOH is added
to
a) 1.00 L of pure water (pH = 11.00)
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  H3O+ + OAc[HOAc]
[H3O+]
0.698 mol/L 0
-x
+x
Initial(M)
Change (M)
0.698-x
Equil. (M)
(re-established)
© 2009 Brooks/Cole - Cengage
x
[OAc-]
0.600 mol/L
+x
0.600 + x
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Adding a Base to a Buffer
41
What is the pH when 1.00 mL of 1.00 M HCl is added to
a) 1.00 L of pure water (pH = 11.00)
b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-]
= 0.600 M (pH = 4.68, pOH = 9.32)
Solution to Part (b): Step 2—Equilibrium
HOAc + H2O  H3O+ + OAc[HOAc] [H3O+]
[OAc-]
Equilibrium
0.698-x
x
0.600+x
Because [H3O+] = 2.1 x 10-5 M BEFORE adding
NaOH, we again neglect x relative to 0.600 and
0.698.
© 2009 Brooks/Cole - Cengage
Adding a Base to a Buffer
42
What is the pH when 1.00 mL of 1.00 M NaOH is added to
a) 1.00 L of pure water (pH = 11.00)
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  H3O+ + OAc-
pH = 4.74 – 0.07 = 4.67
The pH has not changed much
on adding NaOH to the buffer!
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Preparing a Buffer
You want to buffer a solution at pH = 4.30.
This means [H3O+] = 10-pH =10-4.30= 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]
+
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Preparing a Buffer
You want to buffer a solution at pH = 4.30 or
[H3O+] = 5.0 x 10-5 M
POSSIBLE
ACIDS/CONJ BASE PAIRS
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.
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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.
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Preparing a Buffer
A final point —
CONCENTRATION of the acid and conjugate
base are not as important as
the RATIO OF THE NUMBER OF MOLES of
each.
Result: diluting a buffer solution does
not change its pH
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47
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.
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Preparing a Buffer
Buffer prepared from
8.4 g NaHCO3
(8.4g/84 g/mol)
weak acid
16.0 g Na2CO3 (16.0/106 g/mol)
conjugate base
HCO3- + H2O
 H3O+ + CO32What is the pH?
HCO3- pKa = 10.3
pH = 10.3 + log (1.51/1.0)=10.5
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Buffer Example
Phosphate buffers are used a lot in labs to
simulate blood pH of 7.40. As with any other
buffer, you need an acid and its conjugate base.
In this case the ‘acid’ is NaH2PO4 and the ‘base’
is Na2HPO4. So the equilibrium expression is:
H2PO4-(aq) + H2O(l) ⇌ HPO42-(aq) + H3O+(aq)
A) If the pKa of H2PO4- is 7.21, what should the
[HPO42-]/[H2PO4-] ratio be?
B) If you weigh 6.0 g of NaH2PO4 and make 500
mL of solution, how much Na2HPO4 do you
need add to make your pH 7.40 buffer?
(assume the Na2HPO4 doesn’t affect the volume)
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Buffer Example
Part A
H2PO4-(aq) + H2O(l) ⇌ HPO42-(aq) + H3O+(aq)
Can use H-H: pH = pKa + log[HPO42-]/[H2PO4-]
7.40 = 7.21 + log[HPO42-]/[H2PO4-]
log[HPO42-]/[H2PO4-] = 0.19
[HPO42-]/[H2PO4-] = 100.19 = 1.55
Part B
6.0g NaH2PO4 = 0.050 mol in 0.500 L = 0.10 M
120g/mol
[HPO42-] = 1.55[H2PO4-] = 0.155 M HPO42=0.155 mol/L * 0.500 L * 142 g Na2HPO4/mol
= 11.0 g Na2HPO4
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Preparing a buffer Practice
Using the acetic acid (molar mass 60 g/mol) and
sodium acetate (molar mass: 82 g/mol) buffer
from before, how much sodium acetate should
you weigh to make 1.0 L of a pH 5.00 buffer?
Assume [HOAc] is 1.0 M in the final 1.0 L of
buffer. (pKa = 4.74)
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Buffering Capacity
Chapter 15.3
• The pH of a buffered solution is
determined by the ratio [A-]/[HA].
• As long as the ratio doesn’t change much
the pH won’t change much.
• The more concentrated these two are the
more H+ and OH- the solution will be able
to absorb.
• Larger concentrations → bigger buffer
capacity.
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Buffer Capacity
• Calculate the change in pH that occurs
when 0.010 mol of HCl is added to 1.0L of
each of the following:
• 5.00 M HOAc and 5.00 M NaOAc
• 0.050 M HOAc and 0.050 M NaOAc
-5
• Ka= 1.8x10 (pKa = 4.74)
• Note: Since [OAc-]=[HOAc] for both cases,
the initial pH is 4.74.
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Large Buffer System
H3O+ + OAc-  HOAc + H2O
Stoichiometry Calculation:
Before Reaction (moles)
After Reaction (moles)
© 2009 Brooks/Cole - Cengage
H3O+
OAc-
HOAc
0.01
5.00
5.00
0
4.99
5.01
55
Calculate pH via HendersonHasselbalch Equations
At Equilibrium:
HOAc + H2O  H3O+ + OAc0.501
X
4.99
[base]
pH = pKa +
[acid]
pH = -log (1.8 x 10-5) + log[ 4.99/5.01]
pH = 4.738
Since pKa = 4.74, solution has gotten slightly
more acidic
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Small Buffer System
H3O+ + OAc-  HOAc(aq) + H2O
Stoichiometry Calculation:
Before Reaction (moles)
After Reaction (moles)
© 2009 Brooks/Cole - Cengage
H+
OAc-
HOAc
0.01
0.05
0.05
0
0.04
0.06
Calculating pH for Small Buffer
System
HOAc  H3O+ + OAcIn table form (Equilibrium Re-established):
[HOAc], M
[H3O+], M
[OAc−], M
Initially
0.06
0
0.04
Change
-x
+x
+x
0.06-x
x
0.04 + x
At
Equilibrium
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Calculate pH via HendersonHasselbalch Equations
At Equilibrium:
HOAc + H2O  H3O+ + OAc0.06-x
X
0.04-x
Ignore x’s relative to 0.06 and 0.04 since they will be
small. Calculate via H-H eqn:
[base]
pH = pKa +
[acid]
pH = -log (1.8 x 10-5) + log[ 0.04/0.06]
pH = 4.74 – 0.18 = 4.56
So, solution is considerably more acidic than the pH
4.74 of the big buffer system.
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58
Mixing Acids and Bases
(Stoichiometry & Equilibrium)
Chapter 15.4
If you mix 50.0 mL of 0.10 M HCl and 50.0 mL
of 0.10 NaOH, what is the pH?
Since it is a strong acid/strong base, what
happens stoichiometrically?
H+ +
OH- ⇌ H2O(l)
I(moles)
0.005
0.005
After reaction 0
0
Since you consume all acid and base and make
water, what is the pH? 7 of course.
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60
Mixing Acids and Bases
(Stoichiometry & Equilibrium)
Now, what if you mix 50.0 mL of 0.10 M HCl and
20.0 mL of 0.10 NaOH, what is the pH?
H+ +
OH- ⇌ H2O(l)
I(moles)
0.005
0.002
After reaction (mol) 0.003
0
Total volume of solution is 70.0 mL (0.0700 L)
So the [H+] = 0.003 mol/0.0700 L = 0.0429 M
pH = -log[0.0429] = 1.37
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Mixing Acids and Bases
(Stoichiometry & Equilibrium)
61
Now mix 50.0 mL of 0.10 M HCl and 70.0 mL of
0.10 NaOH, what is the pH?
H+ +
OH- ⇌ H2O(l)
I(moles)
0.005
0.007
After reaction (mol)
0.002
Total volume of solution is 120.0 mL (0.1200 L)
Since there was more OH- added than H+, all of the
H+ is consumed and there is excess OH-.
[OH-] = 0.002/0.120 L = 0.017 M OHpOH = 1.77 so pH = 12.23
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62
Mixing Acids and Bases
(Stoichiometry & Equilibrium)
If you mix 50.0 mL of 0.10 M HOAc and 50.0 mL
of 0.10 NaOH, what is the pH?
Since it is a weak acid/strong base, what
happens stoichiometrically?
HOAc +
OH- ⇌ H2O(l) + OAcI(moles)
0.005
0.005
After reaction 0
0
0.005
Since HOAc is a weak acid and OH- is a strong
base, the OH- will still consume (take that H+) from
the HOAc. (The Stoichiometry part.)
© 2009 Brooks/Cole - Cengage
Mixing Acids and Bases
(Stoichiometry & Equilibrium)
If you mix 50.0 mL of 0.10 M HOAc and 50.0 mL
of 0.10 NaOH, what is the pH?
Now, equilibrium will be re-established.
H2O(l) + OAc- ⇌
HOAc +
OHNeed Conc: 0.005 mol/0.100L
I (conc)
0.05
0
0
C
-x
+x
+x
E
0.050-x
x
x
You can think of this part as when we calculated
pH’s of salts of weak acids. It doesn’t matter how
you get to this point, the calculation is still the same!
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63
Mixing Acids and Bases
(Stoichiometry & Equilibrium)
If you mix 50.0 mL of 0.10 M HOAc and 50.0 mL
of 0.10 NaOH, what is the pH?
Now, equilibrium will be re-established.
H2O(l) + OAc- ⇌ HOAc +
OHE
0.050-x
x
x
Krxn =Kb = Kw/Ka = 1x10-14/1.8x10-5 = 5.56x10-10
x2
= 5.56x10-10
0.050−X
Usual assumption that x is small compared to 0.050
so x =
0.050(5.56𝑥10−10 ) = 5.27𝑥10−6
pOH = -log[5.27x10-6] = 5.28 so pH = 8.72
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65
Mixing Acids and Bases
(Stoichiometry & Equilibrium)
If you mix 50.0 mL of 0.10 M HOAc and 20.0 mL
of 0.10 NaOH, what is the pH?
Since it is a weak acid/strong base, what
happens stoichiometrically?
HOAc +
OH- ⇌ H2O(l) + OAcI(moles)
0.0050
0.0020
0
Reaction -0.0020
-0.0020
+0.0020
After reaction 0.0030
0
0.0020
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Mixing Acids and Bases
(Stoichiometry & Equilibrium)
66
If you mix 50.0 mL of 0.10 M HOAc and 20.0 mL
of 0.10 NaOH, what is the pH?
Now, equilibrium will be re-established. Since
you have an acid and its conjugate base, this is a
buffer and you can use H-H eqn.
H2O(l) + HOAc ⇌
OAc- +
H3O+
E
0.0030
0.0020
x
pH = pKa + log [OAc-]/[HOAc]
= 4.74 + log(0.002)/(0.003)
= 4.74 + log(0.667)
= 4.74 – 0.18 = 4.56
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Titrations and pH Curves
pH
Titrant volume, mL
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67
Titrations and pH Curves
Adding NaOH from the buret to acetic acid in the flask,
a weak acid. In the beginning the pH increases very
slowly.
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68
69
Titrations and pH Curves
Additional NaOH is added. pH rises as equivalence
point is approached.
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70
Titrations and pH Curves
Additional NaOH is added. pH increases and then
levels off as NaOH is added beyond the equivalence
point.
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71
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
pH at
half-way
point?
Benzoic acid
+ NaOH
© 2009 Brooks/Cole - Cengage
What is the
pH at
equivalence
point?
pH of solution of
benzoic acid, a
weak acid
72
Titrations and pH Curves
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 the
equivalence point?
HBz + NaOH  Na+ + Bz- + H2O
C6H5CO2H = HBz
© 2009 Brooks/Cole - Cengage
Benzoate ion = Bz-
73
Titrations and pH Curves
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 the
equivalence point?
HBz + NaOH  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
Titrations and pH Curves
The product of the titration of benzoic acid is the
benzoate ion, Bz- .
Bz- is the conjugate base of a weak acid.
Therefore, solution is basic at equivalence point.
Bz- + H2O  HBz + OH-
Kb = 1.6 x 10-10
+
+
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75
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
Titrations and pH Curves
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 the
equivalence point?
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
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76
Titrations and pH Curves
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 the equivalence point?
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
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77
Titrations and pH Curves
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 the equivalence point?
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
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78
Titrations and pH Curves
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  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
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79
Titrations and pH Curves
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  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
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81
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
82
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
Titrations and pH Curves
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 ⇌ 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
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83
Titrations and pH Curves
You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH.
What was the pH when 6.0 mL of
doing these types of problems,
NaOH was added? When
start from initial moles and go to the
point you are considering.
OH- +
Stoichiometry
Before rxn, moles
After rxn, mole
After rxn, [ ]
0.0006
0
0
0
HBz
⇌
0.0025
0.0019
0.0019mol/0.106L
0.018
H2O +
Bz-
-0.0006
0.0006/0.106
0.0057
Equilbrium:
HBz + H2O
⇌
H3O+ + Bz- pKa = 4.20
Equilibrium
Now can use H-H equation:
pH = 4.20 + log (0.0057/0.018)
= 4.20 – .50 = 3.70
Remember: this is essentially a buffer solution since HBz and Bz- are both
in the solution.
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Titrations and pH Curves
85
You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH.
Practice: What is the pH when 20.0 mL
of NaOH are added?
OH- +
Stoichiometry
Before rxn, moles
After rxn, mole
After rxn, [ ]
Equilbrium:
0.0020
0
0
0
HBz
H2O +
0.0025
0.0005
0.0005mol/0.120L
0.0042
+ H2O
Equilibrium
Now can use H-H equation:
pH = 4.20 + log (0.017/0.0042)
= 4.20 + 0.61 = 4.81
© 2009 Brooks/Cole - Cengage
⇌
HBz
⇌
H2O +
Bz-0.0020
0.0020/0.120
0.017
Bz- pKa = 4.20
Acetic acid titrated with NaOH
Weak acid titrated with a
strong base
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87
Strong acid titrated with a strong base
See Figure 18.4
© 2009 Brooks/Cole - Cengage
88
Weak diprotic
acid (H2C2O4)
titrated with a
strong base
(NaOH)
See Figure 18.6
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89
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 -->
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90
Weak base (NH3)
titrated with a strong
acid (HCl)
© 2009 Brooks/Cole - Cengage
Acid-Base Indicators
Chapter 15.5
• Weak acids that change color when
they become bases.
• Since the Indicator is a weak acid, it
has a Ka.
• End point - when the indicator
changes color.
• An indicator changes colors at
pH=pKa1, pKa is the acid dissoc.
constant for the indicator
• Want an indicator where pKa pH at
equiv. pt.
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92
Acid-Base
Indicators
© 2009 Brooks/Cole - Cengage
Indicators for Acid-Base Titrations
Since the pH change is
large near the equivalence
point, you want an
indicator that is one color
before and one color after.
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94
Titration of a Base with an Acid
• The pH at the
equivalence point in
these titrations is < 7.
• Methyl red is the
indicator of choice.
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95
Solubility and Complex Ion
Equilibria
Chapter 16
Lead(II) iodide
© 2009 Brooks/Cole - Cengage
Objectives – Chapter 16
• Define equilibrium constant for
‘insoluble’ salts (Ksp) (16.1)
• Manipulate solubility by common ion
effect (16.1)
• Precipitation Reactions – comparing
Q & K of a reaction (16.2)
• Define Complex ion & show how
complex ions affect solubility (16.3)
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Types of Chemical Reactions
• EXCHANGE REACTIONS:
AB + CD  AD + CB
– Acid-base:
CH3CO2H + NaOH  NaCH3CO2 + H2O
– Gas forming:
CaCO3 + 2 HCl  CaCl2 + CO2(g) + H2O
– Precipitation:
Pb(NO3) 2 + 2 KI  PbI2(s) + 2 KNO3
• OXIDATION REDUCTION (Redox – Ch. 18)
– 4 Fe + 3 O2  2 Fe2O3
• Apply equilibrium principles to acid-base and
precipitation reactions.
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98
Analysis of Silver Group
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
© 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)  Ag+(aq) + Cl-(aq)
When equilibrium has been established, no more
AgCl dissolves and the solution is
SATURATED.
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99
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of
Silver Group
100
AgCl(s)  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)  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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101
+
Ag
Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
Analysis of
Silver Group
AgCl(s)  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 16.1 and Appendix A5.4
• Note: The property ‘solubility’ is in mol/L.
(Can also be expressed as g/L or mg/L!)
• Solubility Product has no units.
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103
Solubility Products
In general, for the equilibrium for the reaction
is expressed as
MmAa(s)  mM+(aq) + aA-(aq)
The Ksp equilibrium expression is..
Ksp =[M+]m[A-]a
The higher the Ksp, the more soluble the salt is.
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104
Some Values of Ksp
© 2009 Brooks/Cole - Cengage
Lead(II) Chloride
PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5 = [Pb2+][Cl–]2
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105
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s)  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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106
Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s)  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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107
Solubility of Lead(II) Iodide
Caveat
3. The value we just calculated by
Ksp = 4 (1.30 x 10-3)3 = 8.79 x 10-9
solubility gives a ball-park value.
The actual Ksp of PbI2 is 9.8 x 10-9.
Note: Calculating Ksp from solubility
gives approximate values that can
be off by a few OOMs due to ionic
interactions.
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109
Solubility/Ksp Practice 1
• Calculate the Ksp if the solubility of CaC2O4 is
6.1 mg/L. Note units here. What should they be in
order to determine Ksp? (Molar Mass = 128.1 g/mol)
CaC2O4 (s)  Ca2+(aq) + C2O42-(aq)
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110
Solubility/Ksp Practice 2
• Calculate the solubility of CaF2 in g/L if the
Ksp is 3.9x10-11. (Molar Mass = 78.1 g/mol)
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111
The Common Ion Effect
Adding an ion “common” to an
equilibrium causes the equilibrium to
shift back to reactant.
© 2009 Brooks/Cole - Cengage
Common Ion Effect
PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9 x 10-5
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112
113
Barium Sulfate
Ksp = 1.1 x 10-10
(a) BaSO4 is a
common mineral,
appearing a white
powder or
colorless crystals.
© 2009 Brooks/Cole - Cengage
(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)  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 = 1.1 x 10-5 mol/L
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114
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)  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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115
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)  Ba2+(aq) + SO42-(aq)
Solution
initial
change
equilib.
© 2009 Brooks/Cole - Cengage
[Ba2+]
0.010
+y
0.010 + y
[SO42-]
0
+y
y
116
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)  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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117
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)  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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119
Common Ion Effect Practice
Do you expect the solubility of AgI to be greater in pure
water or in 0.020 M AgNO3? Explain your answer.
Ksp(AgI) = 8.5 x 10-17
AgI(s)  Ag+(aq) + I-(aq)
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Factors Affecting Solubility - pH
• pH
– If a substance has a
basic anion (like OH-),
it is more soluble in a
more acidic solution.
– Substances with
acidic cations (Al3+)
are more soluble in
more basic solutions.
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Solubility in more Acidic Solution
example: solid Mg(OH)2 in water
Equilibrium: Mg(OH)2  Mg2+ + 2OHFor Ksp = 1.8 x 10-11, calculate pH:
1.8 x 10-11 = x(2x)2
x = 1.65 x 10-4 M [solubility of Mg(OH)2]
= [Mg2+] in solution
[OH-] = 2 x solubility of Mg(OH)2
pOH = -log (2 · 1.65 x 10-4) = 3.48
pH = 10.52
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Solubility in more Acidic Solution
example: solid Mg(OH)2 in pH=9
buffer
• Now put Mg(OH)2 in a pH = 9 buffer solution
• Now [OH-] = 1 x 10-5 (if pH =9 then pOH = 5)
• [Mg2+][OH-]2 = 1.8 x 10-11 (Ksp a function of Temp
only)
• [Mg2+] = 1.8x10-11/(1x10-5)2 = 0.18 M
• Considerably more soluble (~1000x) in more
acidic solution
• Solubility in water was 1.65 x 10-4 M
• Essentially removed OH- by reacting with H+
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Acidic solution example: Ag3PO4
• Ag3(PO4) is more soluble in acid solution than neutral
solution. Why?
– H+ of acid solution reacts with PO43- anion to
produce weak acid HPO42Ag3PO4(s)  3Ag+(aq) + PO43-(aq)
Ksp = 1.8x10-18
H+(aq) + PO43- (aq) HPO42-(aq)
K = 1/Ka = 2.8x1012
Ag3PO4(s) + H+  3Ag+(aq) + HPO42(aq) K = 5.0 x 10-6
– Addition of H+ shifts equilibrium of 1st rxn to
right!
– PO43- is actually a moderately strong base:
PO43- + H2O ⇌ HPO42- + OH- Kb3 = 2.8x10-2
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Acidic solution example: AgCl
• What about AgCl in an acidic solution, such
as HNO3 or H2SO4.
AgCl(s)  Ag+(aq) + Cl-(aq)
H+(aq) + Cl- (aq)← HCl(aq)
Ksp = 1.8x10-10
K = very small
• K of reaction remains very small, largely
determined by the Ksp of AgCl.
• Note: Special case of HCl.
• HCl will affect solubility. How? Why?
• It would add chloride, actually make AgCl less
soluble by Le Chatelier.
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Solubility and Acidity
• Other conjugate base anions that will increase in
solubility in an acidic solution:
OH-, F-, S2-, CO32-, C2O42- and CrO42Why? Because all form weak acids.
Examples:
PbS(s) + 2H+(aq)  H2S(g) + Pb2+(aq)
Ca(OH)2(s) + 2H+(aq)  2H2O(l) + Ca2+(aq)
BaCO3(s) + 2H+(aq)  CO2(g) + H2O(l) + Ba2+(aq)
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Precipitation and Qualitative Analysis
Section 16.2
Hg2Cl2(s)  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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Precipitation and Qualitative Analysis
Hg2Cl2(s)  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 concentrations (Q)
EXCEEDS the Ksp.
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Precipitation and Qualitative Analysis
128
Hg2Cl2(s)  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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Precipitation and Qualitative Analysis
Hg2Cl2(s)  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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Precipitation and Qualitative
Analysis– Practice
• If a solution is 0.020 M in Cl- ions, at what
concentration of Pb2+ does the PbCl2 (Ksp= 1.7 x 10-5)
precipitate?
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131
Separating Metal Ions
Cu2+, Ag+, Pb2+
Ksp Values
AgCl
PbCl2
PbCrO4
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1.8 x 10-10
1.7 x 10-5
1.8 x 10-14
Separating Salts by
Differences in Ksp
• Add CrO42- to solid PbCl2. The less
soluble salt, PbCrO4, precipitates
• PbCl2(s) + CrO42-  PbCrO4 + 2 Cl• Salt
Ksp
PbCl2
1.7 x 10-5
PbCrO4
1.8 x 10-14
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Separating Salts by
Differences in Ksp
PbCl2(s) + CrO42-  PbCrO4 + 2 ClSalt
Ksp
PbCl2
1.7 x 10-5
PbCrO4
1.8 x 10-14
PbCl2(s)  Pb2+ + 2 Cl-
K1 = Ksp
Pb2+ + CrO42-  PbCrO4
K2 = 1/Ksp
Knet = (K1)(K2) = 9.4 x 108
Net reaction is product-favored
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Separating Salts by Differences in Ksp
134
The color of the salt silver chromate, Ag2CrO4, is
red. A student adds a solution of Pb(NO3)2 to a
test tube containing solid red Ag2CrO4. After
stirring the contents of the test tube, the student
finds the contents changes from red to yellow.
What is the yellow salt? Which one has a higher
Ksp?
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Separating Salts by Differences in Ksp
Solution
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Separations by Difference in Ksp
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Qualitative
Analysis
Figure 16.2
Group 1 – Insoluble
Chlorides
Group 2 – Sulfides
Insoluble in acid
Group 3 – Sulfides
insoluble in base
Group 4 – Insoluble
Carbonates
Group 5 – Alkali Metal
and ammonium ions
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Equilibria Involving Complex Ions
Chapter 16.3
The combination of metal ions (Lewis acids) with
Lewis bases such as H2O and NH3 leads to
COMPLEX IONS
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139
Reaction of NH3 with Cu2+(aq)
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Solubility and Complex Ions
• Consider the formation of Ag(NH3)2+:
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
• The Ag(NH3)2+ is called a complex ion.
• A Complex Ion is a central transition metal ion that
is bonded to a group of surrounding molecules or
ions and carries a net charge. It happens a lot with
transition metals because they have available dorbitals. Bases, like ammonia, are attracted to that.
(See Chapter 21 for details.)
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Solubility and Complex Ions
• Note that the equilibrium expression is ‘opposite’ to what we
usually write; the individual species are on the left and the
new substance on the right. This leads to Kf’s that are very
LARGE.
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Solubility and Complex Ions
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Dissolving Precipitates
by forming Complex Ions
Formation of complex ions explains why you
can dissolve a ppt. by forming a complex
ion.
AgCl(s) + 2 NH3  Ag(NH3)2+ + Cl© 2009 Brooks/Cole - Cengage
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144
Solubility and Complex Ions
• Complex Ions
– The formation
of these
complex ions
increase the
solubility of
these salts.
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Solubility and Complex Ions
Example
145
Determine the Keq of a reaction if the solution
contains AgBr and NH3.
• In water, solid AgBr exists in equilibrium with its
ions according to
AgBr  Ag+ + BrKsp = 5 x 10-13
• Also, Ag+ make complex ions with ammonia
expressed as
Ag+ + 2NH3  Ag(NH3)2+
Kf = 1.7 x 107
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Solubility and Complex Ions
Example
146
• Therefore, the equilibrium expressed by both reactions is
AgBr  Ag+ + BrKsp = 5 x 10-13
Ag+ + 2NH3  Ag(NH3)2+
Kf = 1.7 x 107
AgBr + 2NH3  Ag(NH3)2+ + Br- Keq = 8.5 x 10-6
Keq = Ksp x Kf
• So, this shows overall that the equilibrium is
more to the right (aqueous ions) than the
dissolution of the AgBr itself in pure water.
• In other words, NH3 makes AgBr more soluble.
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AP Exam Practice
Complex A/B Problems
• 2011 AP Exam #1
• 2006B AP Exam #1
• 2003 AP Exam #1
• 2002B AP Exam #1
• 2002 AP Exam #1
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Ksp Problems
• 2011B AP Exam #1
• 2010 AP Exam #1
• 2006 AP Exam #1
• 2004 AP Exam #1
• 2001 AP Exam #1