17 Lecture

Chemistry102

5/7/2013

Lecture Presentation

Chapter 17
Additional Aspects of Aqueous
Equilibria

John D. Bookstaver
St. Charles Community College
Cottleville, MO

© 2012 Pearson Education, Inc.

Common Ion Effect
HA(aq) + H2O(l) ⇔ A−(aq) + H3O+(aq)
• Adding a salt containing the anion NaA, which
•

is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left
This causes the pH to be higher than the pH of the acid solution
9lowering the H3O+ ion concentration

2

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Common Ion Effect

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The Common-Ion Effect
• Consider a solution of acetic acid:
CH3COOH(aq) + H2O(l)

H3O+(aq) + CH3COO−(aq)

• If acetate ion is added to the solution,
Le Châtelier says the equilibrium will shift to the left.

Aqueous
Equilibria
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Chemistry102

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The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” Aqueous
Equilibria
© 2012 Pearson Education, Inc.

The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Ka for HF is 6.8 × 10−4.

Ka =

[H3O+] [F−]
= 6.8 × 10−4
[HF]
Aqueous
Equilibria
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Chemistry102

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The Common-Ion Effect
H3O+(aq) + F−(aq)

HF(aq) + H2O(l)

Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

Initially

[HF], M

[H3O+], M

[F−], M

0.20

0.10

0

Change
At equilibrium
Aqueous
Equilibria
© 2012 Pearson Education, Inc.

The Common-Ion Effect
H3O+(aq) + F−(aq)

HF(aq) + H2O(l)

Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M

[H3O+], M

[F−], M

Initially

0.20

0.10

0

Change

−x

+x

+x

At equilibrium
Aqueous
Equilibria
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Dr. Ali Jabalameli

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Chemistry102

5/7/2013

The Common-Ion Effect
H3O+(aq) + F−(aq)

HF(aq) + H2O(l)

Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M

[H3O+], M

[F−], M

Initially

0.20

0.10

0

Change

−x

+x

+x

0.20 − x ≈ 0.20

0.10 + x ≈ 0.10

x

At equilibrium

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

The Common-Ion Effect
6.8 × 10−4 =

(0.10) (x)
(0.20)

(0.20) (6.8 × 10−4)
=x
(0.10)
1.4 × 10−3 = x

Aqueous
Equilibria
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Dr. Ali Jabalameli

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Chemistry102

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The Common-Ion Effect
• Therefore, [F−] = x = 1.4 × 10−3
[H3O+] = 0.10 + x = 0.10 + 1.4 × 10−3 = 0.10 M
• So, pH = −log (0.10) pH = 1.00

Aqueous
Equilibria

© 2012 Pearson Education, Inc.

Buffers
• Buffers are solutions that resist changes in pH
•
•
•

when an acid or base is added
They act by neutralizing acid or base that is added to the buffered solution
But just like everything else, there is a limit to what they can do, and eventually the pH changes
Many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion
9blood has a mixture of H2CO3 and HCO3−
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Buffers
• Buffers are solutions of a weak conjugate acid–base pair.
• They are particularly resistant to pH changes, even when strong acid or base is added.
Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Making an Acid Buffer

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How Acid Buffers Work:
Addition of Base
HA(aq) + H2O(l) ⇔ A−(aq) + H3O+(aq)
• Buffers work by applying Le Châtelier’s Principle
•
•

to weak acid equilibrium
Buffer solutions contain significant amounts of the weak acid molecules, HA
These molecules react with added base to neutralize it
HA(aq) + OH−(aq) → A−(aq) + H2O(l)
9 you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium Copyright © 2011 Pearson Education, Inc.

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How Buffers Work
H2O

new
A−

HA
HA

⇔

A−−

+

H3O+

Added
HO−
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16

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How Acid Buffers Work:
Addition of Acid
HA(aq) + H2O(l) ⇔ A−(aq) + H3O+(aq)
• The buffer solution also contains significant
•
•

amounts of the conjugate base anion, A−
These ions combine with added acid to make more HA
H+(aq) + A−(aq) → HA(aq)
After the equilibrium shifts, the concentration of H3O+ is kept constant

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How Buffers Work
H2O

new
HA

HA

⇔

A−−

+

H3O+

Added
H3O+
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Chemistry102

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Buffers

If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the
HF reacts with the OH− to make F− and water.

Aqueous
Equilibria

© 2012 Pearson Education, Inc.

Buffers

Similarly, if acid is added, the F− reacts with it to form
HF and water.
Aqueous
Equilibria
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Chemistry102

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Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
H3O+ + A−

HA + H2O
Ka =

[H3O+] [A−]
[HA]
Aqueous
Equilibria

© 2012 Pearson Education, Inc.

Buffer Calculations
Rearranging slightly, this becomes
Ka = [H3O+]

[A−]
[HA]

Taking the negative log of both side, we get
−log Ka = −log [H3

O+]

[A−]
+ −log
[HA]

pKa pH Dr. Ali Jabalameli

base

acid

Aqueous
Equilibria

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Buffer Calculations
• So

pKa = pH − log

[base]
[acid]

• Rearranging, this becomes pH = pKa + log

[base]
[acid]

• This is the Henderson–Hasselbalch equation. Aqueous
Equilibria

© 2012 Pearson Education, Inc.

Deriving the Henderson-Hasselbalch
Equation

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Henderson-Hasselbalch Equation
• Calculating the pH of a buffer solution can be

•

simplified by using an equation derived from the Ka expression called the HendersonHasselbalch Equation
The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base
9as long as the “x is small” approximation is valid

25

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Example: What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the
[H3O+] from water is ≈ 0

HC2H3O2 + H2O ⇔ C2H3O2− + H3O+ initial change equilibrium 26

Dr. Ali Jabalameli

[HA]

[A−]

[H3O+]

0.100

0.100

≈0

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Chemistry102

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Example : What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC2H3O2 + H2O ⇔ C2H3O2− + H3O+ initial change

[HA]

[A−]

[H3O+]

0.100

0.100

0

−x

+x

+x

equilibrium 0.100 −x 0.100 + x

27

x

Copyright © 2011 Pearson Education, Inc.

Example : What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10−5 determine the value of Ka initial because Ka is very small, approximate change the [HA]eq = [HA]init equilibrium and [A−]eq = [A−]init , then solve for x

28

Dr. Ali Jabalameli

[HA]

[A−]

[H3O+]

0.100

0.100

≈0

−x

+x

+x x 0.100−x 0.100
0.100+x
0.100

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Example : What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10−5

check if the approximation is valid by seeing if x <
5% of
[HC2H3O2]init

initial change equilibrium

[HA]

[A−]

[H3O+]

0.100

0.100

≈0

−x

+x

0.100

0.100

+x x x = 1.8 x 10−5

the approximation is valid
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Example : What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2? substitute x into the equilibrium concentration definitions and solve

initial change equilibrium

[A−]

[H3O+]

0.100

0.100

≈0

−x

+x

+x

0.100−x 0.100
0.100
+ x 1.8E-5 x 0.100

x = 1.8 x 10−5

30

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[HA]

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Chemistry102

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Example : What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2? substitute [H3O+] into the formula for pH and solve

initial change equilibrium

31

[HA]

[A−]

[H3O+]

0.100

0.100

≈0

−x

+x

+x

0.100

0.100

1.8E−5

Copyright © 2011 Pearson Education, Inc.

Example : What is the pH of a buffer that is
0.100 M HC2H3O2 and 0.100 M NaC2H3O2? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

Ka for HC2H3O2 = 1.8 x 10−5 initial change equilibrium [A−]

[H3O+]

0.100

0.100

≈0

−x

+x

+x

0.100

0.100

1.8E−5

the values match 32

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[HA]

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Chemistry102

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Practice − What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?

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Practice − What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF? write the reaction for the acid with water construct an
ICE table for the reaction enter the initial concentrations – assuming the
[H3O+] from water is ≈ 0

HF + H2O ⇔ F− + H3O+ initial change equilibrium 34

Dr. Ali Jabalameli

[HA]
0.14

[A−] [H3O+]
0.071 ≈ 0

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Chemistry102

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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression

HF + H2O ⇔ F− + H3O+ initial change

[HA]

[A−]

[H3O+]

0.14

0.071

0

−x

+x

+x

equilibrium 0.14 −x 0.071 + x

35

x

Copyright © 2011 Pearson Education, Inc.

Practice – What is the pH of a buffer that is
0.14 M and 0.071 M KF?
Ka pK fora HF for =
HF7.0
= 3.15 x 10−4 determine the value of Ka because Ka is very small, approximate the
[HA]eq = [HA]init and [A−]eq = [A−]init solve for x

initial change equilibrium

36

Dr. Ali Jabalameli

[HA]

[A−]

[H3O+]

0.14

0.071

≈0

−x
+x
0.100+x
0.14
−x 0.071
0.012

+x x Copyright © 2011 Pearson Education, Inc.

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Chemistry102

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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10−4 check if the approximation is valid by seeing if x < 5% of
[HC2H3O2]init

initial change equilibrium

[HA]

[A−]

[H3O+]

0.14

0.071

≈0

−x

+x

+x

0.14

0.071

x

x = 1.4 x 10−3

the approximation is valid
Copyright © 2011 Pearson Education, Inc.

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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF? substitute x into the equilibrium concentration definitions and solve initial change equilibrium

[HA]

[A−]

[H3O+]

0.14

0.071

≈0

−x

+x

+x

0.14−x 0.071
0.072
0.14
+ x 1.4E-3 x x = 1.4 x 10−3

38

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Chemistry102

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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF? substitute [H3O+] into the formula for pH and solve

initial change equilibrium

39

[HA]

[A−]

[H3O+]

0.14

0.071

≈0

−x

+x

+x

0.14

0.072

1.4E−3

Copyright © 2011 Pearson Education, Inc.

Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF? check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

Ka for HF = 7.0 x 10−4 initial change equilibrium [A−]

[H3O+]

0.14

0.071

≈0

−x

+x

+x

0.14

0.072

1.4E−3

the values are close enough

40

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[HA]

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Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and
0.10 M in sodium lactate? Ka for lactic acid is 1.4 × 10−4.

Aqueous
Equilibria
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Henderson–Hasselbalch Equation pH = pKa + log

[base]
[acid]

(0.10) pH = −log (1.4 × 10−4) + log (0.12) pH = 3.85 + (−0.08) pH = 3.77
Aqueous
Equilibria
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Chemistry102

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Example : What is the pH of a buffer that is
0.050 M HC7H5O2 and 0.150 M NaC7H5O2? assume the [HA] and HC7H5O2 + H2O ⇔ C7H5O2− + H3O+
[A−] equilibrium
Ka for HC7H5O2 = 6.5 x 10−5 concentrations are the same as the initial substitute into the
HendersonHasselbalch
Equation check the “x is small” approximation 43

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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?

44

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Practice – What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF? find the pKa from the given Ka

HF + H2O ⇔ F− + H3O+

assume the [HA] and [A−] equilibrium concentrations are the same as the initial substitute into the
HendersonHasselbalch
equation check the “x is small” approximation
Tro: Chemistry: A Molecular Approach, 2/e

45

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Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?
• The Henderson-Hasselbalch equation is
•

generally good enough when the “x is small” approximation is applicable
Generally, the “x is small” approximation will work when both of the following are true:
a) the initial concentrations of acid and salt are not very dilute b) the Ka is fairly small

• For most problems, this means that the initial

acid and salt concentrations should be over 100 to 1000x larger than the value of Ka
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pH Range
• The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

How Much Does the pH of a Buffer
Change When an Acid or Base Is Added?
•
•

Though buffers do resist change in pH when acid or base is added to them, their pH does change
Calculating the new pH after adding acid or base requires breaking the problem into two parts
1. a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other
9 added acid reacts with the A− to make more HA
9 added base reacts with the HA to make more A−

2. an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]
Aqueous
Equilibria
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When Strong Acids or Bases Are
Added to a Buffer

When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Addition of Strong Acid or Base to a Buffer
1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Aqueous
Equilibria
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Dr. Ali Jabalameli

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Calculating pH Changes in Buffers
A buffer is made by adding 0.300 mol
HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74.
Calculate the pH of this solution after
0.020 mol of NaOH is added.

Aqueous
Equilibria
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Calculating pH Changes in Buffers
Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8 × 10−5) = 4.74

Aqueous
Equilibria
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Chemistry102

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Calculating pH Changes in Buffers
The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:
HC2H3O2(aq) + OH−(aq) ⎯⎯→ C2H3O2−(aq) + H2O(l)

Before reaction

HC2H3O2

C 2 H 3 O2 −

OH−

0.300 mol

0.300 mol

0.020 mol

After reaction

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Calculating pH Changes in Buffers
The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:
HC2H3O2(aq) + OH−(aq) ⎯⎯→ C2H3O2−(aq) + H2O(l)
HC2H3O2

C 2 H 3 O2 −

OH−

Before reaction

0.300 mol

0.300 mol

0.020 mol

After reaction

0.280 mol

0.320 mol

0.000 mol

Aqueous
Equilibria
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Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log

(0.320)
(0.200)

pH = 4.74 + 0.06 pH = 4.80
Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Example : What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it?
If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−. construct a stoichiometry table for the reaction HC2H3O2 + OH− ⇔ C2H3O2− + H2O
HA
mols before mols added mols after

56

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A−

0.100 0.100
─

─

OH−
0
0.010

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Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? fill in the table
– tracking the changes in the number of moles for each component HC2H3O2 + OH− ⇔ C2H3O2− + H2O
HA
mols before mols added mols after

Tro: Chemistry: A Molecular Approach, 2/e

57

A−

0.100 0.100
─

─

0.090 0.110

OH−
≈0
0.010
≈0

Copyright © 2011 Pearson Education, Inc.

Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
If the added
−
− chemical is a base, HC2H3O2 + OH ⇔ C2H3O2 + H2O write a reaction for
HA
A– OH−
OH− with HA. If the added chemical is mols before 0.100 0.100 0.010 an acid, write a reaction for it with A−. mols change mols end construct a stoichiometry table new molarity for the reaction enter the initial number of moles for each 58

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Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities

HC2H3O2 + OH− ⇔ C2H3O2− + H2O
HA

A−

OH−

0.100 0.100 0.010 mols before mols change −0.010 +0.010 −0.010
0
mols end
0.090 0.110
0
new molarity 0.090 0.110

59

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Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the
[H3O+] from water is ≈ 0, and using the new molarities of the
[HA] and [A−]

HC2H3O2 + H2O ⇔ C2H3O2− + H3O+ initial change equilibrium 60

Dr. Ali Jabalameli

[HA]

[A−]

[H3O+]

0.090

0.110

≈0

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Chemistry102

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Example : What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression

HC2H3O2 + H2O ⇔ C2H3O2− + H3O+ initial change

[HA]

[A−]

[H3O+]

0.090

0.110

≈0

−x

+x

+x

equilibrium 0.090 −x 0.110 + x

Tro: Chemistry: A Molecular Approach, 2/e

61

x

Copyright © 2011 Pearson Education, Inc.

Example : What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10−5 determine the value of Ka
[HA]
[A−]
[H3O+]
because Ka is very initial
0.100
0.100
≈0
small, approximate
−x
+x
+x
the [HA]eq = [HA]init change
−
− and [A ]eq = [A ]init
0.090−x 0.110
0.110+x
equilibrium 0.090 x solve for x

62

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Chemistry102

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Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? check if the approximation is valid by seeing if x < 5% of
[HC2H3O2]init

Ka for HC2H3O2 = 1.8 x 10−5 initial change equilibrium [HA]

[A−]

[H3O+]

0.090

0.110

≈0

−x

+x

0.090

0.110

+x x x = 1.47 x 10−5

the approximation is valid
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63

Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? substitute x into the equilibrium concentration definitions and solve

initial change equilibrium

[A-]

[H3O+]

0.090

0.110

≈0

−x

+x

+x

0.090−x 0.110
0.110+ x
0.090

1.5E-5 x x = 1.47 x 10−5

64

Dr. Ali Jabalameli

[HA]

Copyright © 2011 Pearson Education, Inc.

32

Chemistry102

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Example : What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? substitute [H3O+] into the formula for pH and solve initial change equilibrium

65

[HA]

[A−]

[H3O+]

0.090

0.110

≈0

−x

+x

+x

0.090

0.110

1.5E−5

Copyright © 2011 Pearson Education, Inc.

Example: What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10−5

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

initial change equilibrium

[A−]

[H3O+]

0.090

0.110

≈0

−x

+x

+x

0.090

0.110

1.5E−5 the values match 66

Dr. Ali Jabalameli

[HA]

Copyright © 2011 Pearson Education, Inc.

33

Chemistry102

5/7/2013

or, by using the
Henderson-Hasselbalch
Equation

67

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Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]

HC2H3O2 + H2O ⇔ C2H3O2− + H3O+
[HA]

[A−]

0.090
0.110
initial
−x
+x change equilibrium 0.090 −x 0.110 + x

[H3O+]
≈0

+x x fll in the ICE table

68

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

34

Chemistry102

5/7/2013

Example : What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? find the pKa from the given Ka assume the [HA] and [A−] equilibrium concentrations are the same as the initial initial change HC2H3O2 + H2O ⇔ C2H3O2− + H3O+
Ka for HC2H3O2 = 1.8 x 10−5

[HA]

[A−]

[H3O+]

0.090

0.110

≈0

−x

+x

+x

0.110

x

equilibrium 0.090

69

Copyright © 2011 Pearson Education, Inc.

Example : What is the pH of a buffer that has
0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in
1.00 L that has 0.010 mol NaOH added to it? substitute into the
HendersonHasselbalch
equation check the “x is small” approximation

HC2H3O2 + H2O ⇔ C2H3O2− + H3O+ pKa for HC2H3O2 = 4.745

70

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

35

Chemistry102

5/7/2013

Example : Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water
HC2H3O2 + H2O ⇔ C2H3O2− + H3O+

pKa for HC2H3O2 = 4.745

71

Copyright © 2011 Pearson Education, Inc.

Practice – What is the pH of a buffer that has
0.140 moles HF (pKa = 3.15) and 0.071 moles
KF in 1.00 L of solution when 0.020 moles of
HCl is added?
(The “x is small” approximation is valid)

72

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

36

Chemistry102

5/7/2013

Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added?
If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−. construct a stoichiometry table for the reaction F− + H3O+ ⇔ HF + H2O

mols before mols added mols after

73

F−

H3O+

HF

0.071

0

0.140

–

0.020

–

Copyright © 2011 Pearson Education, Inc.

Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? fill in the table
– tracking the changes in the number of moles for each component F− + H3O+ ⇔ HF + H2O

mols before mols added mols after

74

Dr. Ali Jabalameli

F−

H3O+

HF

0.071

0

0.140

–

0.020

–

0.051

0

0.160

Copyright © 2011 Pearson Education, Inc.

37

Chemistry102

5/7/2013

Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added?
If the added chemical is a base, write a reaction for OH− with
HA. If the added chemical is an acid, write a reaction for
H3O+ with A−. construct a stoichiometry table for the reaction enter the initial number of moles for each F− + H3O+ ⇔ HF + H2O

mols before

F−

H3O+

HF

0.071

0.020

0.140

mols change mols after new molarity

75

Copyright © 2011 Pearson Education, Inc.

Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities

F− + H3O+ ⇔ HF + H2O
F−

H3O+

HF

mols before

0.071

0.020

0.140

mols change

−0.020

mols after

0.051

0

0.160

new molarity

0.051

0

0.160

76

Dr. Ali Jabalameli

−0.020 +0.020

Copyright © 2011 Pearson Education, Inc.

38

Chemistry102

5/7/2013

Practice – What is the pH of a buffer that has
0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added?
HF + H2O ⇔ F− + H3O+ write the reaction for the acid with water construct an ICE table. assume the [HA] and
[A−] equilibrium concentrations are the same as the initial initial change [HF]

[F−]

[H3O+]

0.160

0.051

≈0

−x

+x

+x

0.051

x

equilibrium 0.160

substitute into the
HendersonHasselbalch
equation
77

Copyright © 2011 Pearson Education, Inc.

Basic Buffers
B:(aq) + H2O(l) ⇔ H:B+(aq) + OH−(aq)
• Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−
H2O(l) + NH3 (aq) ⇔ NH4+(aq) + OH−(aq)

78

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

39

Chemistry102

5/7/2013

Henderson-Hasselbalch Equation for
Basic Buffers
•
•

•

The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product
The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H2O ⇔ H:B+ + OH−
To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction
H:B+ + H2O ⇔ B: + H3O+
9

this does not affect the concentrations, just the way we are looking at the reaction
79

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Relationship between pKa and pKb
• Just as there is a relationship between the Ka

of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base
Ka • Kb = Kw = 1.0 x 10−14
−log(Ka • Kb) = −log(Kw) = 14
−log(Ka) + −log(Kb) = 14 pKa + pKb = 14
80

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

40

Chemistry102

5/7/2013

Example : What is the pH of a buffer that is 0.50
M NH3 (pKb = 4.75) and 0.20 M NH4Cl?
NH3 + H2O ⇔ NH4+ + OH−

find the pKa of the conjugate acid (NH4+) from the given Kb assume the [B] and
[HB+] equilibrium concentrations are the same as the initial substitute into the
HendersonHasselbalch equation check the “x is small” approximation 81

Copyright © 2011 Pearson Education, Inc.

Henderson-Hasselbalch Equation for
Basic Buffers
•
•

•

The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product
The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product
B: + H2O ⇔ H:B+ + OH−
We can rewrite the Henderson-Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH

82

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

41

Chemistry102

5/7/2013

Example : What is the pH of a buffer that is 0.50
M NH3 (pKb = 4.75) and 0.20 M NH4Cl? find the pKb if given Kb

NH3 + H2O ⇔ NH4+ + OH−

assume the [B] and
[HB+] equilibrium concentrations are the same as the initial substitute into the
Henderson-Hasselbalch
equation base form, find pOH check the “x is small” approximation calculate pH from pOH
Copyright © 2011 Pearson Education, Inc.

83

Titration
In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base).

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Dr. Ali Jabalameli

42

Chemistry102

5/7/2013

Titration

A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

Aqueous
Equilibria

© 2012 Pearson Education, Inc.

Titration
• In an acid-base titration, a solution of unknown

concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete
9 when the reaction is complete we have reached the endpoint of the titration

• An indicator may be added to determine the endpoint ( if no PH meter)

9 an indicator is a chemical that changes color when the pH changes

• When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point
86

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

43

Chemistry102

5/7/2013

Titration

87

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Different Type of Titrations:
•
•
•
•

Strong Acid vs. Strong Base
Weak Acid vs. Strong Base
Weak Base vs. Strong Acid
Polyprotic Acid vs. Strong Base

Aqueous
Equilibria
88

Dr. Ali Jabalameli

44

Chemistry102

5/7/2013

Titration of a Strong Acid with a
Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly. Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Titration of a Strong Acid with a
Strong Base
Just before
(and after) the equivalence point, the pH increases rapidly.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Dr. Ali Jabalameli

45

Chemistry102

5/7/2013

Titration of a Strong Acid with a
Strong Base
At the equivalence point, moles acid
= moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Titration of a Strong Acid with a
Strong Base
As more base is added, the increase in pH again levels off.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Dr. Ali Jabalameli

46

Chemistry102

5/7/2013

Titration Curve:
Unknown Strong Base Added to Strong Acid

93

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Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

Because the solutions are equal concentration, and
1:1 stoichiometry, the equivalence point is
•
at equal volumes

After Equivalence
(excess base)
Equivalence Point equal moles of
HCl and NaOH pH = 7.00

Before Equivalence
(excess acid)

94

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

47

Chemistry102

5/7/2013

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

•
•
•
•

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
Before equivalence point added 5.0 mL NaOH
5.0 x 10−4 mole NaOH added

mols before

HCl

NaCl

NaOH

2.50E-3

0

5.0E-4

mols change −5.0E-4 +5.0E-4 −5.0E-4 mols end

2.00E-3 5.0E-4

molarity, new 0.0667

95

0.017

0
0

Copyright © 2011 Pearson Education, Inc.

Titration of 25 mL of 0.100 M HCl with
0.100 M NaOH
• HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(aq)
• To reach equivalence, the added moles NaOH =
•
•

initial moles of HCl = 2.50 x 10−3 moles
At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over
Because the NaCl is a neutral salt, the pH at equivalence = 7.00

96

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

48

Chemistry102

5/7/2013

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

•
•
•
•

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
At equivalence point added 25.0 mL NaOH
2.5 x 10−3 mole NaOH added

HCl

NaCl

NaOH

mols before 2.50E-3
0
2.5E-3 mols change −2.5E-3 +2.5E-3 −2.5E-3 mols end

0

2.5E-3

0

molarity, new

0

0.050

0

Copyright © 2011 Pearson Education, Inc.

97

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

•
•
•
•

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Initial pH = −log(0.100) = 1.00
Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3 added 30.0 mL NaOH
After equivalence point
0.100 mol NaOH
L of added NaOH ×
1L
= moles added NaOH

3.0 x 10−3 mole NaOH added

mols before mols change

HCl

NaCl

NaOH

2.50E-3

0

3.0E-3

−2.5E-3 +2.5E-3 −2.5E-3

mols end

0

2.5E-3

5.0E-4

molarity, new

0

0.045

0.0091

Copyright © 2011 Pearson Education, Inc.

Dr. Ali Jabalameli

49

Chemistry102

5/7/2013

Adding 0.100 M NaOH to 0.100 M HCl
30.0
5.0
10.0
25.0 mL added
35.0
mLNaOH
NaOH
25.0 mL
0.100
M
HCl
0.00050
0.00200
0.00150 equivalence 0.00100 mol point
NaOH
0.00250
HCl
1.18
1.37
7.00 pH = 11.96
12.22
1.00 added 15.0
40.0 mL NaOH
0.00150 mol NaOH
0.00100
HCl pH = 1.60
12.36
added 20.0
50.0 mL NaOH
0.00250 mol NaOH
0.00050
HCl pH = 1.95
12.52

99

Copyright © 2011 Pearson Education, Inc.

Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3

100

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

50

Chemistry102

5/7/2013

•
•
•
•

Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
Before equivalence point added 10.0 mL NaOH mols before

HNO3

NaNO3

NaOH

1.25E-2

0

1.5E-3

mols change −1.5E-3 +1.5E-3 −1.5E-3 mols end

1.1E-3

1.5E-3

0

molarity, new

0.018

0.025

0

101

Copyright © 2011 Pearson Education, Inc.

Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M
HNO3 to reach equivalence

102

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

51

Chemistry102

5/7/2013

Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M
HNO3 to reach equivalence

•
•
•
•

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
At equivalence point: moles of NaOH = 1.25 x 10−2

103

Copyright © 2011 Pearson Education, Inc.

Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3

104

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

52

Chemistry102

5/7/2013

•
•
•
•

Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60
Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
After equivalence point added 100.0 mL NaOH
HNO3

NaNO3

NaOH

mols before

1.25E-2

0

1.5E-2

mols change

−1.25E-2 +1.25E-2 −1.25E-2

mols end

0

1.25E-2

0.0025

molarity, new

0

0.0833

0.017

Copyright © 2011 Pearson Education, Inc.

•
•
•
•

Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to
50.0 mL of 0.25 M HNO3
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
Initial pH = −log(0.250) = 0.60 initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
After equivalence point added 100.0 mL NaOH
HNO3

NaNO3

NaOH

mols before

1.25E-2

0

1.5E-2

mols change

−1.25E-2 +1.25E-2 −1.25E-2

mols end

0

1.25E-2

0.0025

molarity, new

0

0.0833

0.017

106

Dr. Ali Jabalameli

Copyright © 2011 Pearson Education, Inc.

53

Chemistry102

5/7/2013

Titration of a Strong Base with a Strong Acid

• If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown 107

Copyright © 2011 Pearson Education, Inc.

Titration of a Weak Acid with a
Strong Base
• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
• At the equivalence point the pH is >7.
• Phenolphthalein is commonly used as an indicator in these titrations. Aqueous
Equilibria

© 2012 Pearson Education, Inc.

Dr. Ali Jabalameli

54

Chemistry102

5/7/2013

Titration of a Weak Acid with a
Strong Base

At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.
Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Titration of a Weak Acid with a
Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Dr. Ali Jabalameli

55

Chemistry102

5/7/2013

Titration of a Weak Acid with a Strong Base
• Titrating a weak acid with a strong base results in
•
•
•
•

differences in the titration curve at the equivalence point and excess acid region
The initial pH is determined using the Ka of the weak acid
The pH in the excess acid region is determined as you would determine the pH of a buffer
The pH at the equivalence point is determined using the Kb of the conjugate base of the weak acid
The pH after equivalence is dominated by the excess strong base
9 the basicity from the conjugate base anion is negligible
111

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Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH

• HCHO2(aq) + NaOH(aq) → NaCHO2(aq) + H2O(aq)
• Initial pH
[HCHO2] [CHO2−] [H3O+] initial change

0.100

0.000

≈0

−x

+x

+x

x

x

equilibrium 0.100 − x

112

Dr. Ali Jabalameli

Ka = 1.8 x 10−4

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56

Chemistry102

5/7/2013

Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH

• HCHO2(aq) + NaOH(aq) → NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence added 5.0 mL NaOH

HA

A−

OH−

mols before

2.50E-3

0

0

mols added

–

–

5.0E-4

mols after

2.00E-3 5.0E-4

≈0

113

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Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH
• HCHO2(aq) + NaOH(aq) → NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence
CHO2−(aq) + H2O(l) ⇔ HCHO2(aq) + OH−(aq)
[HCHO2] [CHO2−] [OH−]
HA
A−
OH−
initial
0
0.0500
≈0
mols before
2.50E-3
0
0
change
+x –
−x –
+x
mols added
2.50E-3
equilibrium x 0 5.00E-2-x mols after
2.50E-3 x ≈ 0

added 25.0 mL NaOH
Kb = 5.6 x 10−11

[OH−] = 1.7 x 10−6 M

Copyright © 2011 Pearson Education, Inc.

Dr. Ali Jabalameli

57

Chemistry102

5/7/2013

Titration of 25 mL of 0.100 M HCHO2 with
0.100 M NaOH

• HCHO2(aq) + NaOH(aq) → NaCHO2(aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence added 30.0 mL NaOH

3.0 x 10−3 mole NaOH added

HA

A−

NaOH

mols before 2.50E-3
0
3.0E-3 mols change −2.5E-3 +2.5E-3 −2.5E-3 mols end

0

2.5E-3 5.0E-4

molarity, new

0

0.045

115

0.0091

Copyright © 2011 Pearson Education, Inc.

Adding NaOH to HCHO2

added 5.0
25.0
mLNaOH
NaOH
30.0
10.0mL
initial HCHO2 solution equivalence point
0.00200
0.00050 molmL NaOH
0.00150
HCHO
2xs
added
35.0
NaOH
0.00250
mol HCHO
− 2
0.00250
mol
CHO
3.14 pH =
11.96
3.56
2 xs
0.00100
mol NaOH pH = 2.37
−]
[CHO
=
0.0500
M
2 init pH =− 12.22
−6
[OH ]eq = 1.7 x 10 added 12.5 mL NaOH pH = 8.23 added 40.0 mL NaOH
0.00125 mol HCHO2
0.00150 mol NaOH xs pH = 3.74 = pKa pH = 12.36 half-neutralization 15.0 mL NaOH added 50.0
0.00100 mol NaOH
HCHO2xs
0.00250
3.92
pH = 12.52 added 20.0 mL NaOH
0.00050 mol HCHO2 pH = 4.34

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Titrating Weak Acid with a Strong Base
• The initial pH is that of the weak acid solution
9calculate like a weak acid equilibrium problem

• Before the equivalence point, the solution becomes a buffer

9calculate mol HAinit and mol A−init using reaction stoichiometry 9calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init

• Half-neutralization pH = pKa

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Titrating Weak Acid with a Strong Base
• At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established
9mol A− = original mole HA
¾calculate the volume of added base as you did in
¾[A−]init = mol A−/total liters

9calculate like a weak base equilibrium problem

• Beyond equivalence point, the OH is in excess
9[OH−] = mol MOH xs/total liters
9[H3O+][OH−]=1 x 10−14

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Example : A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of
KOH at the equivalence point. write an equation for the reaction for B with HA

HNO2 + KOH → NO2− + H2O

use stoichiometry to determine the volume of added B

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Example : A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH.

write an equation for the reaction for B with HA

HNO2 + KOH → NO2− + H2O

determine the moles of HAbefore
& moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A− made

mols before mols added mols after
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HNO2

NO2−

OH−

0.00400

0

≈0

–

–

0.00100

0.00300 0.00100

≈0

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Example : A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH.
HNO2 + H2O ⇔ NO2− + H3O+ write an equation for the reaction of HA with H2O

Ka = 4.6 x 10−4

determine Ka and pKa for HA use the
HendersonHasselbalch
equation to determine the pH HNO2

NO2−

OH−

0
≈0
mols before 0.00400
0.00100
mols added
–
–
0.00300 0.00100 mols after
≈0
121

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Example : A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. write an equation for the reaction for B with HA determine the moles of HAbefore
& moles of added
B
make a stoichiometry table and determine the moles of HA in excess and moles A− made

HNO2 + KOH → NO2− + H2O

at half-equivalence, moles KOH = ½ mole HNO2

HNO2

OH−

0
≈0
mols before 0.00400
0.00200
mols added
–
–
0.00200 0.00200 mols after
≈0

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NO2−

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Example : A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. write an equation for the reaction of
HA with H2O

HNO2 + H2O ⇔ NO2− + H3O+
Ka = 4.6 x 10-4

determine Ka and pKa for HA use the
HendersonHasselbalch
equation to determine the pH

HNO2

NO2−

OH−

0
≈0
mols before 0.00400
0.00200
mols added
–
–
0.00200 0.00200 mols after
≈0
123

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Titration of a Weak Base with a
Strong Acid

• The pH at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea.
• Methyl red is the indicator of choice.

Aqueous
Equilibria

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Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb =
4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl.

• NH3(aq) + HCl(aq) → NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence: after adding 30.0 mL HCl
NH3

NH4Cl

HCl

mols before

2.50E-3

0

0

mols added

-

-

3.0E-3

mols change

−2.5E-3 +2.5E-3 −2.5E-3

mols end

0

2.5E-3

5.0E-4

molarity, new

0

0.045

0.0091

when you mix a strong acid, HCl, with a weak acid, NH4+, you only need to consider the strong acid
125

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Titrations of Polyprotic Acids
When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. Aqueous
Equilibria
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Chemistry102

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Titration of a Polyprotic Acid
• If Ka1 >> Ka2, there will be two equivalence points in the titration

9the closer the Ka’s are to each other, the less distinguishable the equivalence points are

titration of 25.0 mL of 0.100 M
H2SO3 with 0.100 M NaOH

127

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Solubility Equilibria
• All ionic compounds dissolve in water to some degree 9however, many compounds have such low solubility in water that we classify them as insoluble

• We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water

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Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s)

Ba2+(aq) + SO42−(aq)

Aqueous
Equilibria
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Solubility Products
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42−] where the equilibrium constant, Ksp, is called the solubility product.
Aqueous
Equilibria
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Chemistry102

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Solubility Product
• The equilibrium constant for the dissociation of a
•
•
•
•

solid salt into its aqueous ions is called the solubility product, Ksp
For an ionic solid MnXm, the dissociation reaction is:
MnXm(s) ⇔ nMm+(aq) + mXn−(aq)
The solubility product would be
Ksp = [Mm+]n[Xn−]m
For example, the dissociation reaction for PbCl2 is
PbCl2(s) ⇔ Pb2+(aq) + 2 Cl−(aq)
And its equilibrium constant is
Ksp = [Pb2+][Cl−]2
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Solubility Products
• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or
100 mL (g/mL) of solution, or in mol/L (M).
Aqueous
Equilibria
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Factors Affecting Solubility
• The Common-Ion Effect
– If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease:

BaSO4(s)

Ba2+(aq) + SO42−(aq)
Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Factors Affecting Solubility
• pH
– If a substance has a basic anion, it will be more soluble in an acidic solution.
– Substances with acidic cations are more soluble in basic solutions.

Aqueous
Equilibria
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Factors Affecting Solubility
• Complex Ions
– Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Factors Affecting Solubility
• Complex
Ions
– The formation of these complex ions increases the solubility of these salts.
Aqueous
Equilibria
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Chemistry102

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Factors Affecting Solubility
• Amphoterism
– Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
– Examples of such cations are Al3+, Zn2+, and Sn2+.

Aqueous
Equilibria
© 2012 Pearson Education, Inc.

Will a Precipitate Form?
• In a solution,
– If Q = Ksp, the system is at equilibrium and the solution is saturated.
– If Q < Ksp, more solid can dissolve until Q = Ksp.
– If Q > Ksp, the salt will precipitate until
Q = Ksp.

Aqueous
Equilibria
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Chemistry102

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Molar Solubility
• Solubility is the amount of solute that will dissolve in a given amount of solution
9at a particular temperature

• The molar solubility is the number of moles of solute that will dissolve in a liter of solution
9the molarity of the dissolved solute in a saturated solution Copyright © 2011 Pearson Education, Inc.

139

Example : Calculate the molar solubility of
PbCl2 in pure water at 25 °C write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid PbCl2(s) ⇔ Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]

[Cl−]

0

0

Change

+S

+2S

Equilibrium

S

2S

Initial

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Example : Calculate the molar solubility of
PbCl2 in pure water at 25 °C substitute into the Ksp expression find the value of
Ksp from Table
16.2, plug into the equation, and solve for S

Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
[Pb2+]

[Cl−]

0

0

Change

+S

+2S

Equilibrium

S

2S

Initial

141

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Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 °C is 1.05 x 10−2 M

142

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Chemistry102

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Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 °C is 1.05 x 10−2 M write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid PbBr2(s) ⇔ Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2 initial [Pb2+]

[Br−]

0

0

change

+(1.05 x 10−2) +2(1.05 x 10−2)

equilibrium

(1.05 x 10−2)

(2.10 x 10−2)

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Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 °C is 1.05 x 10−2 M substitute into the Ksp expression plug into the equation and solve Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10−2)(2.10 x 10−2)2

initial

[Br−]

0

0

change

+(1.05 x 10−2) +2(1.05 x 10−2)

equilibrium

(1.05 x 10−2)

144

Dr. Ali Jabalameli

[Pb2+]

(2.10 x 10−2)

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Ksp and Relative Solubility
• Molar solubility is related to Ksp
• But you cannot always compare solubilities of
•

compounds by comparing their Ksps
To compare Ksps, the compounds must have the same dissociation stoichiometry

145

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The Effect of Common Ion on
Solubility

• Addition of a soluble salt that contains one of
•

the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt
For example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2
PbCl2(s) ⇔ Pb2+(aq) + 2 Cl−(aq) addition of Cl− shifts the equilibrium to the left
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Chemistry102

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Example : Calculate the molar solubility of
CaF2 in 0.100 M NaF at 25 °C write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid CaF2(s) ⇔ Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]

[F−]

0

0.100

change

+S

+2S

equilibrium

S

0.100 + 2S

initial

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147

Example : Calculate the molar solubility of
CaF2 in 0.100 M NaF at 25 °C substitute into the Ksp expression, assume S is small find the value of
Ksp from Table
16.2, plug into the equation, and solve for S

Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
[Ca2+]

[F−]

0

0.100

change

+S

+2S

equilibrium

S

0.100 + 2S

initial

148

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Chemistry102

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Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M
Ksp of AgCl = 1.77 x 10−10

149

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Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid AgCl(s) ⇔ Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−]
[Ag+]

[Cl−]

0

0.55

change

+S

+S

equilibrium

S

0.55 + S

initial

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Chemistry102

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Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M substitute into the Ksp expression, assume S is small find the value of
Ksp from Table
16.2, plug into the equation, and solve for S

Ksp = [Ag+][Cl−]
Ksp = (S)(0.55 + S)
Ksp = (S)(0.55)
[Ag+]

[Cl−]

0

0.55

Change

+S

+S

Equilibrium

S

0.55 + S

Initial

151

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The Effect of pH on Solubility
• For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide
9 and the lower the pH, the higher the solubility
9 higher pH = increased [OH−]

•

M(OH)n(s) ⇔ Mn+(aq) + nOH−(aq)
For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M2(CO3)n(s) ⇔ 2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq) ⇔ HCO3− (aq) + H2O(l)
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Precipitation
• Precipitation will occur when the concentrations of
•

the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur

9 Q = Ksp, the solution is saturated, no precipitation
9 Q < Ksp, the solution is unsaturated, no precipitation
9 Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate

• Some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions

Tro: Chemistry: A Molecular Approach, 2/e

153

a supersaturated solution will precipitate if a seed crystal is added precipitation occurs if Q > Ksp
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Selective Precipitation
• A solution containing several different cations

•

can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others
A successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different

155

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Example : Will a precipitate form when we mix
Pb(NO3)2(aq) with NaBr(aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively? write the equation for the reaction determine the ion concentrations of the original salts determine the Ksp for any “insoluble” product write the dissociation reaction for the insoluble product

Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)

Pb(NO3)2 = 0.0150 M
∴Pb2+ = 0.0150 M,
NO3− = 2(0.0150 M)

NaBr = 0.0350 M
∴Na+ = 0.0350 M,
Br− = 0.0350 M

Ksp of PbBr2 = 4.67 x 10–6
PbBr2(s) ⇔ Pb2+(aq) + 2 Br−(aq)

calculate Q, using the ion concentrations compare Q to Ksp. If
Q > Ksp, precipitation

Q < Ksp, so no precipitation
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Chemistry102

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Practice – Will a precipitate form when we mix
Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both 0.0175 M?
Ksp of Ca(OH)2 = 4.68 x 10−6

157

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Practice – Will a precipitate form when we mix
Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both 0.0175 M? write the equation for the reaction determine the ion concentrations of the original salts determine the Ksp for any “insoluble” product write the dissociation reaction for the insoluble product

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)

Ca(NO3)2 = 0.0175 M
∴Ca2+ = 0.0175 M,
NO3− = 2(0.0175 M)

NaOH = 0.0175 M
∴Na+ = 0.0175 M,
OH− = 0.0175 M

Ksp of Ca(OH)2 = 4.68 x 10–6
Ca(OH)2(s) ⇔ Ca2+(aq) + 2 OH−(aq)

calculate Q, using the ion concentrations compare Q to Ksp. If
Q > Ksp, precipitation

Q > Ksp, so precipitation
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Example : What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater? Mg(OH)2(s) ⇔ Mg2+(aq) + 2 OH−(aq) precipitating may just occur when Q = Ksp

159

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Practice – What is the minimum concentration of
Ca(NO3)2(aq) that will precipitate Ca(OH)2 from
0.0175 M NaOH(aq)?
Ksp of Ca(OH)2 = 4.68 x 10−6

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Practice – What is the minimum concentration of
Ca(NO3)2(aq) that will precipitate Ca(OH)2 from
0.0175 M NaOH(aq)?
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)

Ca(OH)2(s) ⇔ Ca2+(aq) + 2 OH−(aq)

precipitating may just occur when Q = Ksp

[Ca(NO3)2] = [Ca2+] = 0.0153 M
161

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Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture. Aqueous
Equilibria
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