AcidBase Equilibria

1
Acid-Base Equilibria

• Chapter 16

2
Weak Acids

• The concentration of ions (specifically H) are
  low in weak acids, even though HA is dissolved.

Concentration
HA
H
A-
3
Acid Ionization

• Boric acid, B(OH)3, is used as a mild antiseptic.
  What is the pH of a 0.025 M aqueous solution of
  boric acid? What is the degree of ionization of
  boric acid in this solution?

4
Acid Ionization

• The hydrogen ion arises principally from the
  reaction
• B(OH)3(aq) H2O(l) B(OH)4-(aq)
  H(aq)
• Ka 5.9 x 10-10

5
Acid Ionization

• To solve, assemble a table of starting, change,
  and equilibrium concentrations.
• Use HBo as the symbol for boric acid Bo- as
  the symbol for B(OH)4-.

6
Acid Ionization
To solve, assemble a table of starting, change,
and equilibrium concentrations. Use HBo as the
symbol for boric acid Bo- as the symbol for
B(OH)4-

• Conc. (M) HBo H
  Bo-
• Starting 0.025 0
  0
• Change -x x
  x
• Equilibrium 0.025 - x x
  x

7
Acid Ionization
The value of x equals the value of the
molarity of the H ion, which can be obtained
from the equilibrium-constant expression.
Substitute into the equilibrium-constant
expression and solve for x. HBo-
x2
5.9 x 10-10
Ka
HBo
(0.025 - x)
8
Acid Ionization
Solve the equation for x, assuming that x is much
smaller than 0.025, so that (0.025 - x) is
approximately 0.025. Solve by taking the square
root of (5.9 x 10-10) x (0.025).
x2
5.9 x 10-10
?
(0.025)
?
(5.9 x 10-10 (0.025)) 1.475 x 10-11
x2
x 3.84 x 10-6 M H degree of
ionization (3.84 x 10-6)/0.025
0.000153
9
Acid Ionization

• Check to make sure that the assumption that
  (0.025 - x) ? 0.025 is valid
• 0.025 - (3.84 x 10-6) 0.02499, or ? 0.025 to
  two significant figures.
• pH -logH log (3.84 x 10-6) 5.4156

10
Weak Acids and Bases

• Weak acids and bases involve partial dissociation
  of acid and base salts.
• Consider an aqueous solution of lactic acid
  (HC3H5O3)

11
Weak Acids and Bases

• Weak acids and bases involve partial dissolution
  of acid and base salts.
• Consider an aqueous solution of lactic acid
  (HC3H5O3) with 1.037 moles of acid in 1 liter of
  water.

HC3H5O3 (aq)
H (aq) C3H5O3- (aq)
12
Weak Acids and Bases

• Weak acids and bases involve partial dissolution
  of acid and base salts.
• Consider an aqueous solution of lactic acid
  (HC3H5O3) with 1.037 moles of acid in 1 liter of
  water.

HC3H5O3 (aq)
H (aq) C3H5O3- (aq)
at equilibrium
1.0
0.037
0.037
13
Lactic Acid Equilibrium - pKa
HC3H5O3 (aq)
H (aq) C3H5O3- (aq)
HC3H5O3-
Ka
HC3H5O3
14
Lactic Acid Equilibrium - pKa
HC3H5O3 (aq)
H (aq) C3H5O3- (aq)
HC3H5O3-
0.0370.037
Ka

HC3H5O3
1.0
1.4 x 10-3 M
15
Lactic Acid Equilibrium - pKa
HC3H5O3 (aq)
H (aq) C3H5O3- (aq)
HC3H5O3-
0.0370.037
Ka

HC3H5O3
1.0
1.4 x 10-3 M
pKa is defined as the log of Ka
pKa -log Ka -log1.4 x 10-3 M 2.9
16
Conjugate Acids and Bases

• When any acid dissolves in solution, a conjugate
  base is created.
• e.g. HA A- H

acid
conjugate base
17
Conjugate Acids and Bases

• When any acid dissolves in solution, a conjugate
  base is created.
• e.g. HA A- H

acid
conjugate base
So What if we have NaA dissolved in water?
18
Conjugate Acids and Bases

• When any acid dissolves in solution, a conjugate
  base is created.
• e.g. HA A- H

acid
conjugate base
So, what if we have NaA dissolved in water?
NaA is a weak base
19
Solving Weak Acid Problems

• List the major species in the solution
• Write the equilibrium expression for the dominant
  H -producing reaction.
• Using the initial concentrations, solve for the
  change in concentration necessary to determine
  Ka.
• Calculate H and pH.

20
Base vs. Acid Dissociation Constant

• The product of the acid-dissociation constant
  times the conjugate base-ionization constant is
  equal to the water dissociation
  constant. Ka Kb Kw 1.0 x 10-14

21
A Problem with Base

• A solution contains a mixture of 0.25 M NH3 (Kb
  1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH
  of the solution.
• What are the major species in solution?
• NH3 , NH4 , Cl- , H2O
• What is the key equilibrium?
• NH3 (aq) H2O (l) NH4 (aq) OH- (aq)

22
A Problem with Base

• A buffered solution contains 0.25 M NH3 (Kb 1.8
  x 10-5) and 0.40 M NH4Cl. Calculate the pH of
  the solution.
• What are the major species in solution?
• NH3 , NH4 , Cl- , H2O
• What is the key equilibrium?
• NH3 (aq) H2O (l) NH4 (aq) OH- (aq)

strong base
weak base
weak acid
weak acid
23
Solve for X

• NH3 (aq) H2O (l) NH4 (aq) OH- (aq)

NH4OH-
Kb 1.8 x 10-5
NH3
24
Solve for X

• NH3 (aq) H2O (l) NH4 (aq) OH- (aq)

NH4OH-
Kb 1.8 x 10-5
NH3
25
Algebra and Simplification

• NH3 (aq) H2O (l) NH4 (aq) OH- (aq)

NH4OH-
Kb 1.8 x 10-5
NH3
0.40 XX

0.25 - X
26
Algebra and Simplification

• NH3 (aq) H2O (l) NH4 (aq) OH- (aq)

NH4OH-
Kb 1.8 x 10-5
NH3
X must be very small
0.40 XX

0.25 - X
0.40 XX

0.25 - X
X OH- 1.1 x 10-5 M
27
And Finally

• pOH - log OH- 4.95
• pH 14.00 - 4.95 9.05
• Is this basic or acidic?
• BASIC

28
Polyprotic Acids

• Acids (or bases) that generate more than one
  equivalent upon dissolving in aqueous solution
  typically undergo a series of acid/base
  equilibria.
• Consider H3PO4

H3PO4 H H2PO4-
H2PO4 H HPO42-
HPO4 H PO43-
29
Polyprotic Acids

• Acids (or bases) that generate more than one H
  (OH-) upon dissolving in aqueous solution
  typically undergo a series of acid/base
  equilibria.
• Consider H3PO4

H3PO4 H H2PO4-
Ka1 7.5 x 10-3 Ka2 6.2 x 10-8 Ka3 4.8 x
10-13
H2PO4 H HPO42-
HPO4 H PO43-
30
Strengths of Polyprotic Acids

• Ka1 gt Ka2 gt Ka3

31
Titration Curves

• A 0.400-g sample of propionic acid was dissolved
  in water to give 50.0 mL of solution. This
  solution was titrated with 0.150 M NaOH. What
  was the pH of the solution when the equivalence
  point was reached?

32
Titration Curves
Rounded answer 8.84 (OH- 6.9 x 10-6 M)
Use HPr to symbolize propionic acid and Pr-
to symbolize the propionate anion. At the
equivalence point, equal molar amounts of
Hpr and NaOH react to form a solution of NaPr.
33
Titration Curves
Start by calculating the moles of HPr. Use
this to calculate the volume of NaOH needed to
neutralize all of the HPr (and use the moles
of HPr as the Moles of Pr- formed at the
equivalence point). Add the volume of NaOH to
the original 0.050 L to find the total volume of
solution.
34
Titration Curves

• Mol HPr 0.40 g HPr ??74.08 g HPr/mol HPr
  0.0054 mol HPr
• L NaOH 0.0054 mol NaOH ? 0.150 mol NaOH/L
  0.036 L
• Total volume 0.036 L 0.050 L HPr soln
  0.086 L
• Pr- (0.0054 mol Pr- from HPr) ??0.086 L
  0.063 M

35
Titration Curves

• Because the Pr- hydrolyzes to OH- and HPr, use
  this to calculate the OH-. Start by
  calculating the Kb constant of Pr- from the Ka of
  its conjugate acid, HPr. Then assemble the usual
  table of concentrations, assume x is negligible,
  and calculate OH- and pH.

36
Titration Curves
Kw
(1.0 x 10-14)
7.69 x 10-10
Kb

Ka
(1.3 x 10-5)
Conc (M) Pr H2O
HPr OH-
Starting 0.06279
0 0
Change -x x
x
Equilibrium 0.06279 - x
x x
37
Titration Curves
HPrOH-
(x)2
7.69 x 10-10
Kb
Pr-
(0.06279 - x)
(x)2
? 7.69 x 10-10 x OH- ? 6.948 x 10-6M
(0.06279)
pOH -log OH- -log (6.948 x 10-6) 5.158
pH 14.00 - 5.158 8.841 8.84
38
Buffering

• What happens when we add the salt of a weak acid
  to a solution of a weak acid?
• (i.e. add NaAcetate to Acetic Acid.)

39
A Complex Problem

• Calculate the change in pH that occurs when 0.10
  mol of gaseous HCl is added to 1.0 L of 5 M
  HC2H3O2 and 5 M NaC2H3O2.
• Note Ka 1.8 x 10-5

40
A Complex Problem

• Calculate the change in pH that occurs when 0.10
  mol of gaseous HCl is added to 1.0 L of 5 M
  HC2H3O2 and 5 M NaC2H3O2.
• Note Ka 1.8 x 10-5

Important particles in solution H2O, H,
Cl-, HC2H3O2, C2H3O2-, Na
HC2H3O2 H2O C2H3O2- H
C2H3O2-H
Ka
HC2H3O2
41
The Conventional Solution
HC2H3O2 H2O C2H3O2- H
C2H3O2-H
Ka 1.8 x 10-5
HC2H3O2
42
The Conventional Solution
HC2H3O2 H2O C2H3O2- H
C2H3O2-H
Ka 1.8 x 10-5
HC2H3O2
43
The Conventional Solution
HC2H3O2 H2O C2H3O2- H
5.0 XX
Ka 1.8 x 10-5
5.0 - X
44
The Conventional Solution
HC2H3O2 H2O C2H3O2- H
5.0 XX
X
Ka 1.8 x 10-5
5.0 - X
45
The Henderson-Hasselbalch Solutions
HC2H3O2 H2O C2H3O2- H
5.0 XX
X
Ka 1.8 x 10-5
5.0 - X

• pH -log X
• For all buffers (weak acids/bases and
  conjugate salts)
• pH pKa base/acid

46
The Henderson-Hasselbalch Solutions
HC2H3O2 H2O C2H3O2- H
5.0 XX
X
K 1.8 x 10-5
5.0 - X

• For all buffers (weak acids/bases and
  conjugate salts)
• pH pKa log (base/acid)
• So in this case,
• pH -log1.8 x 10-5 log(1) 4.74

47
And finally... What Happens with Added
Acid?
HC2H3O2 H2O C2H3O2- H

• H when added reacts with the weak base C2H3O2-.
  (i.e. adding H forces the equilibrium back to
  the left.)
• So, more H gets removed to maintain the
  equilibrium constant.

48
Adjusting Buffer Concentrations to Determine
Change in pH
H C2H3O2- H C2H3O2
5.0 M 4.9 M
5.0 M 5.1 M
0.1 M 0.0
Before After
pH -log1.8 x 10-5 log(4.9/5.1) 4.74
- 0.02 4.72
49
What If the Original Buffer Was Only 0.5 M in
Acid and Salt?
H C2H3O2- H C2H3O2
Before After
50
What If the Original Buffer Was Only 0.5 M in
Acid and Salt?
H C2H3O2- H C2H3O2
0.5 M 0.4 M
0.5 M 0.6 M
0.1 M 0.0
Before After
51
What If the Original Buffer Was Only 0.5 M in
Acid and Salt?
H C2H3O2- H C2H3O2
0.1 M 0.0
0.5 M 0.4 M
0.5 M 0.6 M
Before After
pH -log1.8 x 10-5 log(0.5/0.5) 4.74
- 0 4.74
Before
52
What If the Original Buffer Was Only 0.5 M in
Acid and Salt?
H C2H3O2- H C2H3O2
0.1 M 0.0
0.5 M 0.4 M
0.5 M 0.6 M
Before After
pH -log1.8 x 10-5 log(0.5/0.5) 4.74
- 0 4.74
Before
pH -log1.8 x 10-5 log(0.5/0.6) 4.74
- 0.18 4.56
After
53
What If the Original Buffer Was Only 0.5 M in
Acid and Salt and 0.1 Moles of NaOH Were Added?
H C2H3O2- H C2H3O2
Before After
54
What If the Original Buffer Was Only 0.5 M in
Acid and Salt and 0.1 Moles of NaOH Were Added?
H C2H3O2- H C2H3O2
0.5 M 0.4 M
-0.1 M 0.0
0.5 M 0.6 M
Before After
55
What If the Original Buffer Was Only 0.5 M in
Acid and Salt and 0.1 Moles of NaOH Were Added?
H C2H3O2- H C2H3O2
0.5 M 0.4 M
-0.1 M 0.0
0.5 M 0.6 M
Before After
pH -log1.8 x 10-5 log(0.5/0.5) 4.74
- 0 4.74
Before
56
What If the Original Buffer Was Only 0.5 M in
Acid and Salt and 0.1 Moles of NaOH Are Added?
H C2H3O2- H C2H3O2
0.5 M 0.4 M
-0.1 M 0.0
0.5 M 0.6 M
Before After
pH -log1.8 x 10-5 log(0.5/0.5) 4.74
- 0 4.74
Before
pH -log1.8 x 10-5 log(0.6/0.4) 4.74
0.18 4.92
After