### Acid-Base III

```Be able to write the chemical equation for a
chemical species acting as an acid:
HX(aq)
H+(aq) + X-(aq)
Example Write the equation showing the
bicarbonate ion acting as an acid.
HCO3-(aq)
H+(aq) + CO32-(aq)
Be able to write the chemical equation for a
chemical species acting as a base:
X-(aq) + H2O(l)
OH-(aq) + HX(aq)
Example Write the equation showing the
bicarbonate ion acting as a base.
HCO3-(aq) + H2O(l)
OH-(aq) + H2CO3(aq)
The Common Ion Effect
What is the pH of a solution made by adding 0.30 mol of
acetic acid to enough water to make 250 mL of solution?
Ka is 1.8 x 10-5 at 25°C.
CH3COOH(aq)
initial
change
0.30/0.250 M
-xM
equilibrium 1.2 - x M
Ka = [CH3COO-][H+] =
x2
[CH3COOH]
(1.2 - x)
CH3COO-(aq) + H+(aq)
0M
0M
+x M
+x M
xM
xM
= 1.8 x 10-5
assume x<<1.2 x2 = 2.16 x 10-5 x = 0.0046 M (<<1.2)
pH = -log(0.0046) = 2.33
The Common Ion Effect
What is the pH of a solution made by adding 0.30 mol of
acetic acid and 0.30 mol of sodium acetate to enough
water to make 250 mL of solution?
CH3COOH
initial
CH3COO-(aq) + H+(aq)
0.30/0.250 M
change
-xM
equilibrium 1.2 - x M
Ka = [CH3COO-][H+] = (1.2 + x)x
[CH3COOH]
(1.2 - x)
x = 1.8 x
10-5
0.30/0.250 M
0M
+x M
+x M
1.2 + x M
xM
= 1.8 x 10-5
(assume x<<1.2)
M (x<<1.2)
pH = -log(1.8 x 10-5) = 4.74
The Common Ion Effect
Adding acetate ion shifted the equilibrium to the left,
decreasing [H+] and increasing pH.
CH3COOH
CH3COO-(aq) + H+(aq)
Common Ion Effect
The extent of ionization of any weak
electrolyte is decreased by adding to the
solution a strong electrolyte that has an
ion in common with the weak electrolyte.
The Common Ion Effect
a) Find the solubility of calcium phosphate in water at
25°C.
b) Find the solubility of calcium phosphate in 0.100M
sodium phosphate at 25°C.
The Common Ion Effect
a) Find the solubility of calcium phosphate in water at
25°C.
What information do you need?
formula for calcium phosphate
the equation for the equilibrium
the equilibrium constant
source: Appendix D
The Common Ion Effect
a) Find the solubility of calcium phosphate in water at
25°C.
Ca3(PO4)2(s)
3Ca2+(aq) + 2PO43-(aq)
Ksp = 2.0 x 10-29
I
present
C lose x mol
E present
2.0 x 10-29 = 108x5
+3x M
+2x M
3x M
2x M
x = 7.1 x 10-7 M
The Common Ion Effect
b) Find the solubility of calcium phosphate in 0.100M
sodium phosphate at 25°C.
Ca3(PO4)2(s)
3Ca2+(aq) + 2PO43-(aq)
Ksp = 2.0 x 10-29
I
present
C lose x mol
E present
0.100 M
+3x M
3x M
+2xM
(0.100+2x) M
Assume 2x << .100
2.0 x 10-29 = .27x3
x = 4.2 x 10-10 M
Check: 8.4 x 10-10 << .1 (100,000,000 times smaller)
The Common Ion Effect
a) Find the solubility of calcium phosphate in water at
25°C.
7.1 x 10-7 M
b) Find the solubility of calcium phosphate in 0.100M
sodium phosphate at 25°C.
4.2 x 10-10 M
The presence of phosphate from sodium phosphate
decreased the solubility of the calcium phosphate
by a factor of a 1000.
Buffered Solutions
What is the pH of our 1.2M acetic acid solution if 10.0
mL of 1.0 M NaOH is added to it?
CH3COOH
CH3COO-(aq)
+ H+(aq)
If NaOH is added to the solution, it reacts with some of
the acetic acid:
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)
This has the effect of decreasing the acetic acid
concentration and increasing the acetate ion
concentration. PLUS, addition of the NaOH changes
the volume of the system.
Buffered Solutions
What is the pH of our 1.2M acetic acid solution if 10.0
mL of 1.0 M NaOH is added to it?
CH3COOH
CH3COO-(aq)
+ H+(aq)
If NaOH is added to the solution, it reacts with some of
the acetic acid and produces acetate ion:
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)
This is a stoichiometry problem.
Buffered Solutions
What is the pH of our 1.2M acetic acid solution if 10.0
mL of 1.0 M NaOH is added to it?
CH3COOH
initial:
change:
CH3COO-(aq)
(0.30 - 0.010)/0.260 M
0.010/0.260 M
+ H+(aq)
0M
-xM
+xM
+x M
equilibrium: 1.115 - x M
0.0385 + x M
xM
Ka = [CH3COO-][H+] = (.0385 + x)x
[CH3COOH]
(1.115 - x)
= 1.8 x 10-5
(assume x<<0.0385)
x2 + .0385x - 2.01x10-5 = 0 x = 5.2 x 10-4 M (x<<0.0385)
pH = -log(5.2 x 10-4) = 3.29
Buffered Solutions
What is the pH of our acetic acid/sodium acetate solution
if 10.0 mL of 1.0 M NaOH is added to it?
CH3COOH
initial:
CH3COO-(aq)
+ H+(aq)
(0.30-.010)/0.260 M (0.30+.010)/0.260 M 0 M
change:
-xM
+x M
+x M
equilibrium: 1.115 - x M
1.192 + x M
xM
Ka = [CH3COO-][H+] = (1.192 + x)x
= 1.8 x 10-5
[CH3COOH]
(1.115 - x) (assume x<<1.115, 1.192)
x = 1.7 x
10-5
M (x<<1.115 and 1.192)
pH = -log(1.7 x 10-5) = 4.77
Buffered Solutions
HAc is short for acetic acid.
pH
of 10.0 mL
of 1.0 M NaOH
of 10.0 mL
of 1.0 M HCl
250 mL of
1.2 M HAc
250 mL of
1.2 M HAc/Acbuffer soln
2.33
3.29
1.47
4.74
4.77
4.71
7.00
12.59
1.41
Ac- is short for
acetate.
250 mL of
DI water
Buffered Solutions
pH
of 10 mL 1.0 M NaOH
of 10 mL 1.0 M HCl
250 mL of
1.2 M HAc
2.33
3.29
1.47
4.74
4.77
4.71
7.00
12.59
1.41
250 mL of
1.2 M HAc/Ac
250 mL of
DI water
• The pH of the HAc solution ranged 1.47 - 3.20 with the
addition of 10.0 mL of the acid or base.
Buffered Solutions
pH
of 10 mL 1.0 M NaOH
of 10 mL 1.0 M HCl
250 mL of
1.2 M HAc
2.33
3.29
1.47
4.74
4.77
4.71
7.00
12.59
1.41
250 mL of
1.2 M HAc/Ac
250 mL of
DI water
• When the same amount of acid or base was added to the
HAc/Ac solution, its pH ranged 4.71 - 4.77. The addition
of the acetate ion buffers the solution against changes in
pH.
Buffered Solutions
• contain a weak conjugate acid-base pair
such as HAc/Ac-, NH3/NH4+, H2CO3/HCO3-, HCN/CN-.
• resist changes in pH because they contain an
acid to react with OH- and a base to react with
H+.
• have the ability to resist changes in pH, known
as their buffer capacity, which depends on the
concentrations of the members of the pair.
The pH of a buffer solution may be determined
by the Henderson-Hasselbalch equation.
Henderson-Hasselbalch Equation
For the general equilibrium
HX(aq)
H+(aq) + X-(aq)
the pH is given by
[]
=  + (
)
[]
Henderson-Hasselbalch equation
Henderson-Hasselbalch Equation
• Let’s recalculate the pH of our 1.2 M
HAc/Ac solution.
pH = pKa + log ([base]/[acid])
Ka = 1.8 x 10-5, so pKa = 4.74
pH = 4.74 + log(1.2M/1.2M) = 4.74
• After the addition of 10.0 mL 1.0 M NaOH
pH = 4.74 + log(1.192M/1.115M)
= 4.74 + 0.03 = 4.77
Buffered Solutions
You have been asked to prepare 500.0 mL of a buffer
with the pH of blood, 7.40. What chemicals and
amounts will you use?
Buffers most effectively resist a change in pH
in either direction if they contain equal
concentrations of both members of the acidbase pair.
Find a weak acid with a Ka ≈ 10-7 (so the pKa will be 7).
Ka3(citric acid) = 4.0 x 10-7 at 25°C. pKa = 6.40
7.40 = 6.40 + log [base]/[acid]
[base]/[acid] = 10
Buffered Solutions
You have been asked to prepare 500.0 mL of a buffer with the pH
of blood, 7.40. What chemicals and amounts will you use?
We chose citric acid (abbreviated H3Cit) and
found that we need a base-to-acid ratio of 10.
For Ka3, the base is Cit3- and the acid is HCit2-.
HCit2-(aq)
Ka3 = 4.0 x 10-7 at 25°C
pH = pKa3 + log [Cit3-]
[HCit2-]
Cit3-(aq) + H+(aq)
Buffered Solutions
Solution #1:
Put 1.0 mol of Na3Cit and 0.10 mol Na2HCit in a 500-mL
volumetric flask and dilute to the mark with DI water.
Solution #2:
Put 0.010 mol of Na3Cit and 0.0010 mol Na2HCit in a
500-mL volumetric flask and dilute to the mark with DI
water.
Both solutions will produce the correct pH…what is
Their BUFFERING CAPACITY. Solution #2 has
smaller concentrations of the acid-base pair.
Titration of a Strong Acid with a Strong Base
If we neutralize an acid incrementally and monitor the
pH of the solution as we add the base, the resulting
data would give us a titration curve, a plot of pH
Titration of 0.100 M HCl
14.00
12.00
10.00
pH
8.00
6.00
4.00
2.00
0.00
0
20
40
60
80
100
Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of 0.100 M HCl with
0.100 M NaOH, the titration curve would be a plot of pH
Titration of 0.100 M HCl
14.00
12.00
10.00
pH
8.00
6.00
4.00
2.00
0.00
0
20
40
60
80
100
Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of 0.100 M HCl with
0.100 M NaOH,
here is some useful information:
• 50.0 mL of 0.100 M HCl contain 5.00 mmol H+
• 1.0 mL of 0.100 M NaOH contains 0.10 mmol of OH

=

= molarity (M)
Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of 0.100 M
HCl with 0.100 M NaOH, the [H+] before the
equivalence point is the concentration of
unreacted HCl.
After 25.0 mL of NaOH have been added, the
total volume is (50.0 + 25.0) mL, and
[H+] =(5.00mmol HCl - 2.50mmol NaOH)/75 mL
= 0.033 M
pH = 1.48
Titration of a Strong Acid with a Strong Base
For the neutralization of 50.0 mL of 0.100 M
HCl with 0.100 M NaOH, the [OH-] after the
equivalence point is the concentration of
excess NaOH.
After 75.0 mL of NaOH have been added, the
total volume is (50.0 + 75.0) mL, and
[OH-] = (7.50 mmol NaOH - 5.00 mmol HCl)/125 mL
= 0.0200 M
pOH = 1.70
pH = 14.00- 1.70 = 12.30
Titration of a Strong Acid with a Strong Base
Titration of 0.100 M HCl
14.00
12.00
The initial pH depends
solely on the concentration
of the acid. Here it is 1.00.
10.00
8.00
pH
The pH at the equivalence
point is 7. This is true for
the titration of any strong
acid by any strong
base…and vice versa.
6.00
4.00
2.00
0.00
0
20
40
60
80
100
Titration of a Weak Acid with a Strong Base
For the neutralization of 50.0 mL of 0.10 M HAc with 0.100 M
NaOH, the [H+] before the equivalence point is calculated
from the equilibrium expression using the concentration of
unreacted HAc.
After 25.0 mL of NaOH have been added, the total volume is
(50.0 + 25.0) mL, and
HAc
(5.00-2.50)/75.0 M
Ac-(aq)
+ H+(aq)
2.50/75.0 M
We can use the Henderson-Hasselbalch equation:
Ka = 1.8 x 10-5  pKa = 4.74
[Ac-] = [HAc] = 2.50/75.0 M  [base]/[acid] = 1.00
pH = 4.74 + log 1.00 = 4.74
Titration of a Weak Acid with a Strong Base
Titration of 0.100 M Acetic Acid
14.00
The initial pH depends on
the concentration of the
weak acid and its Ka.
12.00
10.00
pH
8.00
6.00
4.00
2.00
0.00
0
20
40
60
80
100
Titration of a Weak Acid with a Strong Base
For the neutralization of 50.0 mL of 0.100 M HAc with 0.100 M
NaOH, the [H+] at the equivalence point is calculated from the
equilibrium expression for the base.
After 50.0 mL of NaOH have been added, the total volume is
(50.0 + 50.0) mL, and all of the HAc has been changed to Ac-.
Ac-(aq)
initial:
change:
equilibrium:
5.00/100.0 M
-xM
0.0500 - x M
+ H2O(l)
HAc(aq) + OH-(aq)
0M
0M
+xM
+xM
xM
xM
Kb = Kw/Ka = 1.0 x 10-14/(1.8 x 10-5) = 5.6 x 10-10
[OH-][HAc] = 5.6 x 10-10 ≈ x2
x = [OH-] = 5.3 x 10-6
[Ac-]
0.0500
pOH = 5.28
pH = 14.00- 5.28 = 8.72
Titration of a Weak Acid with a Strong Base
Titration of 0.100 M Acetic Acid
14.00
12.00
10.00
pH
8.00
6.00
4.00
The pH at the equivalence
point is 8.72.
2.00
0.00
0
20
40
60
80
100
Titration of a Weak Acid with a Strong Base
For the neutralization of 50.0 mL of 0.100 M
HAc with 0.100 M NaOH, the [OH-] after the
equivalence point is the concentration of
excess NaOH.
This is exactly the same as the postequivalence section of the strong acid-strong
base titration curve.
After 75.0 mL of NaOH have been added, the total volume is (50.0
+ 75.0) mL, and
[OH-] = (7.50 mmol NaOH - 5.00 mmol HCl)/125 mL = 0.0200 M
pOH = 1.70
pH = 14.00- 1.70 = 12.30
Titration of a Weak Acid with a Strong Base
Titration of 0.100 M Acetic Acid
14.00
12.00
This part of the curve
is identical to that of
the HCl/NaOH curve,
since in both cases
the NaOH volume
10.00
pH
8.00
6.00
4.00
2.00
0.00
0
20
40
60
80
100
Titration of a Weak Base with a Strong Acid
The titration curve for a weak base with a
strong acid is calculated similarly to that of
the weak acid/strong base.
You must be careful to write the equilibrium
expression correctly and use the correct Kb.
For the titration of aqueous ammonia, use
NH3(aq) + H2O(l)
Kb = 1.8 x 10-5 at 25°C
NH4+(aq) + OH-(aq)
Titration of a Weak Base with a Strong Acid
Titration of 0.100 M Ammonium Hydroxide
12.00
10.00
pH
8.00
6.00
4.00
2.00
0.00
0
20
40
60
80
100
Region of Maximum Buffer Capacity
[]
=  + (
)
[]
Buffering is best when [base] = [acid].
But when [base] = [acid],
pH = pKa
This occurs when the volume of NaOH
added is ½ the NaOH volume needed to
reach the equivalence point.
Region of Maximum Buffer Capacity
Titration of 0.100 M Acetic Acid
14.00
12.00
10.00
pH
8.00
6.00
4.00
pH = pKa = 4.72. Maximum buffering capacity is here.
2.00
0.00
0
20
40
60
80
100
Titration of a Polyprotic Acid with a Strong Base
13.00
2nd equivalence point
pH = 9.9
12.00
11.00
H2A is
completely
neutralized at
2nd eq pt.
10.00
9.00
8.00
1st equivalence
point pH = 4.5
7.00
pH
6.00
5.00
HA- and A2-
4.00
3.00
2.00
1.00
H2A and HA-
0.00
0
10
20
30
40
50
60
70
80
90
100
Titration of a Polyprotic Acid with a Strong Base
13.00
halfway between 40
and 80 mL
12.00
11.00
10.00
9.00
pKa2= 7.2
8.00
7.00
pH
halfway between
0 and 40 mL
6.00
5.00
HA- and A2-
4.00
3.00
pKa1= 2.0
2.00
1.00
H2A and HA-
0.00
0
10
20
30
40
50
60