3. 19-3
Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffers
19.2 Acid-Base Titration Curves
4. 19-4
Goals & Objectives
• See the following Learning Objectives
on page 870.
• Understand these Concepts:
• 19.1-8
• Master these Skills:
• 19.1-6
5. 19-5
Acid-Base Buffers
An acid-base buffer is a solution that lessens the impact of
pH from the addition of acid or base.
An acid-base buffer usually consists of a conjugate acid-
base pair where both species are present in appreciable
quantities in solution.
An acid-base buffer is therefore a solution of a weak acid
and its conjugate base, or a weak base and its
conjugate acid.
6. 19-6
Figure 19.1 The effect of adding acid or base to an unbuffered
solution.
A 100-mL sample of
dilute HCl is adjusted
to pH 5.00.
The addition of 1 mL of strong acid (left)
or strong base (right) changes the pH by
several units.
7. 19-7
Figure 19.2 The effect of adding acid or base to a buffered
solution.
A 100-mL sample of
an acetate buffer is
adjusted to pH 5.00.
The addition of 1 mL of strong acid (left)
or strong base (right) changes the pH very
little.
The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with
1 M CH3COONa (which provides the conjugate base, CH3COO-
).
8. 19-8
Buffers and the Common-ion Effect
A buffer works through the common-ion effect.
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)
acetic acid acetate ion
Acetic acid in water dissociates slightly to produce some
acetate ion:
If NaCH3COO is added, it provides a source of
CH3COO-
ion, and the equilibrium shifts to the left.
CH3COO-
is common to both solutions.
The addition of CH3COO-
reduces the % dissociation of
the acid.
9. 19-9
Buffer Solutions
• Buffer solutions resist changes in pH.
• They are special cases of the common
ion effect.
• The two most common types
– weak acid + soluble salt of the weak acid
• HF + NaF
– weak base + soluble salt of the weak base
• NH3 + NH4Cl
10. 19-10
Table 19.1 The Effect of Added Acetate Ion on the Dissociation
of Acetic Acid
[CH3COOH]init [CH3COO-
]added % Dissociation*
[H3O+
] pH
0.10 0.00 1.3 1.3x10-3
2.89
0.10 0.050 0.036 3.6x10-5
4.44
0.10 0.10 0.018 1.8x10-5
4.74
0.10 0.15 0.012 1.2x1015
4.92
*
% Dissociation =
[CH3COOH]dissoc
[CH3COOH]init
x 100
11. 19-11
How a Buffer Works
The buffer components (HA and A-
) are able to consume
small amounts of added OH-
or H3O+
by a shift in
equilibrium position.
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)
Added H3O+
reacts with
CH3COO-
, causing a
shift to the left.
Added OH-
reacts with
CH3COOH, causing a shift to
the right.
The shift in equilibrium position absorbs the change in
[H3O+
] or [OH-
], and the pH changes only slightly.
12. 19-12
Figure 19.3 How a buffer works.
Buffer has equal
concentrations of A-
and HA.
H3O+
Buffer has more HA after
addition of H3O+
.
H2O + CH3COOH ← H3O+
+ CH3COO-
OH-
Buffer has more A-
after
addition of OH-
.
CH3COOH + OH-
→ CH3COO-
+ H2O
13. 19-13
Relative Concentrations of Buffer Components
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)
Ka =
[CH3COO-
][H3O+
]
[CH3COOH]
If the ratio increases, [H3O+
] decreases.
[HA]
[A-
]
Since Ka is constant, the [H3O+
] of the solution depends
on the ratio of buffer component concentrations.
[H3O+
] = Ka x
[CH3COOH]
[CH3COO-
]
If the ratio decreases, [H3O+
] increases.
[HA]
[A-
]
14. 19-14
Sample Problem 19.1 Calculating the Effect of Added H3O+
or
OH-
on Buffer pH
Calculate the pH:
PLAN: We can calculate [CH3COOH]init and [CH3COO-
]init from the
given information. From this we can find the starting pH. For (b) and (c)
we assume that the added OH-
or H3O+
reacts completely with the
buffer components. We write a balanced equation in each case, set up
a reaction table, and calculate the new [H3O+
].
PROBLEM:
(a) Of a buffer solution consisting of 0.50 M CH3COOH and 0.50 M
CH3COONa
(b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution
(c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in (a).
Ka of CH3COOH = 1.8 x 10-5
. (Assume the additions cause a negligible
change in volume.)
15. 19-15
Sample Problem 19.1
SOLUTION: (a)
Initial 0.50 - 0.50 0
Change -−x +x +x
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)Concentration (M)
Equilibrium -0.50 - x x0.50 + x
1.8x10-5
M
0.50 M
x 100 = 3.6x10-3
% (< 5%; assumption is justified.)
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.50 – x ≈ 0.50 M and [CH3COO-
] = 0.50 + x ≈ 0.50 M
x = [H3O+
] = Ka x
[CH3COOH]
[CH3COO-
]
≈ 1.8x10-5
x 0.50
0.50
= 1.8x10-5
M
Checking the assumption: pH = -log(1.8x10-5
) = 4.74
16. 19-16
Sample Problem 19.1
Initial 0.50 0.020 0.50 -
Change -0.020-0.020 +0.020 -
CH3COOH(aq) + OH-
(aq) → CH3COO-
(aq) + H2O(l)Concentration (M)
(b) [OH−
]added =
0.020 mol
1.0 L soln
= 0.020 M OH−
Equilibrium 00.48 -0.52
Setting up a reaction table for the stoichiometry:
Initial 0.48 - 0.52 0
Change -−x +x +x
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)Concentration (M)
Equilibrium -0.48 - x x0.52 + x
Setting up a reaction table for the acid dissociation, using new initial [ ]:
17. 19-17
Sample Problem 19.1
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.48 – x ≈ 0.48 M and [CH3COO-
] = 0.52 + x ≈ 0.52 M
x = [H3O+
] = Ka x
[CH3COOH]
[CH3COO-
]
pH = -log(1.7x10-5
) = 4.77
≈ 1.8x10-5
x 0.48
0.52
= 1.7x10-5
M
18. 19-18
Sample Problem 19.1
Initial 0.50 0.020 0.50 -
Change -0.020-0.020 +0.020 -
CH3COO-
(aq) + H3O+
(aq) → CH3COOH(aq) + H2O(l)Concentration (M)
(c) [H3O+
]added =
0.020 mol
1.0 L soln
= 0.020 M H3O+
Equilibrium 00.48 -0.52
Setting up a reaction table for the stoichiometry:
Initial 0.52 - 0.48 0
Change -−x +x +x
CH3COOH(aq) + H2O(l) CH3COO-
(aq) + H3O+
(aq)Concentration (M)
Equilibrium -0.52 - x x0.48 + x
Setting up a reaction table for the acid dissociation, using new initial [ ]:
19. 19-19
Sample Problem 19.1
Since Ka is small, x is small, so we assume
[CH3COOH] = 0.52 – x ≈ 0.52 M and [CH3COO-
] = 0.48 + x ≈ 0.48 M
x = [H3O+
] = Ka x [CH3COOH]
[CH3COO-
]
pH = -log(2.0x10-5
) = 4.70
≈ 1.8x10-5
x 0.52
0.48
= 2.0x10-5
M
20. 19-20
The Common Ion Effect
• Determine the [H+
] and the pH of a
solution that is 0.15M in CH3COOH and
0.15M in NaCH3COO (a strong
electrolyte). Ka = 1.8x10-5
• Partial list of soluble salts:
– All K+
, Na+
and NH4
+
salts
– All NO3
-
salts
25. 19-25
Buffer Capacity
The buffer capacity is a measure of the “strength” of the
buffer, its ability to maintain the pH following addition of
strong acid or base.
The greater the concentrations of the buffer
components, the greater its capacity to resist pH
changes.
The closer the component concentrations are to each
other, the greater the buffer capacity.
26. 19-26
Figure 19.4 The relation between buffer capacity and pH change.
When strong base is
added, the pH increases
least for the most
concentrated buffer.
This graph shows the final pH values for four different buffer solutions after
the addition of strong base.
27. 19-27
Buffer Range
The buffer range is the pH range over which the buffer is
effective.
Buffer range is related to the ratio of buffer component
concentrations.
[HA]
[A-
]
The closer is to 1, the more effective the buffer.
If one component is more than 10 times the other, buffering
action is poor. Since log10 = 1, buffers have a usable
range within ± 1 pH unit of the pKa of the acid
component.
28. 19-28
Preparing a Buffer
• Choose the conjugate acid-base pair.
– The pKa of the weak acid component should be close to the
desired pH.
• Calculate the ratio of buffer component concentrations.
• Determine the buffer concentration, and calculate the
required volume of stock solutions and/or masses of
components.
• Mix the solution and correct the pH.
pH = pKa + log
[base]
[acid]
29. 19-29
Sample Problem 19.3 Preparing a Buffer
SOLUTION:
PROBLEM: An environmental chemist needs a carbonate buffer of
pH 10.00 to study the effects of the acid rain on
limsetone-rich soils. How many grams of Na2CO3 must
she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to
make the buffer? Ka of HCO3
-
is 4.7x10-11
.
PLAN: The conjugate pair is HCO3
-
(acid) and CO3
2-
(base), and we
know both the buffer volume and the concentration of HCO3
-
.
We can calculate the ratio of components that gives a pH of
10.00, and hence the mass of Na2CO3 that must be added to
make 1.5 L of solution.
[H3O+
] = 10-pH
= 10-10.00
= 1.0x10-10
M
HCO3
-
(aq) + H2O(l) H3O+
(aq) + CO3
2-
(aq) Ka =
[CO3
2-
][H3O+
]
[HCO3
-
]
30. 19-30
Sample Problem 19.3 Preparing a Buffer
[CO3
2-
] =
Ka[HCO3
-
]
[H3O+
]
=
(4.7x10-11
)(0.20)
1.0x10-10
= 0.094 M
Amount (mol) of CO3
2-
needed = 1.5 L soln x 0.094 mol CO3
2-
1 L soln
= 0.14 mol CO3
2-
105.99 g Na2CO3
1 mol Na2CO3
0.14 mol Na2CO3 x = 15 g Na2CO3
The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 M
NaHCO3 and add more 0.20 M NaHCO3 to make 1.5 L. Using a pH
meter, she can then adjust the pH to 10.00 by dropwise addition of
concentrated strong acid or base.
31. 19-31
• Determine the number of moles of
NH4Cl that must be used to prepare
1.00L of a buffer solution that is 0.10M
in NH3 and has a pH of 9.15. Kb for NH3
= 1.8x10-5
Buffer Solutions
34. 19-34
Acid-Base Indicators
An acid-base indicator is a weak organic acid (HIn)
whose color differs from that of its conjugate base (In-
).
The color of an indicator changes over a specific,
narrow pH range, a range of about 2 pH units.
The ratio [HIn]/[In-
] is governed by the [H3O+
] of the
solution. Indicators can therefore be used to monitor the
pH change during an acid-base reaction.
36. 19-36
Figure 19.6 The color change of the indicator bromthymol blue.
pH < 6.0 pH > 7.5pH = 6.0-7.5
37. 19-37
Acid-Base Titrations
In an acid-base titration, the concentration of an acid (or a
base) is determined by neutralizing the acid (or base) with
a solution of base (or acid) of known concentration.
The equivalence point of the reaction occurs when the
number of moles of OH-
added equals the number of
moles of H3O+
originally present, or vice versa.
The end point occurs when the indicator changes color.
- The indicator should be selected so that its color change
occurs at a pH close to that of the equivalence point.
38. 19-38
Figure 19.7 Curve for a strong acid–strong base titration.
The initial pH is low.
The pH rises very rapidly
at the equivalence point,
which occurs at pH = 7.00.
The pH increases gradually
when excess base has been
added.
39. 19-39
Calculating the pH during a
strong acid–strong base titration
Initial pH
[H3O+
] = [HA]init
pH = -log[H3O+
]
pH before equivalence point
initial mol H3O+
= Vacid x Macid
mol OH-
added = Vbase x Mbase
mol H3O+
remaining = (mol H3O+
init) – (mol OH-
added)
[H3O+
] = pH = -log[H3O+
]
mol H3O+
remaining
Vacid + Vbase
40. 19-40
pH at the equivalence point
pH = 7.00 for a strong acid-strong base titration.
Calculating the pH during a
strong acid–strong base titration
pH beyond the equivalence point
initial mol H3O+
= Vacid x Macid
mol OH-
added = Vbase x Mbase
mol OH-
excess = (mol OH-
added) – (mol H3O+
init)
[OH-
] =
pOH = -log[OH-
] and pH = 14.00 - pOH
mol OH-
excess
Vacid + Vbase
41. 19-41
Example:
40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH.
The initial pH is simply the pH of the HCl solution:
[H3O+
] = [HCl]init = 0.1000 M and pH = -log(0.1000) = 1.00
To calculate the pH after 20.00 mL of NaOH solution has been added:
OH-
added = 0.02000 L NaOH x 0.1000 mol
1 L
= 2.000x10-3
mol OH-
Initial mol of H3O+
= 0.04000 L HCl x 0.1000 mol
1 L
= 4.000x10-3
mol H3O+
H3O+
remaining = 4.000x10-3
– 2.000x10-3
= 2.000x10-3
mol H3O+
The OH-
ions react with an equal amount of H3O+
ions, so
42. 19-42
[H3O+
] =
2.000x10-3
mol
0.04000 L + 0.02000 L
= 0.03333 M
pH = -log(0.03333) = 1.48
To calculate the pH after 50.00 mL of NaOH solution has been added:
The equivalence point occurs when mol of OH-
added = initial mol of
HCl, so when 40.00 mL of NaOH has been added.
OH-
added = 0.05000 L NaOH x 0.1000 mol
1 L
= 5.000x10-3
mol OH-
OH-
in excess = 5.000x10-3
– 4.000x10-3
= 1.000x10-3
mol OH-
[OH-
] =
1.000x10-3
mol
0.04000 L + 0.05000 L
= 0.01111 M
pOH = -log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05
43. 19-43
Figure 19.8 Curve for a weak acid–strong base titration.
The initial pH is higher than
for the strong acid solution.
The curve rises
gradually in the buffer
region. The weak acid
and its conjugate base
are both present in
solution.
The pH increases slowly
beyond the equivalence
point.
The pH at the equivalent
point is > 7.00 due to the
reaction of the conjugate
base with H2O.
44. 19-44
Calculating the pH during a
weak acid–strong base titration
Initial pH
[H3O+
] =
pH = -log[H3O+
]
[H3O+
][A-
]
[HA]
Ka =
pH before equivalence point
[H3O+
] = Ka x or
pH = pKa + log
[base]
[acid]
[HA]
[A-
]
45. 19-45
Calculating the pH during a
weak acid–strong base titration
A-
(aq) + H2O(l) HA(aq) + OH-
(aq)
pH at the equivalence point
[OH-
] =
where [A-
] = and
Kw
Ka
Kb =
mol HAinit
Vacid + Vbase
[H3O+
] ≈ and pH = -log[H3O+
]
Kw
pH beyond the equivalence point
[OH-
] =
pH = -log[H3O+
]
mol OH-
excess
Vacid + Vbase
[H3O+
] =
Kw
[OH-
]
46. 19-46
Sample Problem 19.4 Finding the pH During a Weak Acid–
Strong Base Titration
PROBLEM: Calculate the pH during the titration of 40.00 mL of
0.1000 M propanoic acid (HPr; Ka = 1.3x10-5
) after adding
the following volumes of 0.1000 M NaOH:
PLAN: The initial pH must be calculated using the Ka value for the weak
acid. We then calculate the number of moles of HPr present
initially and the number of moles of OH-
added. Once we know
the volume of base required to reach the equivalence point we
can calculate the pH based on the species present in solution.
(a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL
SOLUTION:
(a) [H3O+
] = = 1.1x10-3
M
pH = -log(1.1x10-3
) = 2.96
47. 19-47
Sample Problem 19.4
(b) 30.00 mL of 0.1000 M NaOH has been added.
Initial amount of HPr = 0.04000 L x 0.1000 M = 4.000x10-3
mol HPr
Amount of NaOH added = 0.03000 L x 0.1000 M = 3.000x10-3
mol OH-
Each mol of OH-
reacts to form 1 mol of Pr-
, so
HPr(aq) + OH-
(aq) → Pr-
(aq) + H2O(l)Concentration (M)
Initial 0.004000 0.003000 0 -
Change −0.003000 -0.003000 +0.003000 -
Equilibrium 0.001000 0 -0.003000
[H3O+
] = Ka x [HPr]
[Pr-
]
pH = -log(4.3x10-6
) = 5.37
0.001000
0.003000
= (1.3x10-5
) x = 4.3x10-6
M
48. 19-48
Sample Problem 19.4
(c) 40.00 mL of 0.1000 M NaOH has been added.
This is the equivalence point because mol of OH-
added = 0.004000
= mol of HAinit.
All the OH-
added has reacted with HA to form 0.004000 mol of Pr-
.
0.004000 mol
0.04000 L + 0.04000 L
[Pr-
] = = 0.05000 M
Pr-
is a weak base, so we calculate Kb =
Kw
Ka
=
1.0x10-14
1.3x10-5
= 7.7x10-10
[H3O+
] ≈
Kw =
1.0x10-14
= 1.6x10-9
M
pH = -log(1.6x10-9
) = 8.80
49. 19-49
Sample Problem 19.4
(d) 50.00 mL of 0.1000 M NaOH has been added.
Amount of OH-
added = 0.05000 L x 0.1000 M = 0.005000 mol
Excess OH-
= OH-
added – HAinit = 0.005000 – 0.004000 = 0.001000 mol
[OH-
] =
mol OH-
excess
total volume
=
0.001000 mol
0.09000 L
= 0.01111 M
[H3O+
] =
Kw
[OH-
]
=
1x10-14
0.01111
= 9.0x10-13
M
pH = -log(9.0x10-13
) = 12.05
50. 19-50
Figure 19.9 Curve for a weak base–strong acid titration.
The pH decreases
gradually in the buffer
region. The weak base
and its conjugate acid
are both present in
solution.
The pH at the equivalence
point is < 7.00 due to the
reaction of the conjugate
acid with H2O.
51. 19-51
Figure 19.10 Curve for the titration of a weak polyprotic acid.
Titration of 40.00 mL of 0.1000 M
H2SO3 with 0.1000 M NaOH
pKa2 = 7.19
pKa1 = 1.85
52. 19-52
Net Ionic Equations
• Include in the equation only those
species that actually change
concentration.
• Molecular reaction:
– NaOH + HCl = H2O + NaCl
• Net Ionic Equation:
– H+
+ OH-
= H2O
54. 19-54
Net Ionic Equations
• Rules for writing:
– A. List predominant species
• All soluble salts, strong acids and strong bases are
written as their component ions. All others are
written as the molecule.
– B. Combine ions of opposite charge and look
for one or more of the following:
• 1. formation of a weak acid
• 2. formation of a weak base
• 3. formation of water
55. 19-55
Net Ionic Equations
• Rules (continued):
– C. If no reaction in B. above, then look at
each of the following in the order given as a
source of secondary species:
• 1. weak acid
• 2. weak base
• 3. water, if no other possibility
56. 19-56
Net Ionic Equations
• Write net ionic equations for each of the
following for 0.10M solutions being mixed.
Write the form of the equilibrium constant
using Ka, Kb and Kw
– 1. NaOH + HCl
– 2. KF + HNO3
– 3. NH4NO3 + Ca(OH)2
– 4. HF + NaOH
– 5. H2S + KOH
60. 19-60
Hydrolysis Equilibria
• Refers to the reaction of a substance with
water or its ions.
• Write net ionic equations for each of the
following:
– 1. NaF + H2O
– 2. NH4Cl + H2O
– 3. NaCl + H2O
– 4. AlCl3 + H2O
63. 19-63
Hydrolysis Equilibria
Determine the [H+
]. pH and percent
hydrolysis for a 0.10M solution of NH4Br.
Kb for NH3 is 1.8x 10-5
Calculate the pH of a 0.10M solution of
NaCN. Ka for HCN is 4.0x10-10
66. 19-66
Buffering Action
If 0.20 mole of HCl is added to 1.00L of a
solution that is 0.100M in NH3 and
0.200M in NH4Cl, determine the final pH
of the solution. Assume no volume
change from the addition of HCl(g). Kb for
NH3 is 1.8x10-5
