Chapter 19. Ionic Equilibria
Section 19.1. Acid-Base Buffer Solutions
In everyday English, a buffer is something that lessens the impact of
an external force.
** An acid-base buffer is a solution that lessens the change in
[H3O+] that would result when a strong acid or base is added **
A buffer is a concentrated solution of a weak acid (or base), together
with a salt containing the conjugate base (or acid).
i.e., a weak acid and its conjugate base
or a weak base and its conjugate acid.
How does a Buffer work? The Common-Ion Effect
(example of Le Chatelier’s Principle)
= the shift in an equilibrium caused by the addition (or removal) of
one of the species participating in the equilibrium.
Example: addition of sodium acetate (CH3COONa or NaAc) to acetic
acid (CH3COOH or HAc) solution
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)
acetic acid (HAc)
acetate ion (Ac-)
if we add CH3COOSHIFT
Net Effect:
[H3O+] decreases  pH increases
Also, [OH-] increases  pOH decreases
Example: addition of NH4Cl to NH3 solution.
NH3 (aq) + H2O (l) ⇌ NH4+(aq) + OH- (aq)
if we add NH4+
Net Effect:
[OH-] decreases, pOH increases
[H3O+] increases, pH decreases
1
New type of problem to solve! – solutions with two things dissolved
Example: Calculate [H3O+] in a solution that is 0.10 M in HF and 0.20 M
in NaF. Also calculate % ionization.
Problem: Use HF (aq) ⇌ H+ (aq) + F- (aq) ?
or F- (aq) + H2O (l) ⇌ HF (aq) + OH- (aq) ?
Answer:
Since both include HF, F-, H+, OH- (the last two are related
by [H+][OH-] = 1 x 10-14), we can use either equation !!
[init]
[change]
[equil]
HF (aq) ⇌
0.10 M
-x
(0.10-x)
H+ (aq)
0
+x
x
+
F- (aq)
0.20 M
+x
(0.20 + x)
[H  ][F - ]
x(0.20  x)
0.20 x
Ka =
= 6.8 x 10-4 =

[HF]
0.10
(0.10 - x)
 x = (0.10) (6.8 x 10-4)/0.20 = 3.4 x 10-4 M
(x << 0.10)
+
+
-4
 [H ] = [H3O ] = 3.4 x 10 M
Compare:
in pure 0.1 M HF, [H3O+] = 8.5 x 10-2 M
in pure 0.2 M NaF, [H3O+] = 6.0 x 10-9 M
[H  ]
(3.4 x 10-4 )
100% =
100% = 0.34%
[HF]
0.10
Compare: For pure 0.1 M HF, % ionization is 8.5%
% ionization of HF =
 the presence of NaF in the solution has decreased the % ionization
of HF (equilibrium shifted to the left)  less [H3O+]  less acidic.
2
Now, let’s go back and see how a buffer works: It is a solution
containing high concentrations of a weak acid (base) and its conjugate
base (acid).
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)
acetic acid (HAc)
acetate ion (Ac-)
If strong acid added: H+ + Ac-  HA (K>>1  assume rxn goes ~100%)
If strong base added: OH- + HA  H2O + Ac- (K>>1  assume rxn goes
~100%)
Reactions go strongly to right  [H+] and [OH-] in the solution change
little. Similar logic for weak base (e.g., NH3) and salt (e.g., NH4+Cl-).
** As long as not too much strong acid or base is added, it will be
“mopped-up” by the buffer and pH will change only slightly **
Figure 19.2 change of pH on addition of strong acid or base.
Buffer Calculations: Two types we must be able to handle:
A.
B.
Calculate the pH (or pOH, [H3O+], [OH-]) of a buffer solution
Calculate new pH (or pOH, …) after something is added
Example: Sample Problem 19.1
(1) Calculate pH of a solution that is 0.50 M HAc and 0.50 M NaAc.
Ka = 1.8 x 10-5.
HAc (aq)
[init]
0.50 M
change -x
[equil] (0.50-x)
⇌
H+ (aq)
0
+x
x
+
Ac- (aq)
0.50 M
+x
(0.50+x)
3
x(0.50  x)
0.50x

= x = 1.8 x 10-5
0.50 - x
0.50
 [H3O+] = 1.8 x 10-5 M  pH = 4.74
Ka =
(2)
What is the new pH after 0.020 mol of solid NaOH is dissolved in
1.0 L of the buffer solution?
0.020 mol
[NaOH] =
= 0.020 mol/L = 0.020 M
1.0 L
NaOH is a strong base  dissociates 100%  [OH-] = 0.020 M
What will happen when we add the NaOH? All the added OH- will react
with HAc to give more Ac-. Keep track of changes with a table:
[init]
[addition]
[equil]
HAc (aq) + OH- (aq)  Ac- (aq) + H2O (aq)
0.50 M
0.50 M
+ 0.020 M
0.48 M
0
0.52 M
 new [HAc] = 0.48 M , new [Ac-] = 0.52 M
Now calculate new pH
[init]
[change]
[equil]
HAc (aq)
0.48 M
-x
(0.48-x)
⇌
H+ (aq)
0
+x
x
+
Ac- (aq)
0.52 M
+x
(0.52+x)
(0.52  x)x 0.52x

= 1.8 x 10-5
0.48 - x
0.48
-5
(0.48) (1.8x10 )
x=
= 1.7 x 10-5 M (assumption okay)
(0.52)
Ka =
 [H3O+] = 1.7 x 10-5 M
 pH = 4.77
pH changed only slightly – BUFFERS WORK!!
4
(3)
What is the new pH if 0.020 mol HCl had been added to (1)
(instead of 0.020 mol NaOH)?
[HCl] = 0.020 M  [H3O+] = 0.020 M (100% dissociation)
What will happen when we add the H+ (= H3O+)? All of it will react with
Ac- to give more HAc. Keep track of changes with a table:
Ac- (aq) + H3O+ (aq)  HAc (aq) + H2O (l)
[init]
0.50 M
0.50 M
addition
0.020 M
[equil]
0.48 M
0
0.52 M
New [HAc] = 0.52 M
New [Ac-] = 0.48 M
Now calculate new pH
HAc (aq)
[init]
change
[equil]
Ka =
0.52 M
-x
(0.52-x)
⇌
H+ (aq)
+
0
+x
x
Ac- (aq)
0.48M
+x
(0.48-x)
x(0.48 - x)
0.48x

= 1.8 x 10-5
0.52
(0.52 - x)
 x = (0.52)(1.8 x 10-5)/0.48 = 2.0 x 10-5 M
 [H3O+] = 2.0 x 10-5
 pH = 4.70
Again, the pH has changed only slightly, even though a significant
amount of strong acid was added. BUFFERS WORK!!
5
Henderson-Hasselbalch Equation
**This allows us to calculate the pH without calculating [H+] first**
It can be shown for a solution of a weak acid HA that the pH, [HA] and
[A-], and the pKa of the acid are related by
 [A- ] 

pH = pKa + log 
 [HA] 
or, for any weak acid/conjugate base, or weak base/conjugate acid
 [base] 
pH = pKa + log 

[acid]


Let’s apply H-H eqn. to our sample problem 19.1:
pKa = -log (1.8 x 10-5) = 4.74
(1)
[Ac- ]
 0.50 
pH = pKa + log
= 4.74 + log 

[HAc]
 0.50 
 pH = 4.74
(2)
(3)
(since log 1 = 0)
 0.52 
pH = 4.74 + log 
 = 4.77
 0.48 
 0.48 
pH = 4.74 + log 
 = 4.70
 0.52 
Two methods give same answers (within rounding errors on sig. figs.)
Calculations also involving volume changes:
-- introduces an extra complication we must be able to deal with.
Example: A buffer is prepared from 60.0 mL of 0.100 M NH3 and 40.0
mL of 0.100 M NH4Cl. What is the pH of the buffer solution? Kb = 1.8
x 10-5 for NH3.
6
Answer: The buffer involves the NH3/NH4+ conjugate acid/base pair.
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Conc’s: Since two solutions are mixed, both have been diluted.
Use this: ViMi = VfMf
Vi = initial volume, Mi = initial molarity
Vf = final volume, Mf = final molarity
Vf = 60.0 mL + 40.0 mL = 100.0 mL
[NH3]f = ViMi/Vf = (60.0 mL)(0.100 M)/100.0 mL = 0.0600 M
[NH4+]f = ViMi/Vf = (40.0 mL)(0.100 M)/100.0 mL = 0.0400 M
Now, we can calculate pH:
[init]
[change]
[equil]
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
0.0600
0.0400
0
-x
+x
+x
(0.0600-x)
(0.0400+x) x
x(0.0400  x) 0.0400x

0.0600
(0.0600 - x)
(1.8 x 10-5 )(0.0600)
x=
= 2.7 x 10-5 M
(0.0400)
 pOH = -log (2.7 x 10-5) = 4.57  pH = 9.43
Kb = 1.8 x 10-5 =
If we now add a certain volume of strong acid or base, we would again
have to calculate new total volume and new [NH3] and [NH4+] and [acid
or base] before we could calculate new pH.
7
Buffer Capacity and Buffer Range
(A) Buffer Capacity is how much added acid or base a buffer solution
can “mop up” before it can no longer resist changes in the pH.
** The more concentrated the components of the buffer, the
greater the buffer capacity **.
Thus, a buffer containing 0.10 M NH3/0.10 M NH4+ is good, but a
buffer containing 1.0 M NH3/1.0 M NH4+ is much better.
Note: Since [NH3]/[NH4+] is 1 in both cases, they both have pH = 4.74,
but very different buffer capacities.
Also, the more similar are the conc’s of the two components, the higher
is the buffer capacity. The greater the difference, the greater will be
the pH change when strong acid or base is added.
** buffer capacity highest when [HA] = [A-] **
From the H-H eqn:
[A- ]
pH = pKa + log
[HA]
[A- ]
log
= log 1 = 0
[HA]
 pH = pKa gives the highest buffer capacity
 when [A-] = [HA],
** Summarizing: Buffer capacity is highest when:
(1)
(2)
(3)
[HA] and [A-] are large
[HA] = [A-]
The pH is equal to (or very near) the pKa of the HA used.**
8
Example: Which is the best acid to use to make a buffer solution with
a pH = 2.0. Acetic acid (pKa = 4.74); chlorous acid (pKa = 1.95) or formic
acid (pKa = 3.74).
To ensure high capacity, we need [HA]  [A-]. It is possible to get this
and have a pH = 2.0 only if we use chlorous acid since its pKa is close to
the target pH we need.
 the answer is chlorous acid (HClO2).
(B) Buffer Range is the pH range over which the buffer acts
effectively. The further the [A-]/[HA] ratio from 1, the less effective
is the buffer. Useful range is when pH of solution = pKa  1, but the
closer to the acid’s pKa, the better.
Section 19.2. ACID/BASE TITRATION CURVES
**(NOT COVERING IN CLASS - GO OVER YOURSELVES AND IN
DISCUSSION PERIODS)**
Definitions
i) Titration: Technique for determining the amount of acid (or
base) in a solution by measuring the volume of base (or acid) of
KNOWN CONCENTRATION (“standard solution”) that will
neutralize it.
ii) Amount of sample (number of moles):
n = VM
V = volume, M = molarity
iii) Titration Curve: plot of pH vs Vo of added acid or base (Fig 19.7)
iv) Equivalence Point:
The stage in a titration at which
stoichiometric amount of acid/base has been added.
e.g., n (acid in sample) = n (base in standard solution added)
v) Indicator: Compound that undergoes a color change at (or very
near) the equivalence point.
Figs. 19.4 and 19.6
9
acidic
[HIn
H+
+
In-]
basic
different colors
Indicators are weak organic acids (HIn) that change color on
deprotonation to In-.
There are three types of Acid/Base Titration
1. Strong Acid/Strong Base Titration Curves
e.g., NaOH and HCl
Both are 100% dissociated to give H+ (aq), Cl- (aq), Na+ (aq) and
OH- (aq). Na+ (aq) and Cl- (aq) do nothing, but
1
H+ (aq) + OH- (aq)  H2O (l) K =
= 1014
Kw
Example: 40.0 mL of 0.100 M HCl is titrated with 0.100 M NaOH.
i) Original Solution: [H+] = [Cl-] = 0.100 M  pH = 1.000
Note: When we calculate pH, give to the correct Sig. Figs.
But when we measure pH (with a pH meter), we measure it to
only 2 decimal places)  measured pH = 1.00
n (H+) = VM = (40.0 mL)(0.100M) = (0.0400 L)(0.100 M)
 n (H+) = 0.00400 mol (or 4.00 x 10-3 mol)
ii) Add 10.0 mL of NaOH (0.100 M)
Two things happen: - the volume changes
- some of the acid is neutralized
n (OH- added) = VM = (10.0 x 10-3 L)(0.100 M) = 1.00 x 10-3 mol
 1.00 x 10-3 mol of OH- will neutralize 1.00 x 10-3 mol of H+,
because
OH+
H+

H2O
n(init)
0.00400 mol
addition 0.00100 mol
n(equil)
0
0.00300 mol
0.00100 mol
 new n(H+) = 0.00300 M
new volume = 40.0 mL + 10.0 mL = 50.0 mL
= 50.0 x 10-3 L
10
0.00300 mol
n
=
= 0.0600 M
0.0500 L
V
pH = 1.222 measured pH = 1.22
new [H+] =
 new
iii) Add another 10.0 mL of NaOH (0.100 M) - repeat as in ii).
New n(H+) = 0.00200 mol
New volume = 60.0 mL. New [H+] = 0.0333 M pH = 1.478 =1.48
iv) pH = 7.00 at the equivalence point: 40.0 mL of 0.100 M
NaOH will neutralize the 40.0 mL of 0.100 M HCl, because
n(OH- added) = n(H+ originally present)
OH- + H+  H2O
Only Na+ (aq) and Cl- (aq) are left, and they do not change the pH.
 pH = 7.00
i.e., [H+] = [OH-] = 10-7 M from Kw
v) Add another 10.0 mL of NaOH (50.0 mL total)
n(OH-) added (after equivalence point) = VM
= (10.0 mL)(0.100 M) = (10.0 x 10-3 L)(0.100 M) = 1.00 x 10-3 mol
new volume = 80.0 mL + 10.0 mL = 90.0 mL = 90.0 x 10-3 L
 new [OH-] =
 pOH = 1.955
n(OH- ) 1.00 x 10-3 mol
=
V
90.0 x 10-3 L
 pH = 12.045
= 1.11 x 10-2 M
pH = 12.05
We can thus calculate the pH at any point of the titration curve.
** Summary
(1) Calculate original moles of acid (or base)
(2) Calculate moles of added base (or acid)
(3) Calculate new moles of acid (or base) left
(4) Calculate new volume
(5) Calculate new [H+] (or [OH-])
(6) Calculate new pH
**
11
Weak Acid/Strong Base Titration Curves
e.g., propanoic acid (CH3CH2COOH) with NaOH (Figure 19.8)
Three differences with strong acid/strong base case:
1. The initial pH is higher: weak acid is less dissociated than strong
acid (of same conc).  less [H3O+],  higher pH (less acidic)
2. A gently rising stage is seen before equivalence: this is known as
the buffer region. As some conjugate base (CH3CH2COO-) is
formed, we have both HPr and Pr- present  a buffer solution.
At the midpoint, pH = pKa of weak acid.
3. pH > 7.00 at the equivalence point: At the equivalence point,
all propanoic acid has been neutralized.
HPr (aq) + OH- (aq)  Pr- (aq) + H2O (l)
The only thing in solution is Pr- (aq) and Na+ (aq). A salt of Na+ and
Pr- gives a basic solution because Pr- is a weak base.
Pr- (aq) + H2O (l) ⇌ HPr (aq) + OH- (aq)
------------------------------------------------------------------As for strong acid/strong base case, we can calculate the pH at any
stage of the titration.
a)
Original solution of HPr
HPr (aq) ⇌ H+ (aq) +
[init]
0.1000 M
0
change -x
+x
[equil]
(0.1000-x)
x
x2
x2
Ka =

x=
(0.1000  x)
0.1000
x=
Pr- (aq)
0
+x
x
(0.1000)(Ka )
(0.1000)(1.3 x 10-5 ) = 1.1 x 10-3 M = [H3O+]
pH = 2.96
12
b)
Addition of 10.0 mL of 0.1000 M NaOH
orig. moles of HPr = n(HPr) = MV = (0.1000 M)(40.0 mL)(10-3L/mL)
= (0.1000 M)(40.0 x 10-3L) = 40.0 x 10-4 mol
added moles of OH- = n(OH-) = MV = (0.100 M)(10.0 x 10-3L)
= 10.0 x 10-4 mol
OH- will react with HPr to give PrHPr (aq) +
0.00400 mol
n(orig)
n(added)
n(final) 0.00300 mol
OH- (aq)

0.00100 mol
0
Pr- (aq)
0
+ H2O (l)
0.00100 mol
Now calculate new pH: Use Henderson-Hasselbalch equation and
you don’t have to calculate new [HPr] and [Pr-]; can use ratio of
n(HPr) and n(Pr-). Why?
n(Pr- )/Vf
[Pr  ]
pH = pKa + log
= pKa + log
[H Pr]
n(H Pr) / Vf
Vf’s cancel
 pH = pKa + log
n(Pr- )
n(HPr)
 0.00100 
 new pH = pKa + log 

 0.00300
Ka = 1.3 x 10-5  pKa = 4.89  pH = 4.89 + log (0.333) = 4.41
Similarly for 20.0 mL of NaOH, etc. (Sample Problem 19.3)
c)
At the equivalence point: ALL the HPr has been converted to Pr-.
 n(Pr-) = 0.00400 mol because 0.00400 mol of NaOH have been
added to 0.00400 mol of HPr.
n
Volume of 0.100 M NaOH that contains 0.00400 mol of OH- =
M
(remember, n = MV)
0.00400 mol
=
= 0.0400 L  Volume of NaOH = 40.0 mL
0.1000 M
 New total volume = 40.0 + 40.0 mL = 80.0 mL
13
 Final [Pr-] =
n
0.00400 mol
=
= 0.0500 M
Vf
80.0 x 10-3 L
Pr- (aq) + H2O (aq) ⇌ HPr (aq) + OH- (aq)
[init] 0.0500 M
0
0
change -x
+x
+x
[equil] (0.0500-x)
x
x
x2
x2
Kw
1.0 x 10-14
Kb =
=

=
= 7.7 x 10-10
5
(0.0500  x) (0.0500)
Ka
1.3 x 1
 x = [OH-]
(0.0500)(7.7 x 1010 ) = 6.2 x 10-6 M
 pOH = 5.21
 pH = 8.79
Weak Base/Strong Acid Titration Curves
Same shape as weak acid/strong base titration, but inverted (Fig. 19.9).
We start with weak base and add strong acid
 pH starts high (basic) and keeps dropping.
a) Original solution: A weak base  pH > 7. We can calculate the
pH at any point, just as for weak acid/strong base curve.
b) In the buffer region, significant amounts of both base (NH3) and
its conjugate acid (NH4+) are present.
c) pH < 7.00 at the equivalence point: sharp drop because all NH3
has been converted to NH4+.
d) Beyond equivalence point: strong acid is being added, therefore
pH drops steadily.
At the equivalence point, the pH < 7.00. This is because all the NH3 is
gone and the only thing in solution is NH4+ (aq) and Cl- (aq). The Cl- (aq)
does nothing; the NH4+ makes the solution acidic.
NH4+ (aq) ⇌ H+ (aq) + NH3 (aq)
Titration Curves of Polyprotic Acids: Similar to monoprotic
acids, except that two equivalence points are seen. See Fig. 19.10
14