DISCLAIMER: these notes are provided to assist you with mastering the course material but they are not intended as a replacement of the
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conjunction with your own notes.
Ch 10 Acid – Base Equilibria
10.1 Classification of Acids & Bases
 aqueous solutions involving H+(aq) and OH-(aq)
 Arrhenius (from ch 6.3): in water, acids increase [H+], bases increase [OH-]
Bronsted - Lowry Acids & Bases
 consider as proton-transfer reactions:
2O
HCl(g) H

 H + (aq) + Cl- (aq)
as HCl(g) + H 2O(l)  H 3O+ (aq) + Cl- (aq)
 Bronsted - Lowry: acids are proton donors, bases are proton acceptors (this is the usual
treatment of acid and base reactions in aqueous solution)
 also applicable to non-aqueous reactions, eg. gas phase reaction:
H
..
:Cl
.. H + : N
H
H
+
H
.. :Cl:
+ H N H
..
H
 another eg. (a Bronsted base)
NH 3 (aq) + H 2O(l)  NH 4+ (aq) + OH - (aq)
 identify acid(s) and base(s); note formation of hydroxide (OH-), a strong base
 terminology:
 monoprotic acids: single proton donors, eg. HCl, HNO3, acetic acid
 polyprotic acids: donors of two or more protons, eg. H2SO4, H3PO4, carbonic acid
(Table 10.2)
 likewise, mono- , polyprotic bases
 intermediate forms said to be amphiprotic (previously: amphoteric): capable of acting
as either acid (with another base) or base (with another acid), eg. HSO4-, H2PO4-,
HCO3Lewis Acids & Bases
 Lewis acid - an electron pair acceptor (eg. H+)
 Lewis base - an electron pair donor (eg. :NH3 and H2O)
 more general as to what can be an acceptor
 eg. H3N: + BF3  H3N-BF3
Chem 59 - 110 (’02)
2
 note absence of H+ transfer and of water
 Lewis acids generally have vacant orbitals
Cationic Lewis Acids (Hydrolysis of Metal Cations)
 many metal ions (not the common alkali and akaline earth ions) show acidic properties in
water, explained by Lewis acidity
 positively-charged metal ions, eg. Fe3+, dissolve through hydration, electron donation from
water O to metal
 this polarizes O-H bonds of water even more, causes them to dissociate more easily, i.e.become more acidic
 eg.
Fe(H2O)63+(aq)  Fe(H2O)5(OH)2+(aq) + H+(aq)
 amphoteric behavior of some metal hydroxides, (those in the middle of the Periodic Table)
 non-metal oxides, eg. CO2, SO2, known as acid anhydrides
 Group I and II metal oxides, base anhydrides, eg. CaO + H2O  Ca(OH)2
10.2 Acids & Bases in Aqueous Solution: the Bronsted – Lowry Scheme
Conjugate Acid - Base Pairs (extra)
add H+
NH 3 (aq) + H 2 O(l)  NH +4 (aq) + OH - (aq)
base
acid
conjugate
acid
conjugate
base
lose H+
 these pairs of species related by presence or absence of a proton are conjugate acid-base pairs
 generalized: HA and A- or B: and BH+
 can be neutral or charged, Table 10.2
 warning: charge consistency!
 note that in egs. above water has acted as a base in the presence of an acid (eg. HCl, HNO3)
and as an acid in the presence of a base (eg. NH3), said to be amphiprotic (including
autoionization reaction)
 practice recognizing and writing conjugates
Conjugate Acid - Base Strengths (a survey)
 arrange acids in order of ability to donate protons, the more readily an acid gives up a proton
(i.e.- the stronger the acid), the less readily does its conjugate base accept a proton (i.e.- the
weaker the conjugate base)
 conversely, the stronger the base, the weaker is the conjugate acid
59 - 110 (’02), ch 10, Acid – Base Equilibria
3
 define strong acids: [H3O+]  initial concentration of acid
 eg. HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
 whereas weak acids: [H3O+] << initial concentration of acid
 eg. CH3CO2H(aq) + H2O(l)  H3O+(aq) + CH3CO2-(aq)
 conversely, strong base: [OH-]  initial concentration of base
 eg. O2-(aq) + H2O(l)  2 OH-(aq)
 and weak base: [OH-] << initial concentration of base
 eg. ammonia, above, or CO32-(aq) + H2O(l)  HCO3-(aq) + OH-(aq)
 acid - base reaction is necessarily a reaction of two conjugate pairs, direction dominated by
stronger acid - base pair:
HCl(aq) +
H2O(l)  H3O+(aq)
+
Cl-(aq)
stronger acid
stronger base weaker acid
weaker base
than H3O+
than Clthan HCl
than H2O
CH3CO2H(aq)
weaker acid
than H3O+
+
H2O(l)

H3O+(aq)
+
weaker base
stronger acid
than CH3CO2- than CH3CO2H
CH3CO2-(aq)
stronger base
than H2O
 Table 10.2 gives relative strengths of common acids and their conjugates (note, inverse
rankings)
 Later, use this relative information to predict the direction of an acid - base reaction, eg.
page 326-7
The Hydronium Ion & Autoionization of Water
 autoionization:
H 2O(l) H + (aq) + OH - (aq)
K =
[H + ][OH - ]
[ H 2O]
 very little dissociation, [H2O] constant, 55 M
K[H2O] = Kw = [H+][OH-]
Kw = ion-product constant
Kw = 1 x 10-14 at 25C (Table 10.1)
 in neutral solution: [H+] = [OH-] = 1 x 10-7 M
 note: Kw a constant, regardless of whether solution neutral
The Proton in H2O
59 - 110 (’02), ch 10, Acid – Base Equilibria
4
 H+ unlikely; more likely is H3O+, hydronium ion or higher order H5O2+ or H9O4+
 equation above is more likely:
H 2O(l) + H 2O(l) H 3O+ (aq) + OH - (aq)
 H+(aq) and H3O+(aq) used interchangably in terminology; Kw unaffected
Strong Acids & Bases
 H+(or, H3O+) is the strongest proton donor that can exist in water; similarly OH- strongest base
(levelling effect)
 strong electrolytes, fully dissociated
 most common strong acids: HCl, HBr, HI, HNO3, HClO4, H2SO4
 eg.
HNO 3 (aq) + H 3O(l)  H 3O + (aq) + NO -3 (aq)
 simply calculate [H+] from original concentration of the acid
 most common strong bases, metal hydroxides, eg. NaOH, KOH, Ca(OH)2; fully dissociated
 some bases stronger than OH-, dissolve to form OH- (eg. O2-, H-, N3-)
Water and the pH Scale
 water Ionization Constant, Kw
 in pure water, Kw = 1 x 10-14 = [H3O+][OH-]; and [H3O+] = [OH-] = 1 x 10-7 M (at 25oC)
 this is a neutral solution
 disturb this equilibrium by adding acid or base, Kw = 1 x 10-14 = [H3O+][OH-], but [H3O+] 
[OH-]
 in acidic solution, [H3O+] > 1 x 10-7 M
 in basic solution, [OH-] > 1 x 10-7 M
 do Example 10.2
The pH Scale
 [H+] in many aqueous solutions of interest is often quite small, expressed in terms of pH:
pH = - log [H+]
 note: “p” as an operator; log10 implied
 neutral solution: [H+] = 1 x 10-7 M,  pH = 7.0
 pH decreases as [H+] increases
 pH of some common aqueous solutions, Fig. 10.4
 recall, [H+] and [OH-] related by Kw
59 - 110 (’02), ch 10, Acid – Base Equilibria
5
 use of pOH and, autoionization of water:
pH + pOH = - log Kw = 14.00
Measuring pH
 pH meter (electrode theory in Chap 12)
 estimates using colour changes of acid - base indicators, Fig. 10.9
 equilibrium: HIn + H2O  H3O+ + In- , and two forms of indicator have different
colours (note: this is also an acid-base equilibrium)
10.3 Acid & Base Strength (Weak Acids & Bases)
 weak acid, partial dissociation:
HA(aq) + H2O(l)  H+(aq) + A-(aq)
 equilibrium constant = acid-dissociation constant
[H + ][A - ]
Ka =
[ HA]
 the larger Ka, the stronger the acid (note: Ka’s of weak acids always < 1 (strong acids, Ka >>
1)
 examples in Table 10.2
 can be neutral molecules (eg. formic acid, HF, HCN and others), cations (eg. ammonium,
NH4+) or anions (eg. H2PO4-)
 similarly for weak bases:
B(aq) + H2O(l)  BH+(aq) + OH-(aq)
[BH + ][OH  ]
[ B]
 the larger Kb, the stronger the base (note: Kb’s of weak bases always < 1 (strong bases, Kb >>
1)
 can be neutral molecules (eg. CH3NH2) or anions (eg. CN-)
Kb =
Connection Between Ka and Kb for a Conjugate Pair
 use NH4+/NH3 conjugate pair as eg.
NH4+(aq)  NH3(aq) + H+(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
[NH 3 ][H + ]
[NH 4+ ][OH - ]
;
K
=
b
[NH 4+ ]
[NH 3 ]
 add the above two equilibria, net is:
Ka =
H 2O(l) H + (aq) + OH - (aq)
59 - 110 (’02), ch 10, Acid – Base Equilibria
6
 two equilibria are consecutive reactions leading to the overall above, equ’m constant for
equlibria thus added is product of individual equ’m constants
 [NH 3 ][H + ]  [NH 4+ ][OH - ]
Ka x K b = 
 

 [NH 4+ ]   [NH 3 ] 
= [H + ][OH - ] = K w
 this result is general: for a conjugate pair
Ka x Kb = Kw
 example, p. 326-7, of reaction between HF (aq) and CN-(aq) by consideration of relative acid
strengths from Table 10.2
Weak Acid & Base Equilibria on the pH-Scale
 rather than using Ka and Kb, express these on the logarithmic pH-scale:
 define: pKa = - log Ka (lower, stronger acid) and pKb = - log Kb (lower, stronger
base)
 (later: simplify by treating all in terms of acid dissociations, only deal with pKa’s)
 from: Ka x Kb = Kw ; hence: pKa + pKb = 14 (at 25oC)
 pKa’s also given in Table 10.2 and Fig. 10.5
10.4 Equilibrium Calculations for Weak Acids & Bases
 two types:
 calculate Ka or Kb from initial concentration(s) and measured pH
 calculate equilibrium concentrations and pH from initial concentrations and known Ka or
Kb
 four steps:
 write the ionization equilibrium equation
 write the equilibrium expression
 construct a table as in chapter 9 (initial, change equilibrium concentrations for all
species)
 solve for unknown (may or may not be quadratic)
Calculating Ka or Kb from Measured pH
 using a known solution:
 0.10 mol propanoic acid made up to 1.0 L in H2O; measured pH = 2.94
1. write ionization equilibrium:
C3H5O2H (aq)  H+(aq) + C3H5O2-(aq)
2. write equilibrium expression:
Ka
59 - 110 (’02), ch 10, Acid – Base Equilibria
[H + ][C 3 H 5O -2 ]
=
= ?
[HC 3 H 5O 2 ]
7
3. construct table as in chapter 9:
 2.94 = - log [H+]; hence, [H+] = 1.15 x 10-3 M (assume none of this comes from
water ionization)

C3H5O2H (aq)
(aq)
initial
change
H+(aq)
0.10 M
- x M (= - 1.15 x 10-3)
equilibrium
0
+ x M (1.15 x
10-3)
1.15 x 10-3 M
(0.10 - 1.15 x 10-3)  0.10
M
+
C3H5O2-
0
+ x M (1.15 x
10-3)
1.15 x 10-3 M
4. substitute equilibrium concentrations in expression
1.15 x 10-3 
[H + ][C 3 H 5O -2 ]
Ka =
=
[C 3 H 5O 2 H]
010
.
2
= 1.32 x 10-5 ;  pK a = 4.88
Calculating Equilibrium Concentrations and pH for Solutions of Weak Acids
 using Ka and a known solution, 4 steps; use 0.30 M acetic acid as example (similar to
Examples 10.3 and 10.4):
1. write ionization equilibrium:
C2H3O2H(aq)  H+(aq) + C2H3O2-(aq)
2. write equilibrium expression:
[H + ][C 2 H 3O -2 ]
= 1.8 x 10-5
[C 2 H 3O 2 H ]
Ka =
3. construct table as in chapter 9:
initial
change
equilibrium

C2H3O2H(aq)
0.30 M
-xM
(0.30 - x) M
H+(aq)
0
+xM
xM
+
C2H3O2-(aq)
0
+xM
xM
4. substitute equ’m conc’ns in expression
[H + ][C 2 H 3O -2 ]
(x)(x)
Ka =
=
= 1.8 x 10-5
[C 2 H 3O 2 H ]
0.30 - x
 quadratic in x , avoid as long as (0.30 - x)  0.30; start by assuming this, then:
Ka = x2/0.30 = 1.8 x 10-5
[H+] = x = 2.3 x 10-3 M
59 - 110 (’02), ch 10, Acid – Base Equilibria
8
pH = - log(2.3 x 10-3) = 2.64
 now check assumption, also calculate % ionization of acetic acid:
% ionization = (0.0023/0.30)100 = 0.77%
 as long as < 5%, assumption OK (alternatively, OK if [HA]o  100Ka); if not, solve
quadratic (or, use method of successive approximations, p. 331 & Appendix C)
 compare with strong acid (eg. HCl, same amount): pH = - log(0.30) = 0.52
 results of above calculations also show that [H+] much lower than [HA] added - properties
reflect this, eg. conductivity
 another example, if stronger acid (formic acid, Ka = 1.8 x 10-4) and lower initial acid
concentration (1 mM) given, hence must solve quadratic
 similarly for calculating equilibrium concentrations and pH for an aqueous solution of a weak
base, eg. pyridine and ammonia (Example 10.5)
 same rules of thumb for when quadratic must be solved
 note that calculations first furnish [OH-] and then pOH; convert latter to pH
Acid-Base Properties of Salt Solutions: Hydrolysis
 strong electrolytes, dissolve readily, but many have anions and/or cations that react with
water, i.e.- undergo hydrolysis
 eg. anions from weak acids produce OH-:
A-(aq) + H2O(l)  HA(aq) + OH-(aq)
 Example 10.6 for sodium acetate (NaCH3COO); pH of 8.9 calculated for 100 mM
solution
 cations such as ammonia derivatives dissociate in water to give acidic solutions:
NH4+(aq)  NH3(aq) + H+(aq)
 eg. exercise: pH of 0.50 M aqueous ammonium chloride?
1. write ionization equilibrium (see above):
2. write equilibrium expression:
[H + ][NH 3 ]
Ka =
= 5.6 x 10 -10 (Table 10.2)
+
[NH 4 ]
3. construct table as in chapter 9:
59 - 110 (’02), ch 10, Acid – Base Equilibria
9
initial
change
equilibrium

NH4+(aq)
0.50 M
-xM
(0.50 - x) M
NH3(aq)
0
+xM
xM
+
H+(aq)
0
+xM
xM
4. substitute equ’m conc’ns in expression
[H + ][C 2 H 3O -2 ]
(x)(x)
Ka =
=
= 5.6 x 10-10
[C 2 H 3O 2 H ]
0.50 - x
 quadratic in x , avoid as long as (0.50 - x)  0.50; start by assuming this, then:
Ka = x2/0.50 = 5.6 x 10-10
[H+] = x = 1.7 x 10-5 M
pH = - log(1.7 x 10-5) = 4.8
 now check assumption, also calculate % ionization of ammonium cation:
% ionization = (1.7 x 10-5 /0.50)100 = 0.0034%
 some generalizations:
1. salts from a strong acid and a strong base - neutral solution
2. salts from a strong base and a weak acid - basic solution
3. salts from a weak base and a strong acid - acidic solution
4. salts from a weak base and a weak acid - solution pH depends on relative extents of
hydrolysis of the two ions
Acid - Base Reactions
 more complex solutions: additional solutes
 eg. add strong base of known concentration to solution of acid until equivalence point
(stoichiometric amounts of acid and base)
 for weak acid - weak base reaction can predict direction of reaction from relative strengths in
Table 10.2
 eg. acetic acid plus ammonia:
59 - 110 (’02), ch 10, Acid – Base Equilibria
10
overall: CH 3CO 2 H(aq) + NH 3 (aq)  NH 4+ (aq) + CH 3CO -2 (aq)
contributing reactions:
CH 3CO 2 H(aq) + H 2 O(l)  H 3O + (aq) + CH 3CO -2 (aq) ; K a = 1.8 x 10 -5
NH 3 (aq) + H 2 O(l)  NH 4+ (aq) + OH - (aq) ; K b = 1.8 x 10-5
H 3O + (aq) + OH - (aq)  2 H 2 O(l) ; K =
1
Kw
= 1.0 x 1014
Ka K b
= 3.2 x 10 4
Kw
note that the equilibrium for the overall process is favorable due to coupling of unfavorable
ionizations of acid and base with the very favorable union of hydronium and hydroxide ions
for overall reaction: K net =

Strong Acid - Strong Base Reactions
 both completely ionized, solution pH dictated by excess component; if equal molar amounts,
eg.:
H 3O + (aq) + Cl - (aq) + Na + (aq) + OH - (aq)  2 H 2 O(l) + Na + (aq) + Cl - (aq)
net ionic equation: H 3O + (aq) + OH - (aq)  2 H 2 O(l) ; K = 1 K = 1 x 1014
w
 equivalence point only at pH 7.00 for these titrations, not all (see weak acids and bases,
later)
Strong Acid - Weak Base Reactions
 eg. ammonia + HCl
NH 3 (aq) + H 2 O(l)  NH 4+ (aq) + OH - (aq) ; K b = 1.8 x 10-5
H 3O + (aq) + OH - (aq)  2 H 2 O(l) ; K = 1 K
= 1.0 x 1014
w
H 3O + (aq) + NH 3 (aq)  H 2 O(l) + NH 4+ (aq)
Kb
= 1.8 x 109
Kw
 mix equimolar quantities of strong acid and weak base, get quantitiative conversion to
NH4+, which forms an acidic solution
 overall reaction is reverse of acid dissociation of NH4+, hence Knet = 1/Ka
 what is the pH at the equivalence point? pH of a solution of that [NH4+]
 Example: 100 mL of 0.10 M HCl plus 50 mL of 0.20 M NH3
 treat as 0.067 M solution of NH4Cl , calculate pH of 5.21
 similarly: organic amine plus HCl
for overall reaction: K net =
59 - 110 (’02), ch 10, Acid – Base Equilibria
11
General Approach to These Problems
 extension of our general approach to equilibrium calculations
 stoichiometric calculation, from strong acid or base reaction (a titration, later)
 from quantity of conjugate formed in first stage calculate its concentration
(i.e.- volume correction)
 equilibrium calculation to find resulting pH (whether equivalence point or not)
Strong Base - Weak Acid Reactions
 eg. weak acid, formic acid + NaOH
HCO 2 H(aq) + H 2 O(l)  H 3O + (aq) + HCO -2 (aq) ; K a = 1.8 x 10-4
H 3O + (aq) + OH - (aq)  2 H 2 O(l) ; K = 1 K
= 1.0 x 1014
w
HCO 2 H(aq) + OH - (aq)  H 2 O(l) + HCO -2 (aq)
for overall reaction: K net =
Ka
Kw
=
1
= 1.8 x 1010
Kb
 mix equimolar quantities of strong base and weak acid, get quantitiative conversion to
HCO2-, which form a basic solution
 overall reaction is reverse of base association of HCO2-, hence Knet = 1/Kb
 what is the pH at the equivalence point? pH of a solution of that [HCO2-]
 Example: 50 mL of 0.10 M NaOH plus 50 mL of 0.10 M HCO2H
 treat as 0.050 M solution of HCO2Na , calculate pH of 8.23
 hence, appropriatness of phenolphthalein as indicator
Weak Acid - Weak Base Reactions
 eg. acetic acid plus ammonia at start of chapter




overall: CH 3CO 2 H(aq) + NH 3 (aq)  NH 4+ (aq) + CH 3CO -2 (aq)
Ka K b
for overall reaction: K net =
= 3.2 x 104
Kw
for equimolar mixture of the two, expect complete conversion to the salt
is resulting solution acidic or basic?
consider hydrolysis of the two conjugate ion partners in the salt:
CH 3CO -2 (aq) + H 2 O(l)  CH 3CO 2 H(aq) + OH - (aq) ; K b = 5.6 x 10-10
NH 4+ (aq) + H 2 O(l)  NH 3 (aq) + H 3O + (aq) ; K a = 5.6 x 10-10
 i.e.- equal tendencies to make the solution acidic and basic, therefore neutral solution
more difficult situation:
59 - 110 (’02), ch 10, Acid – Base Equilibria
12


50 mL 0.10 M acetic acid plus 50 mL 0.10 M pyridine (an even weaker base than
ammonia)
is solution acidic or basic? (for the conjugates: Ka for pyridinium ion, C5H5NH+, 6.7
x 10-6; Kb for acetate ion 5.6 x 10-10)
overall: CH 3CO 2 H(aq) + C5 H 5 N(aq)  C5 H 5 NH + (aq) + CH 3CO -2 (aq)
contributing reactions:
CH 3CO 2 H(aq) + H 2 O(l)  H 3O + (aq) + CH 3CO -2 (aq) ; K a = 1.8 x 10 -5
C5 H 5 N(aq) + H 2 O(l)  C5 H 5 NH + (aq) + OH - (aq) ; K b = 1.6 x 10-9
H 3O + (aq) + OH - (aq)  2 H 2 O(l) ; K =
1
Kw
= 1.0 x 1014
Ka K b
= 2.7
Kw
here, in contrast to the ammonium acetate situation, the equilibrium constant is much
smaller and conversion to pyridinium acetate is about 16% (calculation in class; think
about how much dissociation acetic acid alone would undergo, about 10 times less)
consider hydrolysis of the two component conjugate ions:
for overall reaction: K net =


CH 3CO -2 (aq) + H 2 O(l)  CH 3CO 2 H(aq) + OH - (aq) ; K b = 5.6 x 10-10

C5 H 5 NH + (aq) + H 2 O(l)  C5 H 5 N(aq) + H 3O + (aq) ; K a = 6.7 x 10-6
second equation has the largest equ’m constant, hence its production of hydronium
ion would dominate the first one’s production of hydroxide ion; predict an acidic
solution
Common Ion Effect (also see p. 370-2)
 eg. salt of conjugate base added to soln of weak acid decreases [H+], i.e.- increases pH
 eg.
C2 H 3O2 H(aq)  H + (aq) + C2 H 3O-2 (aq)
 generate common ion from acid - base reaction: dissociation of weak electrolyte decreases
due to addition of strong electrolyte that has an ion in common (Le Chatelier; we will see this
later in the shape of titration curves before the equivalence point; presence of acetate, above,
acts to suppress further dissociation of acid)
 Example: calculate pH of 2.67 for 0.25 M acetic acid, pH of 4.35 of same in presence of 0.10
M sodium acetate
 for a weak base, common ion lowers pH, eg.
NH 3 (aq) + H 2O(l)  NH 4+ (aq) + OH - (aq)
59 - 110 (’02), ch 10, Acid – Base Equilibria
13
10.5 Buffer Solutions
 mixture of weak acid - base conjugate pairs
 resists change in pH due to added H+ or OH- (will see later with titration curves)
 Example: compare addition of strong acid to water with addition to acetate buffer
 1.0 mL of 1.0 M HCl added to 1.0 L water or to 1.0 L of acetic acid/sodium acetate
buffer in which [CH3CO2H] = 0.70 M, [CH3CO2-] = 0.60 M. Calculate pH’s
 calculate pH of HCl in water = 3.0; pH of buffer 4.68, same with added HCl
 this is the buffering effect
 opposite example shown in Figure 10.12
 do Exercises 10.7 & 10.8
 analogous example: 0.50 L of a buffer composed of 0.50 M formic acid and 0.70 M sodium
formate; pH before and after adding 10.0 mL of 1.0 M HCl? (formic acid Ka = 1.8 x 10-4)
 NB: above example atypical, since buffer concentration high; pH not usually held absolutely
constant on addition of acid or base
General Expressions for Buffer Solutions
HA(aq.) H + (aq.) + A - (aq.)
K 
a
[H + (aq.)][A - (aq.)]
[HA(aq.)]
[H + (aq.)] = K
[HA(aq.)]
a [A (aq.)]
 take (-logarithm) of each side to get:
key equation: Henderson - Hasselbalch
[ -]
pH = pK a + log A
[HA]
 NB: at [base] = [acid], pH = pKa, hence the importance and utility of pKa’s
59 - 110 (’02), ch 10, Acid – Base Equilibria
14
 best buffering at pH near pKa of conjugate pair, ie.- capacity to "mop up" added acid or
base best
 capacity also increased by concentration of conjugate pair
 important properties of a buffer are its pH and capacity
 quantitative aspects:
 use of the equation to calculate pH
 use of the equation to prepare buffers of known pH, Exercise 10.10
 note: from the equation it is the mole ratio of the conjugate pair that is important, not their
absolute concentrations; hence, pH of a buffer does not change with dilution (but capacity
does)
 could equally use Ka or Kb for these buffering equilibria but since Henderson – Hasselbalch
equation is defined in terms of pKa it is convenient and simpler to treat all of these acid –base
equilibria as acid dissociations
10.6 Acid - Base Titration Curves


pH as a function of added strong base or acid, not just at the equivalence point
especially when a weak acid or base titrated with a strong base or acid, respectively
Strong Acid - Strong Base Titrations
 eg. Fig. 10.13, HCl and NaOH
 rapid change through equivalence point for small additions of base
 excess ion dictates solution pH
 eg. 100.0 mL of 0.10 M HCl + 0.10 M NaOH, stages of titration, p. 342-3
Weak Acid - Strong Base Titrations
 eg. acetic acid plus NaOH, Fig. 10.14
 four regions of interest:
1. pH before the titration begins, the weak acid solution itself (pH = 2.87 for 0.10 M
acetic acid)
2. pH at the mid-point of the titration, the pKa (for acetic acid, 4.74); near mid-point,
shallow slope, i.e.- relatively insensitive to added strong acid or strong base =
buffering region
59 - 110 (’02), ch 10, Acid – Base Equilibria
15



3. pH at the equivalence point; here basic due to hydrolysis of conjugate base of weak
acid
4. pH when base added beyond the equivalence point
Example 10.11
steps in calculations:
 stoichiometric calculation to give concentrations of conjugate pair
 equilibrium calculation, usually with Henderson - Hasselbalch equ’n before the
equivalence point
as acid becomes weaker, the equivalence point region of the titration curve is less easy to
discern
Weak Base - Strong Acid Titrations
 eg. ammonia plus HCl; acidic end-point, hence, methyl red appropriate indicator
10.7 Polyprotic Acids & Bases
 more than one H+ can dissociate, eg. (all aq)

H2SO3  H+ + HSO3- ; Ka1 = 1.7 x 10-2 ; pKa1 = 1.8
HSO3-  H+ + SO3-2 ; Ka2 = 6.4 x 10-8 ; pKa2 = 7.2
similarly phosphoric acid: Ka1 = 7.5 x 10-3 ; Ka2 = 6.2 x 10-8 ; Ka3 = 3.6 x 10-13 (pKa’s 2.1,
7.2, 12.4, respectively)
 note: second dissociation much weaker (more difficult to remove second positive charge from
a molecule already negative), typically 104 – 106 times weaker (4 – 6 “log units”, i.e.- pH/pK
units)
 in determining solution pH of a polyprotic acid (fully protonated form) or its “full”
conjugate base (fully deprotonated), it is the first equilibrium which dominates
 eg. the Kb1 process for Na2CO3 (the CO32- to HCO3- step), this generalization falls down
when the two pK’s are 3 or less units apart
 Examples 10.12 and 10.13
Weak Diprotic Acid - Strong Base Titrations
 eg. oxalic acid, first equivalence point not easily discerned from graph
 carbonic acid detail in Example 10.12, Fig. 10.16
 eg. carbonic acid (dissolve CO2 in water),
 neutralization stages:
H 2CO3  OH -  HCO-3  H 2O
HCO-3 + OH-  CO23 + H O
2
 here, first ionization practically complete before second starts (pKa's 6.4 and 10.3)
59 - 110 (’02), ch 10, Acid – Base Equilibria
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 titration curve shows discrete waves (as long as pKa's not closer than about 3 pH units)
 other polyprotic acids: H2SO4 , H3PO4 (see Fig. 10.17 for titration with NaOH) , amino
acids H3N+CHRCO2H , proteins and nucleic acids (polymers)
Suggested Problems
1, 3, 5; 13 – 67, odd; or, first few odd in each sub-section
59 - 110 (’02), ch 10, Acid – Base Equilibria