Chapter 15
Aqueous Equilibria:
Acids and Bases
Everyday Acids and Bases.

Acids: vinegar, lemon juice, sulfuric acid
 Bases: antacids, ammonia
A. Acid-Base Concepts: The Bronsted-Lowry
Theory

So far, we’ve discussed the Arrhenius theory
of acids and bases:
– Acids dissociate to produce H+ (examples: HCl,
H2SO4)
– Bases dissociate to produce OH- (examples:
NaOH, Ba(OH)2 )
But -- as some bases don’t contain OH, this is
limiting.
Bronsted-Lowry Theory of Acids and Bases
An acid is a substance that can transfer H+
A base is a substance that can accept H+
HA + B
BH+ + A-
acid
acid
base
base
Note conjugate
acid/base pairs.
Conjugate Pairs…
In the previous reaction, A- is the conjugate
base of the acid HA
B is the conjugate base of the acid HB+
HA + B
BH+ + AIn an acid/base reaction, keep your eye on the
proton!
examples
Write balanced equations for the dissociation of
the following Bronsted-Lowry acids in water:
a. H2SO4
b. H3O+
Examples -- answer
a.
H2SO4 + H2O
HSO4- + H3O+
b.
H3O+ + H2O
H2O
+ H3O+
BOOKKEEPING -- be able to label acid, base,
conjugate acid, conjugate base! (by convention,
CA/CB are on the product side)
Examples
What is the conjugate acid of…
a. HCO3b. CO32-
Examples -- answers
a.
b.
H2CO3
HCO3-
B. Acid Strength and Base Strength
Think of the acid-base reaction as a “tug of war” between the two
bases for a proton:
HA + H2O
H3O+ + A-
Which is the stronger base: H2O or A-? Which wants the
proton more? This will determine whether the equilibrium
lies more to the right or to the left.
Acid/Base Strength cont.


If H2O is a stronger proton acceptor than A-, H2O will
get the protons, and the solution will mostly contain
H3O+ and A-.
Conversely, if A- is the stronger proton acceptor,, the
solution will mostly contain HA and H2O.
**The proton is always transferred to the stronger
base.
Acid/Base Strength: Equilibrium
The strength of the acids and bases in an acid-base
reaction will dictate the position of the equilibrium in
an acid-base reaction.
HA + H2O
A- + H3O+
if this is the stronger
acid
the equilibrium will lie to the right
Because a strong acid will dissociate more thoroughly/completely.
What does it mean to be a strong acid?



Almost completely dissociated in water
Equilibrium almost entirely to the right
Solution consists almost entirely of H3O+ and A- ions - almost no HA molecules
Strong acids have very weak conjugate bases. If HA
has a strong tendency to lose its proton, A- will not be
a good proton acceptor.
What does it mean to be a weak acid?




Only partially dissociated in water.
Solution contains mostly undissociated HA.
Not much H3O+ and A- ion present in solution.
Equilibrium lies toward left.
Weak acids have strong conjugate bases. If HA
does not have a strong tendency to lose its proton, Awill be a good proton acceptor.
C. Hydrated Protons and Hydronium Ions
H+ does not exist by itself in aqueous solution, despite
the fact that you frequently will see H+ (aq) in acidbase equilibria.
In aqueous solution, H+ binds to a water molecule to
form H3O+ ==> hydronium ion.
D. Dissociation of Water
Water can act either as an acid or base, depending on
the situation.
If acid is present, water is a base:
HA + H2O
A- + H3O+
If base is present, water is an acid:
B + H2O
HB + OH-
Dissociation of Water cont.
Or, water can act as both an acid and a base!
H2O + H2O
acid
base
H3O+ + OHacid
base
This process is referred to as the dissociation of
water, and has a dissociation constant, Kw.
Dissociation of Water cont.
For the dissociation of water,
Kw = [H3O+][OH-]
Remember that pure liquids, such as H2O, are not included in
equilibrium expressions.
= equilibrium constant for the dissociation of
water (also referred to as the ion product
constant for water)
Dissociation of Water cont.
Very little of water is ionized; the equilibrium lies far to
the left.
At 25oC, [H3O+] = [OH-] = 1.0 * 10-7 M
so
Kw = [H3O+][OH-] = 1.0 * 10-14 M
**This is true for any aqueous solution at 25oC.
[H3O+] vs. [OH-]
Defining acidic/basic/neutral solutions:
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]
example
The concentration of OH- in a sample of 25oC seawater
is 5.0 * 10-6 M. Calculate the concentration of H3O+
ions, and classify the solution as acidic, neutral, or
basic.
Example -- answer
Kw = [H3O+][OH-]
At 25oC, Kw = 1.0 * 10-14 M
Given [OH-] = 5.0 * 10-6 M
Solve for [H3O+]
[H3O+] = Kw/[OH-]
= (1.0 * 10-14)/(5.0 * 10-6)
= 2 * 10-9 M
Since [OH-] > [H3O+] -- basic
E. The pH Scale
pH is a more convenient way to express the
concentration of hydronium ion in a solution.
pH = -log[H3O+]
pH < 7 -- acidic
pH > 7 -- basic
pH = 7 -- neutral
example
Calculate the pH of a solution that has an
[H3O+] of 6.0 * 10-5 M.
Example -- answer
If pH = -log (6.0 * 10-5)
then pH = 4.22
This solution is acidic.
F. Measuring pH
Color indicators are often used for approximation
Examples: phenolphthalein (changes from colorless
--> pink when going from acidic --> basic),
bromothymol blue
However, pH meters give you a more precise number,
measuring the electrical potential of the solution
(chapter 18)
G. Equilibria in Solutions of Weak Acids
Beware: a weak acid is not the same thing as a dilute
solution of a strong acid!
Even when dilute, the equilibrium for the strong acid
will lie to the right (not for the weak acid)
Equilibrium Constant for a Weak Acid Dissociation
For the reaction
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
Ka = [H3O+][A-]/[HA]
In dilute aqueous solution, [H2O] is essentially
constant.
pKa = -log Ka
example
The pH of 0.10 M HOCl is 4.23. Calculate Ka for
hypochlorous acid. How close did you come to the
true value?
Example -- answer
Ka = [H3O+][-OCl]/[HOCl]
[H3O+] = 10-pH = 10-4.23 = 5.89 * 10-5 M
= [-OCl]
[HOCl] = 0.10 - 5.89 * 10-5 = 0.0999 M
Ka = (5.89 * 10-5)2/0.0999 = 3.47 * 10-8
H. Calculating Equilibrium Concentrations in
Solutions of Weak Acids
How are Ka values useful?
Allows us to calculate the pH of an acidic solution, as
well as equilibrium concentrations of all species
present
Example: Calculate concentrations of all species
present, and the pH of, a 0.10 M HCN solution.
Concentration of all species in 0.10 M HCN soln.
Step 1 List species present before any dissociation
happens, and identify them as either acid or base.
HCN
acid
H2O
acid or base
Concentration of… in 0.10 M HCN cont.
Step 2 What proton transfer reactions can occur, given
the aforementioned molecules?
HCN + H2O
H3O+ + CNKa = 4.9 * 10-10
H3O+ + OHKw = 1.0 * 10-14
**Ka values in Table 15.2
H2O + H2O
Concentration of… in 0.10 M HCN cont.
Step 3 Label the reaction that proceeds farther to the
right (larger equilibrium constant) as the principal
reaction; the other reaction(s) is the subsidiary
reaction.
HCN reaction: principal
dissociation of water: subsidiary
Concentration of… in 0.10 M HCN cont.
Step 4 Create an ICE table, expressing changes in
concentration in terms of x.
Principal rxn
HCN
+
H2O
H 3 O+
+
CN-
----------------------------------------------------------------------------------------------------------------------
Initial (M)
Change (M)
Equil. (M)
0.10
-x
0.10 - x
~0
+x
x
0
+x
x
H2O is not part of the equilibrium expression, and is present in
excess.
Concentration of… in 0.10 M HCN cont.
Step 5 Place the equilibrium values into the Ka
expression.
Ka = 4.9 * 10-10 = [H3O+][CN-]/[HCN]
= (x)(x)/(0.10 - x)
Concentration of… in 0.10 M HCN cont.
Step 5 continued In this particular case -- Ka is small.
This means the reaction does not proceed very far to
the right.
If this is the case, x is small, and to simplify our math,
we can say that (0.10 - x) ~ 0.10
and 4.9 * 10-10 ~ x2/0.10
Thus x2 = 4.9 * 10-11 and x = 7.0 * 10-6
Concentration of… in 0.10 M HCN cont.
Step 6 Now, you can use x to find all equilibrium
concentrations.
[H3O+] = [CN-] = x = 7.0 * 10-6 M
[HCN] = 0.10 - x = 0.10 M
**x was small/negligible here. This is not always the
case!
Concentration of… in 0.10 M HCN cont.
Step 7 The only concentration left to calculate is [OH-]
from the subsidiary reaction.
[OH-] = Kw/[H3O+] = 1.0*10-14/7.0*10-6
= 1.4 * 10-9 M
Since [H3O+] from the dissociation of water is also
1.4*10-9 M, our assumption in the ICE table that
[H3O+] = 0 was valid.
Concentration of… in 0.10 M HCN cont.
Step 8 Finally… calculate pH!
pH = -log [H3O+]
= -log (7.0*10-6)
=5.15
J. Percent Dissociation in Solutions of Weak Acids


Another way to measure/express acid strength:
percent dissociation.
The stronger the acid, the more dissociated it will be
in aqueous solution.
percent dissociation =
([HA] dissoc./[HA] initial) * 100%
K. Polyprotic Acids




Are acids that contain more than one dissociable
proton
Examples: H2SO4 H3PO4
Dissociate in a stepwise manner, and each
dissociation step has its own Ka
Note that each successive Ka value decreases. After
the first proton is removed, the remaining conjugate
base will have a negative charge, making the next
proton harder to remove.
L. Equilibria in Solutions of Weak Bases
Consider the reaction
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
There is a base-dissociation constant similar to
that for an acid:
Kb = [BH+][OH-]/[B] = [NH4+][OH-]/[NH3]
Weak Bases… cont.


Weak bases are frequently amines
Amines are derivatives of ammonia where one or
more of the H has been replaced by a hydrocarbon
group
example
Calculate the pH and the concentrations of all species
present in 0.40 M NH3 (Kb = 1.8 * 10-5).
Step 1 What species are present prior to
dissociation?
NH3
base
H2O
acid or base
Example -- cont
Step 2/3 Seeing that Kb for NH3 is 1.8 * 10-5, this is
considered the principal reaction (as opposed to Kw)
Step 4
NH3
+ H2O
NH4
+
OH-
----------------------------------------------------------------------------------------------------------
Initial(M) 0.40
Change(M) -x
Equil.(M) 0.40-x
0
+x
x
~0
+x
x
Example -- cont.
Step 5
Kb = [NH4+][OH-]/[NH3] = x2/0.40-x ~x2/0.40
=1.8 * 10-5
x = 2.7 * 10-3 M = [NH4+] = [OH-]
[NH3] = 0.40 - x = 0.40 M
Example -- cont
[H3O+] = Kw/[OH-]
= 1.0 * 10-14/2.7 * 10-3
= 3.7 *10-12 M
Use this information to determine pH
pH = -log[H3O+] = -log(3.7 * 10-12) = 11.43
makes sense… NH3 is basic!
M. Relationship Between Ka and Kb

When dealing with a conjugate pair, you can
calculate one from the other. Consider the
following…
NH4+(aq) + H2O(l)
H3O+(aq) + NH3(aq)
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
-------------------------------------------------------------------2 H2O(l)
H3O+(aq) + OH-(aq)
Ka and Kb cont.

Also consider equilibrium constants
Ka = [H3O+][NH3]/[NH4+] = 5.6 * 10-10
Kb = [NH4+][OH-]/[NH3] = 1.8 * 10-5
and Kw = [H3O+][OH-] = 1.0 * 10-14
Net equilibrium constant of two reactions when
added: Ka * Kb
Ka and Kb cont.
In general, when you add two chemical reaction
together, the net equilibrium constant is the product
of the two individual equilibrium constants.
Ka * Kb = (5.6 * 10-10)(1.8 * 10-5) = 1.0 * 10-14
= [H3O+][NH3]/[NH4+] * [NH4+][OH-]/[NH3]
In aqueous solution, Ka * Kb = Kw
N. Factors That Affect Acid Strength
What makes one acid stronger than another?
Often determined by strength and polarity of the H-A
bond.
How easily is the H-A bond broken? The more easily
the bond is broken, the stronger the acid.
Acid Strength cont.

How easily is the H-A bond broken?
– The weaker the H-A bond, the more easily it is broken.
– The more polar the H-A bond, the more easily the bond is
broken. In a more polar bond, A is more electronegative.
After the H-A bond is broken, A bears the negative charge.
The more electronegative A is, the more stable A is in
bearing the negative charge.
The more easily the H-A bond is broken, the stronger the acid.
Acid Strength cont.



Bond strength is the prevalent factor in
groups(columns).
As you go down a group on the periodic table, acid
strength increases (HI is a stronger acid than HF,
etc).
This is due to atomic radius increasing and bond
strength decreasing farther down the periodic table.
Acid Strength cont.



Polarity is the prevalent factor comparing within a
row.
Within the same row of the periodic table, atomic
radius does not appreciably change, but
electronegativity increases going right.
When H is bonded to a more EN atom, the acid is
stronger. An -OH containing molecule is more acidic
than an -NH containing molecule.
Strength of Oxoacids


Oxoacid: contains an -OH bond, which also contains
the acidic H
Anything that might weaken the O-H bond increases
the strength of the acid.
Case 1 Increase EN of Y, increase acid strength.
Shifts electron density toward Y.
|
--Y--O--H
|
O will “feel” less negative charge
when Y is more EN = greater
stability.
Strength of Oxoacids cont.
Case 1 cont.
Examples… HOCl > HOBr >HOI
Case 2
If the identity of Y stays the same but more bonds are
added to O, acid strength will also increase.
perchloric acid > hypochlorous acid
Strength of Oxoacids cont.
To continue the example…
H-O-Cl is weaker than H-O-Cl-O which is weaker
than
O
|
The additional EN O’s draw
H-O-Cl-O
electron density away from
the site of deprotonation.
Acid Strength



Any time you’re considering the strength of an acid,
think about the stability of the corresponding anion.
If the corresponding anion (conjugate base) is
particularly stable, the acid is more likely to
dissociate.
If there are more EN atoms nearby, there are more
places to distribute the negative charge resulting from
deprotonation. This results in a more stable anion.
O. Lewis Acids and Bases
Another, more generalized, acid/base definition.
Lewis acid: electron pair acceptor
Lewis base: electron pair donor
Lewis acids are frequently metal cations.
Lewis Acids and Bases
Example:
Cu2+ + 4 :NH3 --> Cu(NH3)42+
Lewis
acid
Lewis
base
complex ion
(deep blue)