Acids
and Bases:
Three models:
- Arrhenius
acid: produces H+ in aqueous solution
Arrhenius
base: produces –OH
in aqueous solution
This
model is good for calculating pH and pOH.
But:
only one type of acid and one type of base
- BrØnsted-Lowry
acid: a proton donor
BrØnsted-Lowry base: a proton acceptor
But:
no transfer of protons
- Lewis
acid: an electron-pair
acceptor
Lewis
base: an electron-pair donor
This
is the most general model
In general, we can
describe the reaction when an acid is dissolved in water:
HA(aq) + H2O(l) ↔
H3O+(aq) + A-(aq)
Ka, the acid
dissociation constant, represents the reaction in which a proton is
removed from HA to form A-, the conjugate base.
Ka = ([H3O+][A-])
/ ([HA]) = ([H+][A-]) / ([HA])
 | conjugate
acid: the result of the base that gained a proton (species +)
|
| conjugate
base: what remains of an acid that lost a proton (species -)
|
For dilute solutions,
we can assume [H20] (liquid state) remains constant when an acid is
dissolved so we do not include [H20] in the equation.
Acid Strength:
1.
strong acid
·
equilibrium lies far to the
right (100% dissociation)
·
large Ka (because
[products] »
[reactants])
·
[H+] ≈ [HA]0 at equilibrium (because 100%
dissociation)
·
yields a very weak conjugate base A-, weaker base than water
and has a low affinity for proton
2.
weak acid
·
equilibrium lies far to the
left (minimal dissociation)
·
small Ka
·
[H+] « [HA]0
at equilibrium
·
yields a very strong conjugate base A-, stronger base than
water and has a high affinity for proton
Strong
Acids
|
HCl
|
HNO3
|
|
HBr
|
H2SO4
|
|
HI
|
HClO4
|
Strong Bases
|
LiOH
|
CsOH
|
|
NaOH
|
Ca(OH)2
|
|
KOH
|
Sr(OH)2
|
|
RbOH
|
Ba(OH)2
|
Remember that
“strong” means 100% dissociation. Thus
a 1.0 M HCl solution contains H+
and Cl- ions rather than HCl molecules.
*Memorize! If we know the
above, we can assume everything else is weak.
Some terms:
| monoprotic
acid: has one acidic proton
|
| diprotic
acid: has two acidic protons
|
(ex)
H2SO4(aq) →
H+(aq) + HSO4-(aq)
strong acid
HSO4-(aq)
↔ H+(aq) +
SO42-(aq) weak acid
| oxyacids:
an acidic proton attached to an oxygen atom
|
| amphoteric:
a substance that acts as both an acid AND a base.
Water is the most common amphoteric substance.
|
Dissociation constant
(the ion-product constant) for water:
Experiment shows that at
25°C, no matter what water contains,
Kw = [H+][-OH]
= 1.0 x 10-14 and [H+] = [-OH] = 1.0 x 10-7
The pH scale is a
way to represent solution acidity.
| pH
= -log[H+]
|
| pOH
= -log[-OH]
|
Bases:
| In
general, B(aq) + H20(l) ↔
BH+(aq) + -OH(aq) and the
equilibrium constant is Kb = ([BH+][-OH]) /
[B]
|
| What
does Kb mean? Kb
refers to the reaction of a base with water to form the conjugate acid and
hydroxide ion.
|
| B
competes with -OH for the H+ ion.
|
| A
base does NOT have to contain an -OH ion.
In the following reaction, water is the acid and ammonia is the base:
|
NH3(aq) + H20(l)
↔ NH4+(aq)
+ -OH(aq)
Ammonia
accepts a proton because it has an unshared pair of electrons and thus functions
as a base.
Calculating pH of
strong acids:
(ex) Calculate the pH of
0.10 M HNO3.
Since HNO3 is
a strong acid it completely dissociates in water and so the main species in
solution are H+, NO3-, and H20.
Since we want to calculate pH we want to know [H+].
| [H+]
from HNO3 = 0.10 M
|
| [H+]
from H20 is negligible because so small (1.0 x 10-7 M)
|
*Whenever you have a
strong acid or strong base in water, the dissociation of water (the H+
or –OH) is SO SMALL that it is negligible.
So HNO3 dissolved
is the only important source of H+ ions.
pH = -log[H+]
= -log[0.10] = 1.00
Calculating pH of weak
acids:
(ex)
Calculate pH of 0.100 M HOCl (Ka = 3.5 x 10-8).
Since HOCl is a weak acid
it mostly remains undissociated. The
main species in solution are HOCl and H20. Both species can produce H+:
| HOCl
↔ H+(aq)
+ -OCl(aq)
Ka = 3.5 x 10-8
|
| H20
↔ H+(aq) + -OH (aq)
Kw = 1.0 x
10-14
|
HOCl is a much stronger
acid than H20. Most of
the H+ will come from HOCl. Therefore
the equilibrium expression is:
Ka = ([H+][-OCl])
/ [HOCl] = 3.5 x 10-8
Making the ICE table:
|
|
HOCl
|
H+
|
-OCl
|
|
initial
|
0.100 M
|
0
|
0
|
|
Δ
|
-x
|
+x
|
+x
|
|
at equilibrium
|
0.100 – x
|
x
|
X
|
Plugging these values
into the Ka expression:
Ka = (x)(x) /
(0.100 – x) = 3.5 x 10-8
IMPORTANT: Since Ka
is so small, x is also very small. We
can assume that [HA]0 – x ≈ [HA]0 so Ka
now equals:
3.5 x 10-8 = x2
/ 0.100
Solving for x, x = 5.9 x
10-5 = [H+]. Now
we can calculate pH.
pH = -log[5.9 x 10-5]
= 4.23
Calculating pH of weak
acid mixtures:
(ex) Calculate the pH of
a solution that contains 1.00
M HCN (Ka = 6.2 x 10-10) and
5.00 M HNO2 (Ka = 4.0 x 10-4).
Also, calculate [CN-] in solution at equilibrium.
| The
major species in solution are HCN, HNO2, and H20, all
of which produce H+.
|
| HCN(aq)
↔ H+(aq)
+ CN-(aq)
Ka = 6.2 x 10-10
|
| HNO2(aq)
↔ H+(aq)
+ NO2-(aq)
Ka = 4.0 x 10-4
|
| H20(l)
↔ H+(aq)
+ -OH (aq)
Kw
= 1.0 x 10-14
|
| The
K values tell us that HNO2 is the main producer of H+
ions.
|
Ka = ([H+][
NO2-]) / [HNO2] = 4.0 x 10-4
|
|
HNO2
|
H+
|
NO2-
|
|
initial
|
5.00 M
|
0
|
0
|
|
Δ
|
-x
|
+x
|
+x
|
|
at equilibrium
|
5.00 –x
|
x
|
X
|
| Ka
= (x)(x) / (5.00 –x) = 4.0 x 10-4 ≈ x2 / 5.00
|
x
= 4.5 x 10-2 = [H+]
| pH
= -log[H+] = 1.35
|
| The
CN- ions come from dissociation of HCN.
Therefore, the Ka is
|
Ka = [H+][CN-]
/ [HCN] = 6.0 x 10-10
We know [H+]
because we just calculated it. What
is [HCN] at equilibrium? Since
[HCN]0 = 1.00 M and Ka is very small, a negligible amount
(x) of HCN dissociates. At
equilibrium, [HCN] is 1.00 M.
| Rearranging
the equation and solving for [CN-]:
|
[CN-] = (Ka[HCN]) / [H+]
= (6.2 x 10-10)(1.00) / 4.5 x 10-2 = 1.4 x 10-8 M
Does this make sense?
Yes! Since the Ka for HCN is very small, and HCN is the only
source of CN- ions, the amount of CN- dissociated should
also be very small.
How do we know how much a
weak acid dissociates? Percent
dissociation.
| %
dissoc. = (amount dissociated / initial concentration) x 100%
|
| The
more dilute the acid is, the greater is the percent dissociation (i.e., the
percent dissoc. of acetic acid is greater in a 0.10 M solution than in a 1.0
M solution).
|
| As
[HA]0 decreases:
|
- [H+]
decreases (because less amount of acid to start with)
- percent
dissociation increases
Calculating pH of
strong bases:
(ex) Calculate the pH of
a 5.0 x 10-2 M NaOH solution.
| The
major species in solution are Na+, -OH, H20.
|
| Because
NaOH is a strong base, it dissociates completely and
[-OH] = 5.0 x 10-2 M
|
| pOH
= -log[-OH] = 1.30. Therefore
pH = 14.00 – 1.30 = 12.70
|
Does this make sense?
Yes—notice that pH > 7 so
it is a basic solution. Also, [-OH]
> [H+].
*Remember that [H+][-OH]
= Kw.
Calculating pH of weak
bases:
(ex)
Calculate the pH of a 15.0 M solution of NH3 (Kb =
1.8 x 10-5).
| The
major species in solution are NH3 and H20.
|
| By
the K values, NH3 is the main producer of –OH ions,
since Kb »
Kw
|
| NH3(aq)
+ H20(l) ↔
NH4+(aq) + –OH(aq)
|
| ICE
table:
|
|
|
NH3
|
NH4+
|
–OH
|
|
initial
|
15.0 M
|
0
|
0
|
|
Δ
|
-x
|
+x
|
+x
|
|
at equilibrium
|
15.0 -x
|
x
|
X
|
*H20
not included because it is the solvent.
| Kb
= ([NH4+][–OH]) / [NH3] = (x)(x)
/ (15.0 – x) = x2/ 15.0
|
=
1.8 x 10-5
x
= 1.6 x 10-2 = [–OH]
pOH
= -log[–OH] = 1.80
Therefore
pH = 14.00 – 1.80 = 12.20
Polyprotic acids:
| Polyprotic
acids such as H2CO3 and H2SO4
can furnish more than one proton. However,
such acids dissociate ONE proton at a time:
|
| H2CO3(aq)
↔ H+(aq)
+ HCO3-(aq)
Ka = 4.3 x 10-7
|
| HCO3-(aq)
↔ H+(aq) + CO 32-(aq)
Ka = 5.6 x 10-11
|
Typically in each step of
the dissociation, the acid becomes weaker and thus the K values (or dissociation
constants) become smaller. A
smaller K value means that the proton dissociates LESS readily.
As the negative charge on the acid increases, it becomes harder to remove
a proton. Therefore, only the FIRST dissociation step greatly
contributes to [H+].
| What
is unique about sulfuric acid? It
is a STRONG acid in its first dissociation but a weak acid in its second
dissociation. This is true for
concentrations less than 1.0 M. However,
for DILUTE solutions (greater than 1.0 M) of sulfuric acid, the second step
DOES make a significant contribution of + ions, and so the
quadratic equation must be used.
|
Salts:
A salt is an ionic
compound that when dissolved, its ions behave either as an acid or a base.
Salts that produce
NEUTRAL solutions:
| The
salt of a strong acid and a strong base yields a neutral solution because
neither the cation nor the anion have an affinity for H+ and
therefore will have no effect on pH. Why
do strong acids completely dissociate in aqueous solution?
The conjugate base has no affinity for protons.
|
Salts that produce BASIC
solution:
| A
basic solution is formed if the anion of the salt is the conjugate base of a
weak acid. The conjugate base of a WEAK acid has a high affinity
for a proton and will affect pH.
|
| The
cation must have neutral properties (i.e., Na+, K+).
It is neither an acid nor a base.
|
Salts that produce ACIDIC
solution:
| An
acidic solution is formed if the cation of the salt is the conjugate acid of
a weak base.
|
| The
salt has a highly charged metal ion. The
charge polarizes the
O—H bonds in the water molecules and makes those hydrogens more
acidic.
|
How do we know if H—X
will behave as an acid? The strength
and polarity of a bond determines this.
| H—F
is a weak acid because its bond energy is very high.
H—F is reluctant to dissociate in water.
|
| As
the number of oxygen atoms increases, acid strength increases.
HOCl is a weak acid, but HClO4 is a strong acid. Why? Oxygen
is electronegative and is able to draw electrons away from other atoms as
well as the O—H bond. A
proton now dissociates more readily. This type of behavior is also seen in
hydrated metal ions.
|
Acid-Base properties
of oxides:
| Molecule
H—O—X can behave as an acid and the acid strength depends on the
electronegativity of X.
|
| Molecule
H—O—X can also behave as a base.
|
| What
determines which behavior will occur?
|
- If
the O—X bond is strong and covalent, it will remain intact when the
molecule is dissolved in water but the O—H bond will be relatively weak
and break. An acidic solution
results.
- If
the O—X bond is ionic and weak because X has low electronegativity, the
O—X bond will tend to break when dissolved in water.
Ex. NaOH, KOH
| acidic
oxides: a covalent oxide that, when dissolved in water, produces an acidic
solution.
|
| basic
oxides: an ionic oxide (Group
IA and IIA metals) that, when dissolved in water, produces a basic solution.
|