Acid-base equilibria
Chemistry 321, Summer 2014
Goals of this lecture
•
•
•
•
Quantify acids and bases as analytes
Measure [H+] in solution  pH
Control/stabilize [H+] in solution  buffers
Predict/control the form of acid or base in
solution  ligands
Brønsted-Lowry definition of an acid
A substance that ionizes in water to give hydrogen
ions and a conjugate base
acid
(proton donor)
or, more simply:
hydronium ion
conjugate
base
where Ka is the
acid dissociation
constant
The acid dissociation constant
• [H+] = [H3O+]; we will use H+ consistently instead of H3O+
• In the expression, the activities (e.g., aH+) are replaced by
concentrations (e.g., [H+]) because the activity coefficients for all
species are basically = 1
• For a strong acid, Ka is large (often ∞ for acids that
complete dissociate into ions)
• For a weak acid, Ka is small and can be found in tables
Brønsted-Lowry definition of a base
A substance that ionizes in water to give hydroxide
ions and a conjugate acid
• For a strong base, Kb is large
• For a weak base, Kb is small
Measuring [H+]: the pH scale
Introduced in 1909 by Soren Sorensen at the Carlsberg
Laboratory as a means of converting measured electrical
potential of a solution into a hydrogen ion concentration
(the technology of early pH meters), pH (potenz Hydrogen)
is defined by the equation:
pH = – log10 [H+]
The pH is based on the self-ionization of
water
The equilibrium expression K can be expressed in terms of
species activities, or, assuming the species’ activities are close
to one, in terms of concentrations:
K = aH O+ aOH –
3
a
2
H 2O
+
–
[H 3O ] [OH ]
= 2
[H 2O]
The pH of pure water
Since the [H2O] is constant in most (dilute) solutions, that
term can be multiplied by K and the resulting constant is
the self-ionization constant of water.
+
–
-14
Kw = [H3O ] [OH ] = 1 ´ 10 (at 25°C)
So, for pure water, [H+] = [OH–] = 1.0 × 10–7 M,
thus pH = 7.00 (2 significant figures).
In fact, is the pH of pure water at 25°C really 7.00?
The pH of pure water
Since the [H2O] is constant in most (dilute) solutions, that
term can be multiplied by K and the resulting constant is
the self-ionization constant of water.
+
–
-14
Kw = [H3O ] [OH ] = 1 ´ 10 (at 25°C)
So, for pure water, [H+] = [OH–] = 1.0 × 10–7 M,
thus pH = 7.00 (2 significant figures).
In fact, is the pH of pure water at 25°C really 7.00? If the
water has not been degassed, then CO2 in the
atmosphere will set up a bicarbonate equilibrium and
decrease the pH (i.e., acidify the water).
Strong acids dissociate completely
Therefore, [HA] (written on the outside of the bottle) = [H+]
i
1M
0M
0M
c
–1 M
+1 M
+1 M
e
0M
1M
1M
Thus, the pH of a 1 M HCl solution is – log (1) = 0.0
Strong bases dissociate completely
Thus, a 1 M solution of NaOH will produce 1 M OH– ions.
Using the water self-ionization equation, Kw = [H+] [OH–], the
[H+] = 1 × 10–14 M, which is a pH of 14.0.
Another way to get this is to calculate the pOH = - log10 [OH–],
then use the expression pH + pOH = 14 to calculate the pH. In
this case, pOH = 0.0, so pH = 14.0.
Weak acids and weak bases do not
dissociate completely
In this case, you
cannot assume the
concentration
written on the bottle
will be the [H+].
Instead, [HA] + [A–] =
concentration on
bottle.
A weak acid example
Consider acetic acid (HAc), which dissociates into a hydrogen ion
and an acetate ion (Ac–). The Ka for this dissociation is 1.8 × 10–5
(at 25°C). A typical concern is the pH of a particular acetic acid
solution: “What is the pH of a 0.20 M acetic acid solution?”
Set up the equilibrium reaction equation and the ICE table:
init
0.20 M
0M
0M
change – x M
+xM
+xM
equil
xM
xM
0.20 – x M
The calculation
[HCOO – ] [H + ]
(x) (x)
x2
-5
K a = = » = 1.8 ´ 10
[HCOOH ]
0.20 - x 0.20
The approximation 0.20 – x ≈ 0.20 is useful because it allows
the solving for x without using the quadratic formula. It is
accurate because the acid will not dissociate to a large extent;
in other words, [H+] = [HCOO–] << 0.20 M.
Thus, x = 0.0019 M, which is indeed significantly less
than 0.20 M. Since x = [H+], then pH = 2.72.
In comparison, a strong acid of 0.20 M concentration has a pH
of 0.70.
Soluble salts of weak acid conjugate base ions
Consider the soluble salt sodium chloride (NaCl). It dissolves
completely into sodium ions and chloride ions. The chloride ion is
the conjugate base of hydrochloric acid (a strong acid). Sodium
chloride dissolved in water does not change the pH of the water.
NaCl(s)  Na+ (aq) + Cl– (aq)
Consider the soluble salt sodium acetate (CH3COONa). It dissolves
completely (up to its solubility, which is quite high) into sodium
ions and acetate ions. The acetate ion is the conjugate base of
acetic acid (a weak acid). Sodium acetate dissolved in water
results in an alkaline (pH>7) solution.
CH3COONa (s)  Na+ (aq) + CH3COO– (aq)
How are the two scenarios different?
The pH of a salt solution
The acetate ion generated in the dissolution of sodium acetate
will then participate in the equilibrium above because there is
always a hydrogen ion present in a water molecule. Thus, the
acetate ion will react with the hydrogen ion to form undissociated
acetic acid, effectively removing H+ ions from solution, leaving
hydroxide ions and making the solution alkaline.
In the case of sodium chloride, the generation of chloride ion
will not induce any undissociated hydrochloric acid to form.
Thus, there is no effect on the hydrogen ion concentration so
the pH remains the same.
The pH of a salt solution
What is the pH of a 0.10 M sodium acetate solution?
First, since sodium acetate is soluble, [CH3COO–] = 0.10 M.
Then, write the acetate equilibrium reaction:
CH3COO– (aq) + H2O (l)
CH3COOH (aq) + OH– (aq)
Note that this will have a slightly different equilibrium constant:
you generated a hydroxide ion, so the equilibrium expression
will generate the base equilibrium constant, Kb.
[CH 3COOH ] [OH – ]
K b = [CH 3COO – ]
Next, set up the ICE table to show how the acetate
equilibrium affects that concentration of acetate ion.
–x
+x
+x
0.10 – x
x
x
[CH 3COOH ] [OH – ] K w
1 ´ 10 –14
–10
Kb = = = = 5.6 ´ 10
[CH 3COO – ]
K a 1.8 ´ 10 –5
Make the approximation that x << 0.10
2
x
K b = 5.6 ´ 10 –10 = 0.10
so x = [OH–] = 7.5 × 10–6 M and
pOH = 5.13, so pH = 8.87.
Buffers allow solutions to maintain a
constant pH upon addition of acid or base
Buffers are solutions that contain a weak acid and the salt of
its conjugate base; for instance, an acetate buffer contains
acetic acid and (usually) sodium acetate, which is highly
soluble and thus is an excellent source of acetate ions.
The buffer maintains its pH because of LeChatelier’s
principle: when an acid is added, the equilibrium above shifts
to the left (generating undissociated acid). When a base is
added, it will react with the H+, and the equilibrium will shift
to the right to restore H+ concentration.
To make a buffer, use the Henderson-Hasselbalch equation
Two physicians, Lawrence Henderson (1908) and Karl Hasselbalch
(1916), studying metabolic acidosis, developed the equation that
allowed the prediction of a buffer pH relying only on species
concentrations and the pKa of the weak acid involved.
–
[A ]0
pH = pK a + log [HA]0
where HA, the weak acid, dissociates into A–. [HA]0 is the
initial concentration of HA and [A–]0 is the initial concentration
of A–.
A buffer example
To make an acetate buffer of pH 5.00, use the equation to
determine the initial concentrations of acetic acid and acetate
ion.
–
[CH 3COO ]0
5.00 = 4.74 + log [CH 3COOH ]0
so
[CH 3COO – ]0
= 1.82
[CH 3COOH ]0
so if the acetic acid is 0.10 M, then the concentration of acetate
ion must be 0.18 M.
Choosing the right weak acid for making a buffer
Could other weak acids be used to make a pH 5.00 buffer?
Yes, as long as they have a pKa near 5.00. Why? Calculate [A–]0/[HA]0.
acid
acetic
salicylic
Ka
1.8 × 10-5
1.0 × 10-3
pKa
4.7
3.0
[A-]0/[HA]0
2
100
Note that salicylic acid would be a bad choice to use for the
buffer since there’s hardly any undissociated salicylic acid at pH
5.00 – this means that this solution will not buffer against the
addition of base very well.
Graphic behavior of weak acids
Note a couple of things:
• The equivalence
point pH is not 7 (it is
higher than 7)
• Near the halfequivalence point, the
pH is not affected by
the addition of
significant amounts of
titrant (this is the
buffering behavior of a
weak acid solution)
Challenge problem
Formic acid (HCOOH) is a weak acid with a Ka = 1.8
× 10–4, and dissociates into H+ and the formate ion,
HCOO–. The formate ion can also be introduced to a
solution by dissolving a soluble salt like sodium
formate, NaHCOO.
What is the pH of a solution that is labeled 0.20 M
NaHCOO? Note that you must take into account the
formic acid equilibrium in your calculation.