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Chapter 8 Monoprotic Acid-Base Equilibria
8-1 Strong Acids and Bases
HBr is a strong acid, so the reaction goes to completion.
HBr + H2O  H3O3+ + BrpH = -log[H+] = -log(0.10) = 1.00
Box 8-1 Concentrated HNO3 Is Only Slightly Dissociated2
The figure shows a Raman spectrum of solutions of nitric acid of increasing
concentration.
KOH is a strong base (completely dissociated), so [OH-] = 0.10 M.
pH = -log[H+] = 13.00
Relation between pH
and pOH:
pH + pOH = -logKW = 14.00 at 250C
(9-1)
The Dilemma
[H+] = KW/(1.0 X 10-8) = 1.0 X 10-6M  pH = 6.00
The Cure
KW
Step 1 Pertinent reactions. The only one is H2O = H+ + OH-.
Step 2 Charge balance. The species in solution are K+, OH-, and H+. So,
[K+] + [H+] = [OH-]
(9-2)
Step 3 Mass balance. All K+ comes from the KOH, so [K+] = 1.0 X 10-8 M.
Step 4 Equilibrium constant expression. KW = [H+][OH-] = 1.0 X 10-14.
Step 5 Count. There are three equations and three unknowns ([H+], [OH-],
[K+]), so we have enough information to solve the problem.
Step 6 Solve. Because we are seeking the pH, let’s set [H+] = x. Writing
[K+] = 1.0 X 10-8 M in Equation 9-2, we get
[OH-] = [K+] + [H+] = 1.0 X 10-8 + x
[H+] = 9.6 X 10-8 M  pH = -log[H+] = 7.02
1.
When the concentration is “high” (≥10-6M), pH is calculated by just
considering the added H+ or OH-. That is, the pH of 10-5.00M KOH is
9.00.
2.
When the concentration is “low” (≤10-8M), the pH is 7.00. We have
not added enough acid or base to change the pH of the water itself.
3.
At intermediate concentrations of 10-6 to 10-8M, the effects of water
ionization and the added acid or base are comparable. Only in this
region is a systematic equilibrium calculation necessary.
8-2 Weak Acids and Bases
Let’s review the meaning of the acid dissociation constant, Ka, for the acid
HA:
Weak-acid
equilibrium:
A weak acid is one that is not completely dissociated. That is, Reaction 9-3
does not go to completion. For a base, B, the base hydrolysis constant, Kb,
is defined by the reaction
Weak-base
equilibrium:
A weak base is one for which Reaction 9-4 does not go to completion.
pK is the negative logarithm of an equilibrium constant:
pKW = -log KW
pKa = -log Ka
pKb = -log Kb
The acid HA and its corresponding base, A-, are said to be a conjugate
acid-base pair, because they are related by the gain or loss of a proton.
Relation between Ka and
Kb for conjugate pair:
Ka · Kb = Kw
(9-5)
Weak Is Conjugate to Weak
The conjugate base of a weak acid is a weak base. The conjugate acid of a
weak base is a weak acid.
Using Appendix G
Appendix G lists acid dissociation constants. Each compound is shown in its
fully protonated form.
Names refer to neutral molecules.
8-3 Weak-Acid Equilibria
A Typical Weak-Acid Problem
Reactions:
Charge balance:
Mass balance:
Equilibrium expressions:
x2 + (1.07 X 10-3) x – 5.35 X 10-5 = 0
x = 6.80 X 10-3 (negative root rejected)
[H+] = [A-] = x = 6.80 X 10-3 M
[HA] = F – x = 0.0432 M
pH = -logx = 2.17
[A-] (from HA dissociation) = 6.8 X 10-3 M
[H+] from HA dissociation = 6.8 X 10-3 M
[OH-] (from H2O dissociation) = 1.5 X 10-12 M
[H+] from H2O dissociation = 1.5 X 10-12 M
Fraction of Dissociation
The fraction of dissociation, α, is defined as the fraction of the acid in the form
A-:
Fraction of dissociation
of an acid:
Weak electrolytes (compounds that are only partially dissociated) dissociate
more as they are diluted.
The Essence of a Weak-Acid Problem
Equation for weak acids:
Box 8-2 Dyeing Fabrics and the Fraction of Dissociation
Dyes are colored molecules that can form covalent bonds to fabric. For
example, Procion Brilliant Blue M-R is a dye with a blue chromophore (the
colored part) attached to a reactive dichlorotriazine ring:
Ka≈10-15
CelluloseㅡCH2OH =
cellulose ㅡ CH2O- + H+
RO-
ROH
Fraction of dissociation
fraction of dissociation
A handy tip: Equation 9-11 can always be solved with the quadratic formula.
However, an easier method worth trying first is to neglect x in the
denominator.
8-4 Weak-Base Equilibria
Kb
B + H2O = BH+ + OH[B] = F – [BH+] = F-x
Equation for weak base:
A Typical Weak-Base Problem
B + H2O = BH+ + OH0.0372-x
x
x
[H+] = KW/[OH-] = 1.0 X 10-14/3.1 X 10-4 = 3.2 X 10-11
pH = -log[H+] = 10.49
What fraction of cocaine has reacted with water? We can formulate α for a
base, called the fraction of association:
Fraction of association
of a base:
Conjugate Acids and Bases-Revisited
Earlier, we noted that the conjugate base of a weak acid is a weak base,
and the conjugate acid of a weak base is a weak acid.
Isomer of hydroxyl benzoic acid
ortho
para
Ka
Kb(= Kw/Ka)
1.07 X 10-3
2.9 X 10-5
9.3 X 10-12
3.5 X 10-10
pH of 0.050 0 M o-hydroxybenzoate = 7.83
pH of 0.050 0 M p-hydroxybenzoate = 8.62
8-5 Buffers
A buffered solution resists changes in pH when acids or bases are added or
when dilution occurs. The buffer is a mixture of an acid and its conjugate
base.
Mixing a Weak Acid and Its Conjugate Base
If you mix A moles of a weak acid with B moles of its conjugate base, the
moles of acid remain close to A and the moles of base remain close to B.
HA = H+ + A0.10-x
x
x
Fraction of dissociation
pKa = 4.00
A- + H2O = HA + OH0.10-x
x
x
pKb = 10.00
Fraction of dissociation
HA dissociates very little, and adding extra A- to the solution will make the
HA dissociate even less. Similarly, A- does not react very much with water,
and adding extra HA makes A- react even less.
Henderson-Hasselbalch Equation
The central equation for buffers is the Henderson-Hasselbalch equation,
which is merely a rearranged form of the Ka equilibrium expression.
Henderson-Hasselbalch
equation for an acid:
Henderson-Hasselbalch
equation for a base:
Challenge
Show that, when activities are included, the Henderson-Hasselbalch
equation is
Properties of the Henderson-Hasselbalch Equation
Regardless of how complex a solution may be, whenever pH = pKa, [A-]
must equal [HA]. This relation is true because all equilibria must be
satisfied simultaneously in any solution at equilibrium. If there are 10
different acids and bases in the solution, the 10 forms of Equation 9-16
must all give the same pH, because there can be only one concentration
of H+ in a solution.
A Buffer in Action
Notice that the volume of solution is irrelevant, because volume cancels in
the numerator and denominator of the log term:
moles of B/L of solution
moles of BH+/L of solution
moles of B
moles of BH+
Box 8-3 Strong Plus Weak Reacts Completely
B + H+ = BH+
Ka(for BH+)
Weak Strong
base acid
OH- + HA = A- + H2O
Strong
base
Weak
acid
Ka(for HA)
H+ + OH- = H2O
Strong
acid
Strong
base
Kb(for A-)
Demonstration 9-2 How Buffers Work
Percentage of reaction
completed
Calculated
pH
Procedure: All solutions should be fresh.
Prepare a solution of formaldehyde by
diluting 9 mL of 37 wt% formaldehyde to
100 mL. Dissolve 1.5 g of NaHSO310 and
0.18 g of Na2SO3 in 400 mL of water, and
add ~1 mL of phenolphthalein solution
(Table 11-4). Add 23 mL of formaldehyde
solution to the well-stirred buffer solution
to initiate the clock reaction. The time of
reaction can be adjusted by changing the
temperature, concentrations, or volume.
We see that the pH of a buffer does not change very much when a limited
amount of strong acid or base is added.
But why does a buffer resist changes in pH? It does so because the strong
acid or base is consumed by B or BH+.
Preparing a Buffer in Real Life!
1.
Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker
containing about 800 mL of water.
2.
Place a calibrated pH electrode in the solution and monitor the pH.
3.
Add NaOH solution until the pH is exactly 7.60.
4.
Transfer the solution to a volumetric flask and wash the beaker a few
times. Add the washing to the volumetric flask.
5.
Dilute to the mark and mix.
Buffer Capacity14
Buffer capacity, β, is a measure of how well a solution resists changes in pH
when strong acid or base is added. Buffer capacity is defined as
Buffer capacity:
Buffer capacity reaches a maximum when pH = pKa.
That is, a buffer is most effective in resisting changes in pH when pH = pKa
(that is, when [HA] = [A-]).
In choosing a buffer, seek one whose pKa is as close as possible to the
desired pH.
The useful pH range of a buffer is usually considered to be pKa ±1 pH unit.
Buffer pH depends on Ionic Strength and Temperature
The correct Henderson-Hasselbalch equation, 9-18, includes activity
coefficients.
When What You Mix Is Not What You Get
Mass balance:
Charge balance:
FHA + FA- = [HA] + [A-]
[Na+] + [H+] = [OH-] + [A-]
[HA] = FHA – [H+] + [OH-]
[A-] = FA- + [H+] – [OH-]
(9-20)
(9-21)
The Henderson-Hasselbalch equation (with activity coefficients) is always
true, because it is just a rearrangement of the Ka equilibrium expression.