Chapter 3: Acid – Base Equilibria
HCl + KOH  KCl + H2O
acid + base  salt + water
What is an acid?
The Arrhenius concept proposed that acids are substances
that produce hydrogen ions (H+) in aqueous solutions.
The Brönstead – Lowry model describes an acid simply as
a proton donor
Consider the following reaction:
HCl(aq) + H2O(l)  H3O+(aq) + ClHere the proton is transferred from the HCl molecule to the
water molecule to for the hydronium ion (H3O+). A general
form of this equation is:
HA(aq) + H2O(l)  H3O(aq) + A-(aq)
Acid + Base
Conjugate + Conjugate
Acid
Base
If this:
HA(aq) + H2O(l)  H3O(aq) + A-(aq)
is the general form of the acid reaction, then we can
calculate an equilibrium constant (Ka) for the reaction. Ka is
called the acid dissociation constant.
Ka = [H3O+][A-] / [HA] which equals [H+][A-] / [HA].
A strong acid is one that undergoes significant dissociation
and has a very large Ka. A weak acid only partially
dissociates and has a relatively small Ka.
pKa = -logKa
Table 3-1 Dissociation Constants for Acids at 25°C
Acid
Formula
pKa1
pKa2
Hydrochloric
HCl
~-3
Sulfuric
H2SO4
~-3
Nitric
HNO3
0
Oxalic
H 2 C 2 O4
1.2
4.2
Phosphoric
H3PO4
2.15
7.2
Hydrofluoric
HF
3.18
Formic
HCOOH
3.75
Acetic
CH3COOH
4.76
Carbonic
H2CO3
6.35
10.33
Hydrosulfuric
H 2S
7.03
>14
Boric
H3BO3
9.27
>14
Silicic
H4SiO4
9.83
13.17
pKa3
1.99
12.35
As you noticed from the previous slide (table), acids can
contain more than one acidic proton. A diprotic acid
contains 2 acidic protons and undergoes a two-step
dissociation.
i.e.
H2SO4  H+ + HSO4HSO4-  H+ + SO42-
A triprotic acid (such as phosphoric acid H3PO4) has
three acidic protons, and must undergo a three-step
dissociation.
Recall:
pKa = -logKa
Acids that undergo significant dissociation have a
negative pKa, and acids that only partially dissociate
have a positive pKa.
What is a base?
The Arrhenius concept proposed that a base is a substance
that produces OH- ions in aqueous solution.
The Brönstead – Lowry model describes a base simply as a
proton acceptor.
Consider the following reaction:
NH3(aq) + H2O(l)  NH4+(aq) + OHHere the OH- ion is derived from the H2O molecule and the
NH3 molecule acts as a proton acceptor. A general form of
this equation is:
B(aq) + H2O(l)  BH+(aq) + OH-(aq)
Base + Acid
Conjugate + Conjugate
Acid
Base
If this:
B(aq) + H2O(l)  BH+(aq) + OH-(aq)
is the general form of the base reaction, then we can
calculate an equilibrium constant (Kb) for the reaction. Kb
is called the base dissociation constant.
Kb = [BH+][OH-] / [B].
A strong base is one that undergoes essentially
complete dissociation and has a large Kb. A weak base
only partially dissociates and has a relatively small Kb.
pKb = -logKb
Table 3-2
Dissociation Constants for Bases at 25°C
Base (Hydroxide)
Formula
pKb1
pKb2
pKb3
Methylamine
CH3NH2
3.36
Ammonium
NH4(OH)
4.7
Magnesium
Mg(OH)2
8.6
Pyridine
C 5H 5N
8.8
Manganese
Mn(OH)2
9.4
3.4
Ferrous
Fe(OH)2
10.6
4.5
Al, amorphous
Al(OH)3
12.3
10.3
9.0
Al, gibbsite
Al(OH)3
14.8
10.3
9.0
Ferric, amorphous
Fe(OH)3
16.5
10.5
11.8
2.6
As it was with acids, base may also require more than one
step to complete dissociation.
The dissociation of water and pH
H2O  H+ + OHand
Kw = [H+][OH-] / [H2O] which equals [H+][OH-]
Where Kw is the equilibrium constant for water. Kw varies as a
function of temperature. See table 3-3.
Kw(25°C) = 10-14 = [H+][OH-]
*Remember that Kw is a constant therefore if you know the Kw
and the concentration of either H+ or OH-, you can find the
remaining unknown concentration.
For example: If [H+] for a solution is 10-3, what is [OH-] in that
solution?
[OH-] = 10-11
The pH scale was designed to simplify the description of the
acidity of a solution. The pH is often called the hydrogen ion
exponent.
pH = -log[H+]
Therefore, if, for example [H+] = 10-3, then the pH = 3
Pure water at 25°C has a pH of 7. This means that the
number of H+ ions and the number of OH- ions are equal at
10-7 a piece.
Here are a few common
solutions and their
corresponding pH. Note
the concentration of H+
compared to pure H2O.
Why is knowing the specific [H+] important?
Let’s do some examples…
Determine the pH of a solution with a [OH-] of 3 X 10-5.
Recall:
Kw(25°C) = 10-14 = [H+][OH-]
10-14 / 3 X 10-5 = [H+]
[H+] = 3.33 X 10-9.  pH = -log(3.33 X 10-9) = 8.48
Is this solution acidic or basic?
What is the easiest way to determine the
concentration of either H+ or OH-?
Titration.
Ct X Vt = Cs X Vs
Where Ct is the concentration of the
titrant, Vt is the volume of the titrant,
Cs is the concentration (acidity or
alkalinity) of the unknown solution
and Vs is the volume of the
unknown solution.
For example:
30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid.
Determine the concentration of the acid.
NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)
NaOH: V = 30mL , M = 0.10M; HCl: V = 25.0mL, M = ?
NaOH: V = 30 x 10-3L , M = 0.10M; HCl: V = 25.0 x 10-3L, M = ?
# of moles(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles
From the balanced chemical equation find the mole ratio
NaOH:HCl equals: 1:1 Moles(NaOH): Moles(HCl) is 1:1
Therefore, at the equivalence point, there are 3 X 10-3 moles(HCl)
Calculate concentration of HCl: M = n / V
n = 3 x 10-3 mol,
V = 25.0 x 10-3L
M(HCl) = 3 x 10-3 / 25.0 x 10-3 = 0.12M or 0.12 mol L-1
Ok, now what is the pH of this .12M(HCl) solution? –before the
titration.
(We know the pH of the tested solution is 7 as the added NaOH
neutrallized the solition.)
0.12mol(HCL) / L
What is the pKa for HCl at 25°C?
-3
Ka = [H+][Cl-] / [HCl] we know that [H+] = [Cl-] and we know
that the amount a hydrochloric acid left over after dissociation
is 0.12 – [H+]. Also if pKa = -3 then Ka = 1000. If we substitute
x for [H+] we could set up a quadratic formula and solve for x.
Huh?
Time for a quick cleansing breath……
Ok, now, where were we…
If Ka = [x][x] / 0.12 – x, then we can solve for x using the
quadratic formula:
Ka = 1000 = x2 / (0.12 – x)  x2 + 1000x – 120 = 0
= -1000 [(10002 – 4(1)(-120)]1/2 / 2(1)
= either –1000 or 0.12
Since –1000 makes no sense, we can assume x = 0.12
Oh, did I forget to tell you that with HCl, the reaction
goes into, for all our purposes, full dissociation…
So, what is the pH of [H+] = 0.12 moles/L solution?
0.92
Now that you have had a review of the math, then
exercise 3-1 in the text shouldn’t seem so insane!
pH of Natural Waters
Not so natural water….acid mine drainage.
CaSiO3 + 3H2O + 2CO2  Ca2+ + 2(HCO3- ) + H4SiO4
wollastonite + water + carbon dioxide  calcium + bicarbonate + silicic acid
From pore spaces
&/or atmosphere
and:
and:
Ca2+ + 2(HCO3- )  CaCO3 + H2O + CO2
CaCO3 + SiO2  CaSiO3 + CO2
calcite + quartz  wollastonite + carbon dioxide
________
CO2(g) + H2O  H2CO3(aq)
where [H2CO3(aq)] = KCO2PCO2
and
Ka1 = [H+][HCO3-] / KCO2PCO2
H2CO3(aq)  H+ + HCO3-  HCO3-  H+ + CO32This implies that the complete dissociation of carbonic acid is a
two-step process.
Such that:
Ka1 = [H+][HCO3-] / [H2CO3(aq)] and
Ka2 = [H+][CO32-] / [HCO3-]
Further at 25°C: [H2CO3(aq) ] / [HCO3-] = [H+] / Ka1 = [H+] / 10-6.35
And
[HCO3-] / [CO32-] = [H+] / Ka2 = [H+] / 10-10.33
Given: [H2CO3(aq) ] / [HCO3-] = [H+] / Ka1 = [H+] / 10-6.35
[HCO3-] / [CO32-] = [H+] / Ka2 = [H+] / 10-10.33
How would you expect H2CO3 to affect the pH of natural waters?
It depends upon the level of dissociation.
Table 3 -5. Examples of processes that control the CO 2 content and pH of surface and ground waters
Process
Reaction
pH
Temperature change
Increase T, decrease solubility of CO 2 (g)
Increases
Decrease T, increase solubility of CO 2 (g)
Decreases
Photosynthesis
6CO2(g) + 6H2O  C6H12O6 + 6O2(g)
Increases
Respiration
C6H12O6 + 6O2 (g)  6CO2 (g) + 6H2O
Decreases
Anaerobic decay
2CH2O  CH4 (g) + CO2 (g)
Decreases
Denitrification
5CH2O + 4NO3- + 4H+  5CO2 (g) + 2N2 (g) + 7H2O
Increases
Dissolution of carbonate
CaCO3 calcite + 2H+  Ca2+ + H2O + CO2 (g)
Increases
Precipitation of carbonate
Weathering of Al-silicate
Ca2+ + H2O + CO2 (g)  CaCO3 calcite + 2H+
2KAlSi3O8 feldspar + 2CO2 (g) + 11H2O 
Decreases
Increases
minerals
Al2Si2O5(OH)4 kaolinite + 2K+ + 2HCO- + 4H4SiO4 (aq)
Lets consider the effects of water in contact with atmospheric
carbon dioxide.
CO2(aq) + H2O(l)  H2CO3(aq)
Given this relationship between CO2 in solution and water,
would you expect the surface waters of the ocean to be more or
less acidic than the ocean waters out of contact with the
atmosphere?
In general, surface waters are more acidic at the surface
and in coastal waters (for another reason…)
With the CO2—H2O system there are two end-member
cases:
An open system: in equilibrium with atmospheric CO2.
A closed system: isolated from atmospheric CO2.
Let’s recall problem 25 from the problem set 2….
What did we discover about the ionic charges in a solution?
The total positive (cation) charge and total negative (anion)
charge in a solution must be equal.
For example: for the CO2—H2O system, the charge balance
equation is written:
mH+ = mHCO3- + 2mCO32- + mOHWhere m is the molar concentration of each species in solution.
The Carbonic acid – Carbonate System is regulated by not
only the presence of H2CO3, but also the presence of the
minerals calcite or aragonite (CaCO3).
CaCO3calcite  Ca2+ +CO32Ksp = [Ca2+][CO32-]
Final charge balance equation becomes:
mH+ + 2mCa2+ = mHCO3- + 2mCO32- + mOH-
Salts
1. strong acid with a strong base. –makes a neutral solution.
2. weak acid with a strong base. –makes a basic solution.
3. weak base with a strong acid. –makes an acidic solution.
4. weak acid with a weak base.
–makes either a basic or
an acidic solution
depending on the relative
strength of the ions.
**Most, but not all, minerals can be considered to be salts of
weak acids and strong bases.
Amphoteric Hydroxides: hydroxides that can behave as either
and acid or a base. This behavior varies as a function of pH.
For example: Al(OH)3 behaves as a base in an acidic solution
and forms aluminum salts. In a basic solution, it behaves as an
acid H3AlO3 and forms salts with AlO33-. Think Le Châtelier’s
principle.
KA is the equilibrium constant for an amphoteric reaction. A
table of these constants is listed on page 75 of the text.
Acidity and Alkalinity
Acidity is the capacity of water to donate protons. Also
described as the ability of a solution to neutralize bases.
Alkalinity is the capacity of water to accept protons. Also
described as the ability of a solution to neutralize acids.
Nonconservative species: species whose abundances vary as
a function of pH or some other intensive variable (i.e.P & T).
Conservative species: species whose abundances do not vary
as a function of pH or some other intensive variable (i.e.P & T).
Buffers
A buffered solution is a solution that resists changes in pH
when either hydrogen or hydroxyl ions are added to the
solution.
A buffer is a weak acid and its salt or a weak base and its salt.
Let’s go back to our buddy LeChâtelier’s principle and look at
the following reactions.
NaC2H3O2  Na+ + C2H3O2HC2H3O2  H+ + C2H3O21. What would happen if a strong acid were added to a solution
containing both NaC2H3O2 and HC2H3O2?
2. What would happen if a strong base were added to a
solution containing both NaC2H3O2 and HC2H3O2?
Question 1:
NaC2H3O2  Na+ + C2H3O2HC2H3O2  H+ + C2H3O2HCl  H+ + Cl-
When you add additional H+ to this buffered solution, what
happens to the concentration of:
HC2H3O2?
Increases
C2H3O2- ?
Decreases
NaC2H3O2?
Decreases
What ultimately happens to the H+ added to the solution?
It combines with C2H3O2- to make HC2H3O2 until all the is
NaC2H3O2 used up.
Question 2:
NaC2H3O2  Na+ + C2H3O2HC2H3O2  H+ + C2H3O2NaOH  Na+ + OH-
When you add additional OH- to this buffered solution,
what happens to the concentration of:
HC2H3O2?
Decreases
C2H3O2- ?
Increases
NaC2H3O2?
Increases
What ultimately happens to the OH- added to the solution?
It combines with H+ to make H2O. More H+ is created until all
of the HC2H3O2 is used up.
Example 3-13
What would happen to the pH of a 1 liter solution, containing
carbonic acid, if 10-4 mol of H+ ions are added to the solution?
At pH = 7, T = 25°C, [HCO3-] = 10-3 mol/L, pKa1 = 6.35,
therefore, [H2CO3] = 10-3.65 mol/L
H+ + HCO3-  H2CO3(aq)
According to LeChâtelier’s principle the concentration of will
decrease by 10-4 mol/L, and the concentration of will increase
by 10-4 mol/L.
pH = pKa1 + log([HCO3-] / [H2CO3])
= 6.35 + log([10-3 – 10-4] / [10-3.65 + 10-4]) = 6.79
If the pH of the solution would have been 4 without the H2CO3;
and with the H2CO3 the pH is 6.79, then it can be said that the
solution was successfully buffered.
The previous exercise demonstrated the HendersonHasselbalch equation.
Consider the generic: H+ + A-  HA.
pH = -logKa + log ([A-] / [HA])
Remember that buffers are important in the natural
environment because they control the impact of acid or
base additions on natural waters.
This is an acidic lake in Norway.
Norway’s geology consists primarily
of crystallines and metamorphites
and little limestone. The acidity was
caused by acid rain.
Buffering Capacities of Natural Waters
Water is an effective buffer only at very high or very low pH.
Buffering capacity: a measure of the amount of H+ or OHions a solution can absorb without significant change in pH.
Buffering Capacity of
H2CO3
Buffering capacity of
CaCO3 – H2CO3
system
Figure 3-9 summarizes the buffering capacities of waters in
contact with minerals
Does the fate of the Norwegian acid lake seem clearer?
I urge you all to read the case studies in this
chapter. It puts things in perspective.
Next week we will start with Chapter 4:
Oxidation – Reduction Reactions.