Strong Acids/ Bases
• Strong Acids more readily release H+ into
water, they more fully dissociate
– H2SO4  2 H+ + SO4
2-
• Strong Bases more readily release OH-
into water, they more fully dissociate
– NaOH  Na+ + OH-
Strength DOES NOT EQUAL Concentration!

Acid-base Dissociation
• For any acid, describe it’s reaction in water:
– HxA + H2O  x H+ + A- + H2O
– Describe this as an equilibrium expression, K (often
denotes KA or KB for acids or bases…)
• Strength of an acid or base is then related to the
dissociation constant  Big K, strong acid/base!
• pK = -log K  as before, lower pK=stronger
acid/base!
]
[
]
][
[
A
H
H
A
K
x
x



pKx?
• Why were there more than one pK for
those acids and bases??
• H3PO4  H+ + H2PO4
- pK1
• H2PO4
-  H+ + HPO4
2- pK2
• HPO4
1-  H+ + PO4
3- pK3

• LOTS of
reactions are
acid-base rxns
in the
environment!!
• HUGE effect on
solubility due to
this, most other
processes
Geochemical
Relevance?

Dissociation of H2O
• H2O  H+ + OH-
• Keq = [H+][OH-]
• log Keq = -14 = log Kw
• pH = - log [H+]
• pOH = - log [OH-]
• pK = pOH + pH = 14
• If pH =3, pOH = 11  [H+]=10-3, [OH-]=10-11
]
[
]
][
[
2O
H
OH
H
Keq



Definition of pH

pH
• Commonly represented as a range between
0 and 14, and most natural waters are
between pH 4 and 9
• Remember that pH = - log [H+]
– Can pH be negative?
– Of course!  pH -3  [H+]=103 = 1000 molal?
– But what’s gH?? Turns out to be quite small 
0.002 or so…

BUFFERING
• When the pH is held ‘steady’ because of
the presence of a conjugate acid/base
pair, the system is said to be buffered
• In the environment, we must think about
more than just one conjugate acid/base
pairings in solution
• Many different acid/base pairs in solution,
minerals, gases, can act as buffers…

Henderson-Hasselbach Equation:
• When acid or base added to buffered system
with a pH near pK (remember that when pH=pK
HA and A- are equal), the pH will not change
much
• When the pH is further from the pK, additions of
acid or base will change the pH a lot
]
[
]
[
log
HA
A
pK
pH




Buffering example
• Let’s convince ourselves of what buffering
can do…
• Take a base-generating reaction:
– Albite + 2 H2O = 4 OH- + Na+ + Al3+ + 3 SiO2(aq)
– What happens to the pH of a solution containing
100 mM HCO3- which starts at pH 5??
– pK1 for H2CO3 = 6.35

• Think of albite dissolution as titrating OH- into
solution – dissolve 0.05 mol albite = 0.2 mol OH-
• 0.2 mol OH-  pOH = 0.7, pH = 13.3 ??
• What about the buffer??
– Write the pH changes via the Henderson-Hasselbach
equation
• 0.1 mol H2CO3(aq), as the pH increases, some of
this starts turning into HCO3-
• After 12.5 mmoles albite react (50 mmoles OH-):
– pH=6.35+log (HCO3-/H2CO3) = 6.35+log(50/50)
• After 20 mmoles albite react (80 mmoles OH-):
– pH=6.35+log(80/20) = 6.35 + 0.6 = 6.95
]
[
]
[
log
HA
A
pK
pH



Greg Mon Oct 11 2004
0 10 20 30 40 50 60 70 80 90 100
5
5.5
6
6.5
7
7.5
8
8.5
Albite reacted (mmoles)
pH

Bjerrum Plots
• 2 D plots of species activity (y axis) and
pH (x axis)
• Useful to look at how conjugate acid-base
pairs for many different species behave as
pH changes
• At pH=pK the activity of the conjugate acid
and base are equal

pH
0 2 4 6 8 10 12 14
log
a
i
-12
-10
-8
-6
-4
-2
H2S0
HS-
S
2-
H+
OH-
7.0 13.0
Bjerrum plot showing the activities of reduced sulfur species as a
function of pH for a value of total reduced sulfur of 10-3 mol L-1.

pH
0 2 4 6 8 10 12 14
log
a
i
-8
-7
-6
-5
-4
-3
-2
6.35 10.33
H2CO3* HCO3
-
CO3
2-
H+
OH-
Common pH
range in nature
Bjerrum plot showing the activities of inorganic carbon species as a
function of pH for a value of total inorganic carbon of 10-3 mol L-1.
In most natural waters, bicarbonate is the dominant carbonate species!

THE RELATIONSHIP
BETWEEN H2CO3* AND HCO3
-
We can rearrange the expression for K1 to obtain:
This equation shows that, when pH = pK1, the activities of
carbonic acid and bicarbonate are equal.
We can also rearrange the expression for K2 to obtain:
This equation shows that, when pH = pK2, the activities of
bicarbonate and carbonate ion are equal.
*
1
3
2
3
CO
H
HCO
H
a
a
a
K 






3
2
3
2
HCO
CO
H
a
a
a
K

pH
0 2 4 6 8 10 12 14
log
a
i
-8
-7
-6
-5
-4
-3
-2
6.35 10.33
H2CO3* HCO3
-
CO3
2-
H+
OH-
Common pH
range in nature
Bjerrum plot showing the activities of inorganic carbon species as a
function of pH for a value of total inorganic carbon of 10-3 mol L-1.
In most natural waters, bicarbonate is the dominant carbonate species!

THE CO2-H2O SYSTEM - I
Carbonic acid is a weak acid of great importance
in natural waters. The first step in its formation is
the dissolution of CO2(g) in water according to:
CO2(g)  CO2(aq)
At equilibrium we have:
Once in solution, CO2(aq) reacts with water to form
carbonic acid:
CO2(aq) + H2O(l)  H2CO3
0
2
2
2
CO
CO
CO
p
a
K 

THE CO2-H2O SYSTEM - II
In practice, CO2(aq) and H2CO3
0 are combined
and this combination is denoted as H2CO3*. It’s
formation is dictated by the reaction:
CO2(g) + H2O(l)  H2CO3*
For which the equilibrium constant at 25°C is:
Most of the dissolved CO2 is actually present as
CO2(aq); only a small amount is actually present
as true carbonic acid H2CO3
0.
46
.
1
*
10
2
3
2
2



CO
CO
H
CO
p
a
K

THE CO2-H2O SYSTEM - III
Carbonic acid (H2CO3*) is a weak acid that dissociates
according to:
H2CO3*  HCO3
- + H+
For which the dissociation constant at 25°C and 1 bar is:
Bicarbonate then dissociates according to:
HCO3
-  CO3
2- + H+
35
.
6
*
1 10
3
2
3 




CO
H
H
HCO
a
a
a
K
33
.
10
2 10
3
2
3 





HCO
H
CO
a
a
a
K

pH
0 2 4 6 8 10 12 14
log
a
i
-8
-7
-6
-5
-4
-3
-2
6.35 10.33
H2CO3* HCO3
-
CO3
2-
H+
OH-
Common pH
range in nature
Bjerrum plot showing the activities of inorganic carbon species as a
function of pH for a value of total inorganic carbon of 10-3 mol L-1.
In most natural waters, bicarbonate is the dominant carbonate species!

SPECIATION IN OPEN CO2-H2O
SYSTEMS - I
• In an open system, the system is in contact with its
surroundings and components such as CO2 can migrate
in and out of the system. Therefore, the total carbonate
concentration will not be constant.
• Let us consider a natural water open to the atmosphere,
for which pCO2
= 10-3.5 atm. We can calculate the
concentration of H2CO3* directly from KCO2
:
Note that M H2CO3* is independent of pH!
2
3
2
2
*
CO
CO
H
CO
p
M
K  2
2
3
2 * CO
CO
CO
H K
p
M 
2
2
3
2
log
log
log * CO
CO
CO
H K
p
M 


SPECIATION IN OPEN CO2-
H2O SYSTEMS - II
• The concentration of HCO3
- as a function of pH is next
calculated from K1:
but we have already calculated M H2CO3*:
so 2
2
3
2 * CO
CO
CO
H K
p
M 
*
1
3
2
3
CO
H
H
HCO
M
a
M
K




 
H
CO
H
HCO
a
M
K
M
*
1 3
2
3

 
H
CO
CO
HCO
a
p
K
K
M 2
2
3
1
  pH
p
K
K
M CO
CO
HCO



2
2
3
1
log
log

SPECIATION IN OPEN CO2-
H2O SYSTEMS - III
• The concentration of CO3
2- as a function of pH is next
calculated from K2:
but we have already calculated M HCO3
- so:
and

 
H
CO
CO
HCO
a
p
K
K
M 2
2
3
1




3
2
3
2
HCO
H
CO
M
a
M
K


 
H
HCO
CO
a
M
K
M 3
2
3
2
2
1
2 2
2
2
3

 
H
CO
CO
CO
a
p
K
K
K
M
  pH
p
K
K
K
M CO
CO
CO
2
log
log 2
2
2
3
1
2 



SPECIATION IN OPEN CO2-
H2O SYSTEMS - IV
• The total concentration of carbonate CT is
obtained by summing:

 

 2
3
3
3
2 * CO
HCO
CO
H
T M
M
M
C
2
2
1
1 2
2
2
2
2
2





H
CO
CO
H
CO
CO
CO
CO
T
a
K
p
K
K
a
K
p
K
K
p
C





















2
2
1
1
1
log
log 2
2
H
H
CO
CO
T
a
K
K
a
K
K
p
C

pH
2 3 4 5 6 7 8 9 10 11 12
log
concentration
(molar)
-8
-6
-4
-2
0
CT
H+
OH-
H2CO3*
HCO3
-
CO3
2-
pK1 pK2
Plot of log concentrations of inorganic carbon species H+ and OH-,
for open-system conditions with a fixed pCO2
= 10-3.5 atm.

pH
2 3 4 5 6 7 8 9 10 11 12
log
concentration
(molar)
-8
-6
-4
-2
0
CT
H+
OH-
H2CO3*
HCO3
-
CO3
2-
pK1 pK2
Plot of log concentrations of inorganic carbon species H+ and OH-,
for open-system conditions with a fixed pCO2
= 10-2.0 atm.

Calcite Solubility?
• CaCO3 -> Ca2+ + CO3
2-
• Log K=8.48
• Ca2+ in Ocean = 0.0106 m