Acid-Base Chemistry
Extremely important in the environment
Examples
• Acid rain
• Biological systems
- cells buffer (i.e., stabilize) their internal pH
- biological processes in nature and in engineered processes need
stable pH
• Air-water interactions
CO2 in atmosphere
↑↓
CO2 in surface waters
↑↓ + H2O
H2CO3
↑↓ - H+
HCO3- bicarbonate ion
↑↓ - H+
CO32- carbonate ion
Thus, adding CO2 lowers the pH (makes system more acidic)
• Acid mine drainage – Coal contains reduced sulfur and acid mine
drainage has long been an environmental problem. Much of this
sulfur is contained in pyrite, FeS2, a common mineral. On exposure to
the air it oxidizes giving ferric, sulfate and hydrogen ions:
FeS2 + 15/4 O2 + 1/2 H2O ↔ Fe3+ + 2 SO42- + 2 H+
The ferric ions undergo a hydrolysis reaction with water liberating more
protons (thus making the system even more acidic) and precipitating ferric
hydroxide (“rust”) on the stream bottom and banks.
Fe3+ + 3 H2O ↔ Fe(OH)3 (s) + 3 H+
• A treatment question – Most metals (e.g., Fe, Cu, Zn etc) come from
open-pit mines using large equipment for processing. After removal
of the metals from the finely crushed rock the residual “tailings” are
disposed in piles or disposal areas. The residual sulfide will, over
time, oxidize causing the material to become acidic and thus leaching
metals into surface and ground water. To prevent this potential
contamination, limestone (CaCO3) is added to neutralize the released
hydrogen ions. Can you calculate the amount of limestone that would
have to be added if, for instance, the waste material had been found to
contain 4% S?
Consider the reaction
H2O ↔ H+ + OHBased on the definition of K
K=
{H }{OH }
+
{H 2 O}
-
1000 g
L = 55.5 mol/L
⇒ concentration of H2O:
18 g/mol
~
But
Define {H2O} = 1
In a dilute system { } = [ ]
Kw = [H+] [OH-]
at 25°C, Kw = 1.0 × 10-14
[H+] [OH-] = 10-14
When are they equal? [H+] = [OH-] = (10-14)1/2 = 10-7
But in an acidified lake
[H+] = 10-4; [OH-] = 10-14/[H+] = 10-14/10-4 = 10-10
Range of concentrations is orders of magnitude.
So – define “p” = “-log10”
pH = -log10 [H+]
pOH = -log10 [OH-]
So we don’t always have to deal with negative values - if we just use logs
Note the – sign: as [H+] increases, pH decreases
pH 7
pH 4
[H+] = 10-7
[H+] = 10-4
[OH-] = 10-7
[OH-] = 10-10
Convert the equilibrium expression
-log10 ([H+] [OH-]) = -log10 [H+] + -log10 [OH-] = -log10 Kw
= pKw = -log10 10-14 = 14
and
-log10 [H+] + -log10 [OH-] = 14
so
pH + pOH =14
It is useful to consider log – log plots of concentration
0
+
log [H ]
2
10
4
- log C = pC
-
log [OH ]
-4
6
pH 7
8
+
-
[H ] = [OH ]
10
10
-10
12
14
0
2
4
6
8
10
12
14
pH --- "M aste r Variable "
We can treat acids and bases in a similar manner
Definitions
Acid can donate H+ (proton)
Base can accept H+
Generic acid “HA”
HA ↔ H+ + A- (dissociation)
A strong acid (HCl – hydrochloric, HNO3 – nitric, H2SO4 – sulfuric, but only
first proton) is one for which the dissociation is complete in water
If the dissociation is <100%, it is a weak acid
Generic base “BOH”
BOH ↔ B+ + OH- If the dissociation is <100%, it is a weak base
Strong bases dissociate completely in water (NaOH – sodium hydroxide,
KOH – potassium hydroxide)
(How does this involve H+?
generate 1 OH-, this accepts 1 H+
OH- + H+ → H2O
net: BOH + H+ → B+ + H2O
The extent of these reactions may be determined using dissociation
constants.
We can so the same thing we did for H2O:
HA ↔ H+ + AH+ AKa =
= " acidity constant"
[HA]
[ ][ ]
(eqn 1)
Example acetic acid CH3COOH = “HA”
and acetate ion CH3COO- = “A-”
pKa = 4.7, Ka = 10-4.7 = 2.0 × 10-5
Put 10-4 mol CH3COOH in 1 L solution. What is the pH?
Is this a strong acid?
From equation 1: 2 × 10
[H ][CH COO ]
=
+
−5
3
-
[CH 3COOH ]
2 × 10 −5 [CH 3 COOH ]
+
[
]
and H =
[CH COO ]
-
3
we could solve for
[CH 3COOH ] but we would still need another equation.
[CH COO ]
3
-
What else do we know? 10-4 M!
[CH3COOH] + [CH3COO-] = 10-4 = CT
(eqn 2)
Combining eqn 1 and eqn 2 gives
[HA] =
CT
and
[A ] =
-
[H ]
K + [H ]
+
+
a
Ka
(eqn 3)
(eqn 4)
CT
Ka + H+
We will use these instead of solving directly for [H+]. The two equations
generate a useful graph (see Appendix p 9 for mathematical details).
[ ]
0
-log C = pC
2
log C
T
CH COO3
4
6
CH COOH
3
8
10
0
2
4 pKa
6
pH
8
10
12
14
This graph shows the concentration of the CH3COOH (protonated) species
and CH3COO- (dissociated) species at any pH. Note that at a low pH, the
protonated species is predominant, but at higher pH, the dissociated species
predominates.
Notice that when the pH equals the pKa, the CH3COOH and CH3COOconcentrations are equal.
Because a strong acid completely dissociates to A-, the pKa would have to be
very low.
At any pH below 4.7, acetic acid is less than 50% dissociated. It is a weak
acid. A strong acid would have to have a very small pKa. Actually, their
values are negative!
So now we know that acetic acid is a weak acid. But we never figured out
where we were on the log C – pH diagram (i.e., the pH after adding 10-4 M
acetic acid to the H2O).
We need another equation. There are two that can be used, either a charge
balance or a proton condition. In this course we’ll just use the charge
balance which simply says that the total negative charge in the system must
equal the overall positive chare (“electroneutrality”).
We add the concentrations of all anions on one side of the equation, cations
on the other. In our example
[H+] = [H3CCOO-] + [OH-]
We finish our problem by combining our acetic acid graph with our previous
graph for [H+] and [OH-], and look for a pH where the [H+], [H3CCOO-], and
[OH-] lines satisfy the above condition.
0
H+
2
CH3COO
-log C = pC
4
-
OH
-
6
8
CH COOH
3
10
12
14
0
2
4
6
pH
8
10
12
14
You could have plotted -log C as is frequently done. The graph would have
been identical except for the sign of the values on the y-axis. If you draw
the graph by hand, you must use graph paper. Preferable is 10 × 10 to the
centimeter (10 × 10 to the inch may also be used).
The solution to the charge balance equation is found by inspection being
careful to consider only the lines of species that are represented in the
electroneutrality equation and remembering that the graph shows
logarithmic, not linear, values. The charge balance is fulfilled when pH ≅
4.5, where [CH3COO-] ≅ 10-4.5, [H+] ≅ 10-4.5, and [OH-] is so small that it can
be neglected (~10-9.5). So the pH of the acetic acid solution will be 4.5.
So the overall approach is this:
1. Get the pKa. It’s –log10 Ka if you have Ka. If you have Kb or pKb (the
base dissociation constant), use Kb = 10-14/ Ka, or p Kb = 14 – p Ka.
2. Get CT and –log10CT (pCT). CT is the sum of [HA] and [A-] already in
the solution plus the amount added. It must be in moles/L.
3. Draw the graph in the following steps:
a. Draw axes as shown in the example.
b. Draw a vertical line at pH = pKa.
c. Draw a horizontal line at –log10CT.
d. Locate the intersection from (b) and (c) and from it, draw lines
downward in each direction at 45°. Also locate a point (x) 0.3
units below this intersection.
e. Draw a line that connects the lower portion of each 45° with
point x and then over onto the horizontal CT line. The one
that’s higher at low pH is protonate; the one that is higher at
high pH is dissociated.
f. Now draw the H+ and OH- lines. These are 45° lines that cross
at pH = 7
4. Write the charge balance (electroneutrality) equation*. Find the
location on your graph where it is satisfied. Usually only one
negatively charged species and one positively charged species will be
significant, and the others will be low enough to neglect.
5. Once you have this point, it gives you the pH and the concentration
(-log10 concentrations) for all species.
We will stop with this application. But this type of graph can be used to
analyze titrations, precipitation, equilibria with the atmosphere or other gas
phases, and mineral equilibria.
*
Any species with more than one + or – charge must include this coefficient
in the equation. Remember, it is moles of charge not moles of compound
that is being considered. For example, adding Ca(OH)2 to H2O, we would
write
2 [Ca2+] + [H+] = [OH-]
Appendix: How the pC-pH graph is obtained from the equilibrium
equations (eqn. 3 and eqn. 4).
1. First consider the low pH region, where pH<< pKa.
Then –log10[H+] << -log10 Ka, and [H+] >> Ka. Equation 4 was
CT K a
A- =
Ka + H+
The denominator can now be approximated by [H+], so taking logs
gives
log10 [A-] = log10 CT + log10 Ka – log [H+]
p[A-] = pCT + log10 Ka + pH
which is the 45° line upward*. Also using [H+] >> Ka, eqn. 3 gives
p[HA] = pCT, the horizontal line.
[ ]
[ ]
2. Now at high pH, we have [H+] << Ka. Equation 3 gives
p[HA] = pCT + pH - pKa, thedownward* sloping 45° line. Equation
4 gives [A-] = CT, horizontal line.
3. At the intersection, where the [HA] and [A-] lines cross, [HA] = [A-]
and equation 1 then gives Ka = [H+] or the pH where the lines cross
must be where pH = pKa.
Then, since [A-] = [HA], both of these must be CT/2.
Then p[A-] = p[HA] = -log10(CT/2)
= -log10CT – log (1/2)
= -log10CT + 0.3
So, the lines cross at 0.3 units below* the CT line.
*
If “upward” and “downward” seem inconsistent, note that the pC axis
increases going downward.