Download Fractional Ionization of a Monoprotic Weak Acid

Acid-Base Equilibrium (Mono- and
Polyprotic) [Chapter 10,11]
THE TRUTH, THE WHOLE TRUTH AND
NOTHING BUT THE TRUTH.
Strong Acids and Bases
pH + pOH = - log Kw = pKw = 14.00
for Kw = 1.0x10-14
pH + pOH = - log Kw = pKw = 13.996
for Kw = 1.01x10-14
at 25oC
EXAMPLE: What is the pH of
a 1.0x10-8M solution of HCl?
MHCl = 1.0 x 10-8 M
[H+]HCl = 1.0 x 10-8 M
pH = 8.00
OH NO! WHAT
SHOULD WE DO !
WRONG!!!!!!!!
HCl is an acid!
Acids have pH less than 7!
What's wrong?
We ignored the fact that water is also an
ACID!
[H+]total = [H+]water + [H+]HCl
near pH 7, the contribution of [H+] from
water becomes the dominate source for
the [H+] in the solution
[H+]total = [H+]water + [H+]HCl
[H+]total = [H+]water + 1.0 x 10-8M
let [H+]water = x = [OH-]
Kw = [H+]total[OH-] = 1.01 X 10-14 at 25oC
[H+]total[OH-] = 1.01 X 10-14
([H+]water + 1.0 x 10-8 M) * [OH-]
=
1.01 X 10-14 ( x + 1.0 x 10-8M) * x = 1.01 X 10-14
( x + 1.0 x 10-8M) * x = 1.01 X 10-14M2
by quadratic
x = 9.51 x 10-8M
[H+]total = 9.51 x 10-8M + 1.0 x 10-8M
[H+]total = 10.5 x 10-8M = 1.05 x 10-7M
pH = 6.98
Fractional Ionization of a
Monoprotic Weak Acid
Weak acids are those that are not leveled, or
completely ionized in the solvent.
Acetic acid is a relatively weak acid. The degree
of ionization in aqueous solution depends on the
formal concentration of HOAc, as well as the
existence of other acid or base species that may
be in solution.
If we divide both sides by volume, and use the definition of formal
Which is a mathematical statement of mass balance, i.e., the sum of the
molar concentrations of all acetic species equals the formula weights
we put into solution.
A Little Algebra….
and substituted into the formal mass balance equation to yield
The fraction of acid in acetate form is
[OAc-]/FHOAc. Solving for this fraction
results in the amount of HOAc in the
form of the acetate
a
&
a
The ratios are often called the "alpha“ (a) of acetate and acetic acid respectively.
Fractional Composition (a) plot for acetic acid
What is the pH when [HA] = [A-]???
1
a A( pH)
a HA( pH)
4.75
0.5
0
0
2
4
6
pH
Which is the pKa….
8
10
12
Weak-Acid Equilibria
Typical Weak-Acid Problem
[HA] = CHA - [A-] = CHA - [H3O+]
substituting into the Ka expression
[H3O+]2
Ka = ----------------CHA - [H3O+]
produces the quadratic equation
[H3O+]2 + Ka[H3O+] - KaCHA = 0
Weak-Acid Equilibria
Typical Weak-Acid Problem
[HA] = CHA - [A-] = CHA - [H3O+]
produces the quadratic equation
[H3O+]2 + Ka[H3O+] - KaCHA = 0
Weak-Acid Equilibria
Typical Weak-Acid Problem
simplifying assumption
if CHA >> [H3O+] and CHA - [H3O+] ~ CHA
[H3O+]2
Ka = ----------- thus
CHA
Weak-Acid Equilibria
Typical Weak-Acid Problem
simplifying assumption
Assume valid if CHA >>> Ka
where each > signifies a power of 10
EXAMPLE: Calculate the
[H3O+] in an aqueous 0.140 M
acetic acid solution.
HC2H3O2 + H2O
H3O+ + C2H3O2-
CHA = 0.140 M
let
[H3O+] ~ [C2H3O2-]
[HC2H3O2] = 0.140M - [H3O+]
EXAMPLE: Calculate the [H3O+]
in an aqueous 0.140 M acetic
acid solution.
HC2H3O2 + H2O
H3O+ + C2H3O2-
CHA = 0.140 M
let[H3O+] ~ [C2H3O2-]; [HC2H3O2] = 0.140M - [H3O+]
using the quadratic equation
EXAMPLE: Calculate the [H3O+] in an
aqueous 0.140 M acetic acid solution.
CHA = 0.140 M
Ka = 1.75 X 10-5 M
using the quadratic equation
CHA = 0.140 M
Ka = 1.75 X 10-5M
using the quadratic equation
CHA = 0.140 M
Ka = 1.75 X 10-5M
using the quadratic equation
EXAMPLE: Calculate the [H3O+] in an aqueous 0.140 M
acetic acid solution.
CHA = 0.140 M
Ka = 1.75 X 10-5M
assume [HC2H3O2] = 0.140M - [H3O+] ~ 0.140M
EXAMPLE: Calculate the [H3O+] in an aqueous
0.140 M acetic acid solution.
CHA = 0.140 M
Ka = 1.75 X 10-5M
assume [HC2H3O2] = 0.140M - [H3O+]
0.140M
[H3O+] = 1.57 x 10-3M
check assumption
[HC2H3O2] = 0.140M - [H3O+]
= 0.140M - 1.57 x 10-3M = 0.138 M
assumption valid
~
Fraction of Dissociation
EXAMPLE: Calculate the fraction of
dissociation in an aqueous 0.140 M acetic
acid solution.
1.11% dissociated
Use of Acid/Base Distributions in
pH Problems
Interesting Features of a plots
• First, notice that the pH where two species
concentrations are the same is around the
pKa for that equilibrium. In fact, for
polyprotic acids with pKa's that differ by
over 3 to 4 units, the pH is equal to the
pKa.
Take for example the point where [H3PO4]=[H2PO4-].
The equilibrium equation relating these two species is
If we take the -log10, or "p", of this equation
Since [H3PO4]=[H2PO4-], and log10(1) = 0, pH=pKa1
pKa1
pKa3
pKa2
Second, you might notice that the concentrations of the conjugate bases
are maximum half-way between the pKa points.
For example, the point where [H2PO4-] is a maximum lies half-way between
between pKa1 and pKa2. Since H2PO4- is the major species present in
solution, the major equilibrium is the disproportionation reaction.
This equilibrium cannot be used to solve for pH because [H3O+] doesn't occur
in the equilibrium equation. We solve the pH problem adding the first two
equilibria equations
+
Note that when we add chemical equilibria, we take the product of the
equilibrium equations. Taking the -log10 of the last equation
Since the disproportionation reaction predicts [H3PO4]=[HPO42-]
What about the “Z” word??
• Zwitterions – (German for “Double Ion”) – a
molecule that both accepts and losses
protons at the same time.
• EXAMPLES???
• How about – AMINO ACIDS
H
H
R
C COOH
NH2
neutral
R
C COOH
NH3
amino-protonated
H
R
C COO
NH3
zwitterion
Both groups protonated
H
R
C COO
NH2
carboxylic-deprotonated
Let’s look at the simplest of the amino acids, glycine
H
H
H
K1
C COOH
H
C COO
NH3
NH3
H2Gly+
HGly
H
K2
H
C COO
NH2
Gly-
glycinate
glycinium
K1  10
 2.35
[ H  ][ HGly ]

[ H 2 Gly  ]
K 2  10
9.78
[ H  ][Gly  ]

[ HGly ]
In water the charge balance would be,
[ H 2 Gly  ]  [ H  ]  [OH  ]  [Gly  ]
Combining the autoprotolysis of water and the K1 and K2 expressions into the charge balance yields:
K [ HGly ] K w
[ H  ][ HGly ]
 [H  ]  2 


K1
[H ]
[H  ]
[H  ] 
K 2 [ HGly ]  K w
[ HGly ] / K1  1
Example
• Calculate the pH and the concentrations of
the various species in a 0.100 M glycine
solution.

[H ] 
K 2 [ HGly ]  K w
10 9.78[0.10]  10 14
 6.08


10
 6.08
 2.35
[ HGly ] / K1  1
[0.10] / 10
1
[ H 2Gly  ] 
[ H ](0.10)
 10  4.73  1.86 x10 5 M
 2.35
10
10 9.78 (0.10)
 4.70
5
[Gly ] 

10

1
.
99
x
10
M

[H ]

What is the most prevalent form of glycine in
This solution?
HGly
Gly-
H2Gly+
Buffers
A buffered solution resists changes in pH
when acids and bases are added or when
dilution occurs.
definition - a buffer solution is composed of:
a weak-acid and its salt (conjugate base)
or a weak-base and its salt (conjugate acid)
Buffers
Henderson-Hasselbalch Equation
[H3O+] = Ka ([HA]/[A-])
pH = pKa + log([A-]/[HA])
when the [A-] = [HA]
pH = pKa
AH HA! REMEMBER THE RESULT
FROM THE a plots
Biological Buffers
•
Biochemical reactions are especially sensitive to pH. Most biological
molecules contain groups of atoms that may be charged or neutral
depending on pH, and whether these groups are charged or neutral has a
significant effect on the biological activity of the molecule.
•
In all multicellular organisms, the fluid within the cell and the fluids
surrounding the cells have a characteristic and nearly constant pH. This pH is
maintained in a number of ways, and one of the most important is through
buffer systems.
Two important biological buffer systems are the dihydrogen phosphate system and
the carbonic acid system.
The phosphate buffer system
•
•
The phosphate buffer system operates in the internal fluid of all cells.
This buffer system consists of dihydrogen phosphate ions (H2PO4-) as
hydrogen-ion donor (acid) and hydrogen phosphate ions (HPO42-) as
hydrogen-ion acceptor (base).
H2PO4-(aq)
H+(aq) + HPO42-(aq)
Ka = [H +] [HPO42-] =6.23 × 10-8 at 25oC
[H2PO4-]
•
If additional hydrogen ions enter the cellular fluid, they are consumed in the
reaction with HPO42-, and the equilibrium shifts to the left. If additional
hydroxide ions enter the cellular fluid, they react with H2PO4-, producing
HPO42-, and shifting the equilibrium to the right.
when the concentrations of H2PO4- and HPO42- are the
same, what will the pH equal?
7.21
Buffer solutions are most effective at maintaining a pH
near the value of the pKa. In mammals, cellular fluid has
a pH in the range 6.9 to 7.4, and the phosphate buffer is
effective in maintaining this pH range.
Carbonate Buffer
Another biological fluid in which a buffer plays an important role in
maintaining pH is blood plasma. In blood plasma, the carbonic acid and
hydrogen carbonate ion equilibrium buffers the pH.
In this buffer, carbonic acid (H2CO3) is the hydrogen-ion donor (acid) and
hydrogen carbonate ion (HCO3-) is the hydrogen-ion acceptor (base).
H2CO3(aq)
H+(aq) + HCO3-(aq)
Additional H+ is consumed by HCO3- and additional OH- is
consumed by H2CO3. Ka for this equilibrium is 7.9 × 10-7, and the
pKa is 6.1 at body temperature. In blood plasma, the concentration
of hydrogen carbonate ion is about twenty times the concentration
of carbonic acid. The pH of arterial blood plasma is 7.40. If the pH
falls below this normal value, a condition called acidosis is
produced. If the pH rises above the normal value, the condition is
called alkalosis.
The concentrations of hydrogen carbonate ions and
of carbonic acid are controlled by two independent
physiological systems. Carbonic acid concentration is
controlled by respiration, that is through the lungs.
Carbonic acid is in equilibrium with dissolved carbon
dioxide gas.
H2CO3(aq)
CO2(aq) + H2O(l)
An enzyme called carbonic anhydrase catalyzes the
conversion of carbonic acid to dissolved carbon
dioxide. In the lungs, excess dissolved carbon dioxide
is exhaled as carbon dioxide gas.
CO2(aq)
CO2(g)
The concentration of hydrogen carbonate ions is
controlled through the kidneys. Excess hydrogen
carbonate ions are excreted in the urine.
The much higher concentration of hydrogen carbonate
ion over that of carbonic acid in blood plasma allows
the buffer to respond effectively to the most common
materials that are released into the blood. Normal
metabolism releases mainly acidic materials: carboxylic
acids such as lactic acid (HLac). These acids react with
hydrogen carbonate ion and form carbonic acid.
HLac(aq) + HCO3-(aq)
Lac-(aq) + H2CO3(aq)
The carbonic acid is converted through the action of the
enzyme carbonic anhydrase into aqueous carbon
dioxide.
H2CO3(aq)
CO2(aq) + H2O(l)
An increase in CO2(aq) concentration stimulates increased breathing,
and the excess carbon dioxide is released into the air in the lungs.
The carbonic acid-hydrogen carbonate ion buffer works throughout the body
to maintain the pH of blood plasma close to 7.40. The body maintains the
buffer by eliminating either the acid (carbonic acid) or the base (hydrogen
carbonate ions). Changes in carbonic acid concentration can be effected
within seconds through increased or decreased respiration. Changes in
hydrogen carbonate ion concentration, however, require hours through the
relatively slow elimination through the kidneys
EXAMPLE: What is the ratio of [H2CO3]/[HCO3-] in the
blood buffered to a pH of 7.40?
Ka = 4.4 x 10-7M
[H3O+] = 10-pH = 10-7.4 = 4.0 x 10-8M
Buffer Capacity
• refers to the ability of the buffer to retard
changes in pH when small amounts of
acid or base are added
• the ratio of [A-]/[HA] determines the pH of
the buffer whereas the magnitude of [A-]
and [HA] determine the buffer capacity