Acid-Base Equilibrium (Monoprotic)
Chapter 9
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
WRONG!!!!!!!!
HCl is an acid!
Acids have pH less than 7!
OH NO! WHAT
SHOULD WE DO !
What's wrong?
We ignored the fact that water is also an ACID!
[H+]total = [H+]water + [H+]HCl
It’s BACK ! Systematic Equilibrium
near pH 7, the contribution of [H+] from water
becomes the dominate source for the [H+] in
the solution
SO HOW DO WE SOLVE THIS ?
[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
What is the pH of a 0.5 M HF solution (at 250C)?
HF (aq)
Initial (M)
Change (M)
Equilibrium (M)
+][F-]
[H
= 7.1 x 10-4
Ka =
H+ (aq) + F- (aq)
[HF]
HF (aq)
H+ (aq) + F- (aq)
0.50
0.00
0.00
-x
+x
+x
0.50 - x
x
x
x2
= 7.1 x 10-4
Ka =
0.50 - x
x2
Ka 
= 7.1 x 10-4
0.50
[H+] = [F-] = 0.019 M
[HF] = 0.50 – x = 0.48 M
Ka << 1
0.50 – x  0.50
x2 = 3.55 x 10-4
x = 0.019 M
pH = -log [H+] = 1.72
When can I use the approximation?
Ka << 1
0.50 – x  0.50
When x is less than 5% of the value from which it is subtracted.
x = 0.019
0.019 M
x 100% = 3.8%
0.50 M
Less than 5%
Approximation ok.
What is the pH of a 0.05 M HF solution (at 250C)?
x2
Ka 
= 7.1 x 10-4 x = 0.006 M
0.05
More than 5%
0.006 M
x 100% = 12%
Approximation not ok.
0.05 M
Must solve for x exactly using quadratic equation or method
of successive approximation.
What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4?
HA (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122
0.00
0.00
-x
+x
+x
0.122 - x
x
x
x2
= 5.7 x 10-4
Ka =
0.122 - x
Ka 
H+ (aq) + A- (aq)
x2
= 5.7 x 10-4
0.122
0.0083 M
x 100% = 6.8%
0.122 M
Ka << 1
0.122 – x  0.122
x2 = 6.95 x 10-5
x = 0.0083 M
More than 5%
Approximation not ok.
x2
= 5.7 x 10-4
Ka =
0.122 - x
ax2 + bx + c =0
x = 0.0081
HA (aq)
Initial (M)
Change (M)
Equilibrium (M)
x2 + 0.00057x – 6.95 x 10-5 = 0
-b ± b2 – 4ac
x=
2a
x = - 0.0081
H+ (aq) + A- (aq)
0.122
0.00
0.00
-x
+x
+x
0.122 - x
x
x
[H+] = x = 0.0081 M
pH = -log[H+] = 2.09
Ionized acid concentration at equilibrium
percent ionization =
x 100%
Initial concentration of acid
For a monoprotic acid HA
Percent ionization =
[H+]
[HA]0
x 100%
[HA]0 = initial concentration
Weak Bases and Base Ionization Constants
NH3 (aq) + H2O (l)
NH4+ (aq) + OH- (aq)
[NH4+][OH-]
Kb =
[NH3]
Kb is the base ionization constant
Kb
weak base
strength
Solve weak base problems like weak acids
except solve for [OH-] instead of [H+].
CHM 112 Summer 2007
Prushan
M.
Ionization Constants of Conjugate Acid-Base Pairs
HA (aq)
A- (aq) + H2O (l)
H2O (l)
H+ (aq) + A- (aq)
OH- (aq) + HA (aq)
H+ (aq) + OH- (aq)
Ka
Kb
Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Kw
Ka =
Kb
Kw
Kb =
Ka
15.7
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
The common ion effect is the shift in equilibrium caused by the
addition of a compound having an ion in common with the
dissolved substance.
The presence of a common ion suppresses
the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s)
CH3COOH (aq)
Na+ (aq) + CH3COO- (aq)
H+ (aq) + CH3COO- (aq)
common
ion
Consider mixture of salt NaA and weak acid HA.
NaA (s)
Na+ (aq) + A- (aq)
HA (aq)
H+ (aq) + A- (aq)
[H+]
Ka [HA]
=
[A-]
-log [H+] = -log Ka - log
[HA]
[A-]
-]
[A
-log [H+] = -log Ka + log
[HA]
[A-]
pH = pKa + log
[HA]
[H+][A-]
Ka =
[HA]
Henderson-Hasselbalch
equation
[conjugate base]
pH = pKa + log
[acid]
pKa = -log Ka
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
HCOOH pKa = 3.77
H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
[HCOO-]
pH = pKa + log
[HCOOH]
[0.52]
= 4.01
pH = 3.77 + log
[0.30]
A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon
the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq)
Add strong base
OH- (aq) + CH3COOH (aq)
CH3COOH (aq)
CH3COO- (aq) + H2O (l)
Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq)
[NH3]
pH = pKa + log
[NH4+]
start (moles)
end (moles)
H+ (aq) + NH3 (aq)
pKa = 9.25
0.029
0.001
NH4+ (aq) + OH- (aq)
0.028
0.0
[0.30]
pH = 9.25 + log
= 9.17
[0.36]
0.024
H2O (l) + NH3 (aq)
0.025
final volume = 80.0 mL + 20.0 mL = 100 mL
0.028
0.025
[0.25]
+
pH = 9.25 + log
[NH4 ] =
[NH3] =
0.10
0.10
[0.28] = 9.20
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