Acid and bases and its strength by using ph scale lecture_4.ppt

BCHS 3304: General Biochemistry I, Section 07553
Spring 2003 1:00-2:30 PM Mon./Wed. AH 101
http://www.uh.edu/sibs/faculty/glegge
Instructor:
Glen B. Legge, Ph.D., Cambridge UK
Phone: 713-743-8380
Fax: 713-743-2636
E-mail: glegge@uh.edu
Office hours:
Mon. and Wed. (2:30-4:00 PM) or by appointment
353 SR2 (Science and Research Building 2)
1

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SR1
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contact email: MYSTIK1775@aol.com

Acids and Bases
January 27 2003

Ionization of water
• Although neutral water has a tendency to
ionize
H2O <-> H+
+ OH-
• The free proton is associated with a
water molecule to form the hydronium ion
H3O+
• High ionic mobility due to proton jumping

Proton Jumping
Large proton and hydroxide mobility
• H3O+ : 362.4 x 10-5
cm2
•V-1
•s-1
• Na+
: 51.9 x 10-5
cm2
•V-1
•s-1
• Hydronium ion migration; hops by switching
partners at 1012
per second

Equilibrium expression
• Described by:
K = [H+
][OH-
]
[H2O]
• Where K is the dissociation constant
• Considering [H2O] constant yields
Kw = [H+
][OH-
]

Kw
Kw = [H+
][OH-
]
• Where Kw is the ionization constant of
water
• For pure water ionization constant is
10-14
M2
at 25º
• For pure water
[H+
] = [OH-
] = (Kw)1/2
= 10-7
M

Acids and bases
• For pure water (neutral)
[H+
] = [OH-
] = (Kw)1/2
= 10-7
M
• Acidic if [H+
] > 10-7
M
• Basic if [H+
] < 10-7
M

Acids and Bases
Lowery definition:
• Acid is a substance that can donate a proton.
• Base is a substance that can accept a proton.
HA + H2O H3O+
+ A-
/OH-
Acid Base Conjugate Conjugate
Acid Base
or
HA A-
+ H+
Acid Conjugate Conjugate
Base Acid

If you establish equilibrium, changes in [H+
] will shift
the ratio of HA and A-
.
By adding more H+
, A-
will be consumed
forming HA.
If there is sufficient [A-
], the extra H+
will also be
consumed and the [H+
] will not change.



 A
H
HA

Acid strength is specified by its dissociation constant
Molar concentration
for: HA + H2O H3O+
+ A-
• reactants products
HA H3O+
A-
H2O
a measure of relative proton affinities for each conjugate acid
base pair.
O]
[H
]
HA
[
]
][A
O
[H
2
-
3


a
K
These ratios are

What to do about the water!
The concentration of H2O remains almost unchanged especially
in dilute acid solutions.
What is the concentration of H2O?
Remember the definition: Moles per liter
1 mole of H2O = 18 g = 18 ml
1000 g/liter
ml
g 1
1 
M
mol
g
g
5
.
55
/
18
1000


]
[
]
][
[
]
[ 2
HA
A
H
O
H
K
Ka




From now on we will drop the a, in Ka
Weak acids (K<1)
Strong acids (K>1)
Strong acid completely dissociates: Transfers all its
protons to
H2O to form H3O+
HA H+
+ A-

Weak Acids
Weak acids do not completely dissociate:
They form an equilibrium:



 H
A
HA
If we ADD more H+
, the equilibrium shifts to
form more HA using up A-
that is present.

Dissociation of H2O



 OH
H
O
H2
Water also dissociates [H2O] = 55.5
]
[
]
][
[
2O
H
OH
H
K



2
14
M
10
]
][
[ 



 OH
H
Kw
Ionization constant for water

Since there is equal amounts of [H+
] and [OH-
]
M
10
x
1
]
[
]
[ 7




 OH
H
This is neutral
At [H+] above this concentration the solution is ACIDIC
At [H+] below this concentration the solution is BASIC
2
10 1 ] [
 

x H
9
10
1
]
[ 

 x
H

[H+
] pH
10-7
= 7
10-3
= 3
10-2
= 2
10-10
= 10
5x10-4
= 3.3
7x10-6
= 5.15
3.3x10-8
= 7.48
pH = -Log[H+
]
It is easier to think in log
of concentrations but it
takes practice!!

Relationship between pH and [H+
] / [OH-
] concentration

Observation
If you add .01 ml i.e 1/100 ml of 1M HCl to 1000 ml
of water, the pH of the water drops from 7 to 5!!
i.e 100 fold increase in H+
concentration: Log = 2
change.
Problem:
Biological properties change with small changes in
pH, usually less than 1 pH unit.
How does a system prevent fluctuations in pH?

Buffers
A buffer can resist pH changes if the pH is at or near a weak
acid pK value.
Buffer range: the pH range where maximum resistance to pH
change occurs when adding acid or base. It is = +
1 pH from
the weak acid pK
If pK is 4.8 the buffering range is 3.8 5.8
Why?

Henderson - Hasselbalch equation
]
[
]
][
[
HA
A
H
K



From
]
[
]
[
]
[ 


A
HA
K
H
Rearrange
Take (-)Log of each
]
[
]
[
log
log
HA
A
K
pH




]
[
]
[
log
HA
A
pK
pH




Above and below this range there is insufficient
amount of conjugate acid or base to combine with
the base or acid to prevent the change in pH.
[HA]
]
[A
log
pK
pH
-


1
10
10
1
from
varies
ratio
]
HA
[
]
[A



For weak acids
HA A-
+ H+
This equilibrium depends on concentrations of each component
.
If [HA] = [A-
] or 1/2 dissociated
Then 0
1
log
]
[
]
[
log 


HA
A
By definition the pK is the pH where [HA] = [A-
] :
50% dissociated
: pH = pK

The buffer effect can be seen in a titration curve.
To a weak acid salt, CH3C00-,
add HC1 while monitoring pH vs. the
number of equivalents of acid added.
or
do the opposite with base.
Buffer capacity: the molar amount of acid which the buffer can
handle without significant changes in pH.
i.e
1 liter of a .01 M buffer can not buffer 1 liter of a 1 M solution of
HCl
but
1 liter of a 1 M buffer can buffer 1 liter of a .01 M solution of HCl

Acid and bases and its strength by using ph scale lecture_4.ppt

Distribution curves for acetate and acetic acid

Table 2-3
Dissociation constants and pK’s of Acids & buffers
Acid K pK
Oxalic 5.37x10-2
1.27
H3PO4 7.08x10-3
2.15
Succinic Acid 6.17x10-5
4.21 (pK1)
Succinate 2.29x10-6
5.65 (pK2)
H2PO4
-
1.51x10-7
6.82
NH4
+
5.62x10-10
9.25
Glycine 1.66x10-10
9.78

You will be asked to solve Henderson -
Hasselbalch type problems:
You may be asked the pH, pK, the ratio of acid or base or solve for
the final concentrations of each.
Exams Problems
[HA]
]
[A
log
pK
pH
-



The 6 step approach
1. Write the Henderson + Hasselbalch equation.
2. Write the acid base equation
3. Make sure either an H+
or OH-
is in the equation.
3. Find out what you are solving for
4. Write down all given values.
5. Set up equilibrium conditions.
6. Plug in H + H equation and solve.

What is the pH of a solution of that contains 0.1M CH3C00-
and 0.9 M CH3C00H?
1) pH = pK + Log [A-
]
[HA]
2) CH3C00H CH3C00-
+ H+
3) Find pH
4) pK = 4.76 A-
= 0.1 M HA = 0.9 M
5) Already at equilibrium
6) X = 4.76 + Log 0.1
0.9

What would the concentration of CH3C00-
be at pH 5.3
if 0.1M CH3C00H was adjusted to that pH.
1) pH = pK + Log [A-
]
[HA]
2) CH3C00H CH3C00-
+ H+
3) Find equilibrium value of [A-
] i.e [CH3C00-
]
4) pH = 5.3; pK = 4.76
5) Let X = amount of CH3C00H dissociated at equilibrium
[A-
] = [X]
[HA] = [0.1 - X]
6) 5.3 = 4.76 + Log [X]
[0.1 - X]
Now solve.

Blood Buffering System
• Bicarbonate most significant buffer
• Formed from gaseous CO2
CO2 + H2O <-> H2CO3
H2CO3 <-> H+
+ HCO3
-
• Normal value blood pH 7.4
• Deviations from normal pH value lead to
acidosis

Acid and bases and its strength by using ph scale lecture_4.ppt

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