Transcript Lecture 1

Lecture 1: Introduction and review
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Quiz 1
Website: http://www.esf.edu/chemistry/nomura/fch530/
Review of acid/base chemistry
Universal features of cells on Earth
Cell types: Prokaryotes and Eukaryotes
Review of pH, acids, and bases
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pH is generally defined as the negative logarithm of the hydrogen ion
activities (concentration) expressed over 14 orders of magnitude
pH = -log10 [H+]
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The pH scale is a reciprocal relationship between [H+] and [OH-]
Because the pH scale is based on negative logarithms, low pH values
represent the highest [H+] and thus the lowest [OH-]
At neutrality, pH 7, [H+] = [OH-]
Review of pH, acids, and bases
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Strong electrolytes dissociate completely in water
– Electrolytes are substances capable of generating ions in solution
– Increase the electrical conductivity of the solution
The dissociation of a strong acid in water
H3O+ + Cl-
HCl + H2O
The equilibrium constant is
K=
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[H3O+][Cl-]
[H2O][HCl]
[H2O] is constant in dilute aq. Solutions and is incorporated into the equilibrium constant.
giving rise to a new term Ka-the acid dissociation constant = K[H2O], [H3O+] is
expressed as [H+]
[H+][Cl-]
Because Ka is large for HCl,
[H+] in solution = [HCl] added to solution.
a
K=
[HCl]
Thus, a 1M HCl solution has a pH of 0, a
1 mM HCl solution has a pH of 3, and so
on. Conversely, 0.1 M NaOH solution has
a pH of 13.
Ka=
[H+][Cl-]
[HCl]
For a strong acid Ka will approach be large because
the nearly all of the protons will be dissociated.
The [H+] at equilibrium is equal to the initial
concentration of the acid.
Calculate the pH of a 1M HCl solution
HCl + H2O
H3O+ + Cl-
0.0004% at equilibrium
99.996% at equilibrium
Since we are at equilibrium, H3O+ is equal to the initial concentration of acid.
[H+] = [H3O+] = [HCl] = 1M
We know that pH is the -log of [H+], therefore for 1M HCl at equilibrium
pH = -log10 [H+]
pH = -log10 (1)
pH = 0
[H+]
pH
[OH-]
0
(100)
1.0
0.00000000000001
(10-14)
1
(10-1)
0.1
0.0000000000001
(10-13)
2
(10-2)
0.01
0.000000000001
(10-12)
3
(10-3)
0.001
0.00000000001
(10-11)
4
(10-4)
0.0001
0.0000000001
(10-10)
5
(10-5)
0.00001
0.000000001
(10-9)
6
(10-6)
0.000001
0.00000001
(10-8)
7
(10-7)
0.0000001
0.0000001
(10-7)
8
(10-8)
0.00000001
0.000001
(10-6)
9
(10-9)
0.000000001
0.00001
(10-5)
10
(10-10)
0.0000000001
0.0001
(10-4)
11
(10-11)
0.00000000001
0.001
(10-3)
12
(10-12)
0.000000000001
0.01
(10-2)
13
(10-13)
0.0000000000001
0.1
(10-1)
14
(10-14)
0.00000000000001
1.0
(100 )
Review of pH, acids, and bases
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Weak electrolytes only slightly dissociate in water
– Acetic acid, CH3COOH
The dissociation of a weak acid in water
CH3COOH + H2O
H3O+ +
CH3COO-
The acid dissociation constant is
Ka =
[H+][CH3COO-]
[CH3COOH]
= 1.74 X 10-5 M
Ka is also called the ionization constant, because Ka is small, most of the acetic acid is
not ionized.
Acid dissociation constant
HA
The general ionization of an acid is as follows: –
H+ + ASo the acid dissociation constant is as follows:
Ka =
[H+][A-]
[HA]
There are many orders of magnitude spanned by Ka values, so pKa is used instead:
pKa = - log10 Ka
The larger the value of the pKa, the smaller the extent of dissociation.
pKa <2 is a strong acid
Henderson-Hasselbalch Equation
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Describes the dissociation of a weak acid in the presence of its
conjugate base
– The general ionization of a weak acid is as follows:
HA
H+ + ASo the acid dissociation constant is as follows:
Ka =
[H+][A-]
[HA]
Rearranging this expression in terms of the parameter of interest [H+] gives the
following:
[H+]
=
Ka [HA]
[A-]
Henderson-Hasselbalch Equation
Take the log of both sides:
log[H+]
=
[HA]
log Ka + log
[A-]
Change the signs and define pKa as -log Ka :
pH = pKa - log
[HA]
[A-]
or
pH = pKa + log
[A-]
[HA]
Titration curves and buffers
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Titration curves can be calculated by the
Henderson-Hasselbalch equation
– As OH- is added to the reaction, it reacts
completely with HA to form A-
[A-] =
x
vol
x = the equivalents of OH- added and V
represents the volume of the solution. If we
let co represent HA equivalents initially
present, then:
[HA] =
(co-x)
vol
We can reincorporate this into the HendersonHasselbalch eqn.
x
(c -x )
pH = pKa + log
o