*
*
*Oxidation
increase in oxidation number
addition of oxygen
removal of hydrogen
*Reduction
decrease in oxidation number
addition of hydrogen
removal of oxygen
*
*Hydrolysis - reaction in the presence
of water, acid, base or enzyme.
*Tautomerism - interconversion of
aldehyde/ketone to alcohol.
*Condensation - reaction of two or
more substances with the removal of
water from the molecules.
*CH3CH2OH + O2 → CH3CHO + H2O
*(CH3)2CHOH + O2 → (CH3)2C=O + H2O
*CH3CHO + H2 → CH3CH2OH
* (CH3)2C=O + H2 → (CH3)2CHOH
*NADH + H+ → NAD + 2H+
*NAD + 2H+ → NADH + H+
*FAD + 2H+ → FADH2
*FADH2 → FAD + 2H+
*
*CH3COOCH3 + H2O → CH3COOH + CH3OH
*CH3COOH + CH3OH → CH3COOCH3 + H2O
*
*pH = -log [H+]
*[H+] = inv. log –pH
*pH = pKw – pOH
*pH = 14 – pOH
*[H+] = Kw/[OH-]
*[H+] = 1 x 10-14/ [OH-]
pOH = - log [OH-]
[OH-] = inv. log – pOH
pOH = pKw – pH
pOH = 14 – pH
[OH-] = Kw/[H+]
[OH-] = 1x10-14/[H+]
Henderson-Hasselbalch Equation:
pH = pKa + log A-/HA
*
*
•From the reaction of two water molecules, the
following equilibrium constant expression can be
written:
K =
[H3O+][−OH]
[H2O]2
•Multiplying both sides by [H2O]2 yields Kw, the
ion-product constant for water.
Kw =
ion-product
constant
[H3O+][−OH]
7
*
•Experimentally it can be shown that
[H3O+] = [−OH] = 1.0 x 10−7 M at 25 oC
Kw
=
[H3O+] [−OH]
Kw
=
(1.0 x 10−7) x (1.0 x 10−7)
Kw
=
1.0 x 10−14
•Kw is a constant, 1.0 x 10−14, for all aqueous
solutions at 25 oC.
8
*
To calculate [−OH] when
[H3O+] is known:
To calculate [H3O+] when
[−OH] is known:
Kw = [H3O+][−OH]
[−OH] =
Kw = [H3O+][−OH]
Kw
[H3O+] =
[H3O+]
−14
1
x
10
[−OH] =
[H3O+]
Kw
[−OH]
−14
1
x
10
[H3O+] =
[−OH]
9
*
If the [H3O+] in a cup of coffee is 1.0 x 10−5 M, then
the [−OH] can be calculated as follows:
[−OH] =
Kw
[H3O+]
=
1 x 10−14
1 x 10−5
=
1.0 x 10−9 M
In this cup of coffee, therefore, [H3O+] > [–OH], and
the solution is acidic overall.
10
*
11
*
pH = −log [H3O+]
The lower the pH, the higher the concentration of
H3O+.
•Acidic solution:
pH < 7  [H3O+] > 1 x 10−7
•Neutral solution:
pH = 7  [H3O+] = 1 x 10−7
•Basic solution:
pH > 7  [H3O+] < 1 x 10−7
12
*
•If [H3O+] = 1.2 x 10–5 M for a solution, what is its pH?
pH = –log [H3O+] = –log (1.2 x 10–5)
= –(–4.92) = 4.92
•The solution is acidic because the pH < 7.
13
*
•If the pH of a solution is 8.50, what is the [H3O+]?
pH = −log [H3O+]
8.50 = −log [H3O+]
−8.50 = log [H3O+]
antilog (−8.50 ) = [H3O+]
[H3O+] = 3.2 x 10−9 M
•The solution is basic because [H3O+] > 1 x 10–7 M.
14
*
15
*
16
*Refer to the lab workbook.
*
*Chemical reactions in the cell
*Ionization of water
*Computations of pH and buffers
*pH of body fluids
*