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equilibrium with acids and bases
Have you read the section about equilibrium yet? If you haven't this
most likely won't make any sense to you. If you have, let's join in on the
fun of acid-base equilibrium.
Ka and Kb
Remember Kc from the equilibrium section? It's back, and
more useful than ever. Now to distinguish between the Kc of
acids and bases we use Ka and Kb. (a for acids and
b for bases) The equilibrium that is calculated in acids is usually the
disaccociation of the H+ ions and the rest of the molecule. The
weak acids and bases are the only ones that have Ka's and
Kb's because in the strong acids dissociation is very close to 100%.
Table of Ka's
| Substance |
Formula |
Ka |
| Acetic Acid |
HC2H3O2 |
1.7 x 10-5 |
| Benzoic Acid |
HC7H5O2 |
6.3 x 10-5 |
| Boric Acid |
H3BO3 |
5.9 x 10-10 |
| Carbonic Acid |
H2CO3 |
4.3 x 10-7
|
|
HCO3- |
4.8 x 10-11 |
| Cyanic Acid |
HCNO |
3.5 x 10-4
|
| Formic Acid |
HCNO2 |
1.7 x 10-4 |
| Hydrocyanic Acid |
HCN |
4.9 x 10-10 |
| Hydrofluric Acid |
HF |
6.8 x 10-4 |
| Hydrogen Sulfate ion |
HSO4- |
1.1 x 10-2 |
| Hydrogen Sulfide |
H2S |
8.9 x 10-8 |
|
HS- |
1.2 x 10-13 |
| Hypochlorous acid |
HClO |
3.5 x 10-8 |
| Nitrous Acid |
HNO2 |
4.5 x 10-4 |
| Oxalic Acid |
H2C2O4 |
5.6 x 10-2
|
|
HC2O4- |
5.1 x 10-5 |
| Phosphoric Acid |
H3PO4 |
6.9 x 10-3 |
|
H2PO4- |
6.2 x 10-8 |
|
HPO4-2 |
4.8 x 10-13 |
| Phosphorous Acid |
H2PHO3 |
1.6 x 10-2 |
|
HPHO3- |
7 x 10-7 |
| Propionic Acid |
HC3H5O2 |
1.3 x 10-5 |
| Pyruvic Acid |
HC3H3O3 |
1.4 x 10-4 |
| Sulfurous Acid |
H2SO3 |
1.3 x 10-2 |
|
HSO3- |
6.3 x 10-8 |
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Table of Ka's
Table of Kb's
| Substance |
Formula |
Kb |
| Ammonia |
NH3 |
1.8 x 10-5 |
| Aniline |
C6H5NH2 |
4.2 x 10-10 |
| Dimethylamine |
(CH3)3NH |
5.1 x 10-4 |
| Ethylamine |
C2H5NH2 |
4.7 x 10-4 |
| Hydrazine |
N2H4 |
1.7 x 10-6 |
| Hydroxylamine |
NH2OH |
1.1 x 10-8 |
| Methylamine |
CH3NH2 |
4.4 x 10-4 |
| Pyridine |
C5H5N |
1.4 x 10-9 |
| Urea |
NH2CONH2 |
1.5 x 10-14 |
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Table of Kb's
weak acid example:
Calculate the pH of a 0.100 M solution of HClO
| HClO | <--> |
H+ | + |
ClO- |
| .100 - x |
|
x |
|
x |
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The x in the denominator can be dropped because Ka/M is less than
10-3. If Ka/M is greater than 10-3 you have to use the
quadratic formula to solve the equation.
Therefore:
3.5 x 10-9 = x2
x = 5.9 x 10-5
[H+] = 5.9 x 10-5 M
[ClO-] = 5.9 x 10-5 M
[HClO] = .100 M - 5.9 x 10-5 ~= .100 M
pH = -log(5.9 x 10-5) = 4.2
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weak base example:
What is the concentration of OH- of a .20 molar solution
of aniline?
| C6H5NH2 | <--> |
C6H5NH3+ | + |
OH- |
| .20 - x |
|
x |
|
x |
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The x in the denominator can be dropped because Kb/M is less than
10-3. If Kb/M is greater than 10-3 you have to use the
quadratic formula to solve the equation.
Therefore:
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x2 = 8.4 x 10-11
x = 9.2 x 10-6
[OH-] = 9.2 x 10-6
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