Themes > Science > Chemistry > Inorganic Chemistry > Acids and Bases > Acids and Bases > 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, lets join in on the fun of acid-base equilibrium.

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

The base Kb's are as follows:
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

Examples

 

Weak Acid Example:

Calculate the pH of a 0.100 M solution of HClO

HClO <--> H+ + ClO-
.100 - x x x
Ka = __x2__
.100 - x

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 10-8 = __x2__
.100

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

 

Weak Base Example:

What is the concentration of OH- of a .20 molar solution of aniline?
C6H5NH2 <--> C6H5NH3+ + OH-
.20 - x x x
Kb = __x2__
.20 - x

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:
Kb = __x2__
.20

x2 = 8.4 x 10-11

x = 9.2 x 10-6

[OH-] = 9.2 x 10-6


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