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Ka, Acid Dissociation Constant and Kb, Base Dissociation Constant (Lesson Recap)

Ka, Acid Dissociation Constant

Ka- is the Keq for the dissociation for a weak acid.

e.g Acetic Acid, when dissolved in H2O
CH3COOH (aq) + H2O (l)
(ACID) (BASE)

CH3COO (aq) + H3O(l)
(CONJ BASE) (CONJ ACID)

CH3COOH (aq) + H2O(l) ↔ CH3COO (aq) + H3O(l)
*ONLY weak acid and back can be reversible

Equation for Ka :
Keq= [CH3COO(aq)] [H3O(l)][CH3COOH(aq)] [H2O(l)]
Keq x [H2O (l)] = Ka
Ka= [CH3COO(aq)] [H3O(l)] [CH3COOH(aq)]

Example Question: Calculate the pH at equilibrium for a 1.0 mol/L Ka= 1.8x10-5

CH3COOH (aq) + H2O(l) ↔ CH3COO (aq) + H3O(l)
| CH3COOH (aq)| CH3COO (aq)| H3O(l)|
I| 1mol/L| 0| 0|
C| -x| x| x|
E| 1-x| x| x|

pH= -log [H+]
pH= -log [H3O]
pH= -log [4.2x10-3]
pH= 2.38
X^2 1-X =1.8x10-5
Use 100 rule
1.8x10-5= X2
1.8x10-5
X= 4.2x10-3

Kb, Base Dissociation Constant

Kb is the equilibrium constant for the reaction in which a weak base is in equilibrium with its conjugate acid in aqueous solution. In other words, it is designed to show how well a base dissociates in water.

It is usually used for a weak base.

This concept is the same as acid dissociation constant, also known as Ka

e.g: Ammonium, when dissolved in water

NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)
Base Acid Conj Acid Conj base
Keq = [NH4+(aq)] [OH-(aq)][NH3(aq)] [H2O(l)]

Like Ka, H2O is multiplied with Keq to make Kb

Equation for Kb:

Keq = [NH4+(aq)] [OH-(aq)][NH3(aq)]

Example Question: Calculate the pH of a base, NH3(aq) with a concentration of 0.5 mol/L. Kb = 1.8x10-5 NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)

| NH3(aq) | NH4+(aq)| OH-(aq)|
I| 0.5| 0| 0|
C| -X| X| X|
E| 0.5-x | X| X|

Kb= [NH4+(aq)] [OH-(aq)][NH3(aq)]
1.8x10-5=x^20.5
0.000009= x2
= 0.003
SUB INTO NH3
0.5 – 0.003= 0.497

POH= -log (0.003)
POH= 2.52...
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