Acid & Base Equilibria

Kb – Base Dissociation Constant

Recall: A strong base:

Attracts/accepts H+ easily“Dissociates completely”Does not form equilibriumOnly two strong bases O2- & NH2

-

A weak base:Does NOT accept or attract H+ easily“Partially Dissociates”Forms equilibrium

Kb will be a measure of the level of dissociation

Consider:

NH3 + H2O NH4+ + OH-

Keq = [NH4+][OH-]

[NH3][H2O]

Recall [H2O] is constant, so…

Keq x [H2O] = [NH4+][OH-]

[NH3]Kb = [OH-][NH4+]

[NH3]

Conjugate Base

Base

Conjugate Acid

The greater the Kb, the stronger the base

If Kb <<<1 then very little of the base accepts/attracts H+, therefore weak base.

Strengths of bases can be compared by Kb.ex. ammonia Kb = 1.8 x 10-5

HCO3- Kb = 2.3 x 10-8

So ammonia is a stronger base

Relationship between Ka & Kb:

Ka & Kb of conjugate acid-base pairs are related:

Ka x Kb = Kw

Ex. NH3 & NH4+

C. acid, NH4+, Ka = 5.59 x 10-10

Base, NH3, Kb = 1.79 x 10-5

1.0 x 10-14

Calculate the Kb for CN- acting as a base in water. The Ka of HCN = 4.8 x 10-10.

CN- + H2O HCN + OH-

Kb = KwKa

Kb = 1.0 x 10 -14

4.8 x 10-10

Kb = 2.08 x 10-5

Ka x Kb = Kw

Calculate the Kb for 0.25M X-, if the [OH-] & [HX] is 2.0 x 10-4 M.

X- + H2O HX + OH-

Kb = [HX][OH-] [HX]

Kb = [2.0 x10-4][2.0 x 10-4] [0.25]

Kb = 1.60 x 10-7

Calculate the pH of a 0.10M NH3 solution. The Ka of NH4

+ is 5.59 x 10-10. NH3 + H2O NH4

+ + OH-

Kb = [NH4+][OH-]

[NH3]

1.79 x 10-5 = [x][x] [0.10 – x]

[OH-] = 1.34 x 10-3M

pOH = 2.87

NH3 is a BASE! Need Kb

Kb = KwKa

Kb = 1.0 x 10-14 5.59 x 10-10

Kb = 1.79 x 10-5 pH = 11.13

Amphoteric – Acid or Base?

Recall some molecules can act as an acid or a base

Comparing the Ka & Kb will determine whether an amphoteric substance is “acting” acidic or basic.

The amphoteric anion HCO3-1 has a Ka

value of 4.7 x 10-11. If HCO3-1 was placed in

water, would the solution be acidic or basic?

Write out the equations as it “acts” like an acid & a base

HCO3-1 + H2O CO3

-2 + H3O+ Ka = 4.7 x 10-11

HCO3-1 + H2O H2CO3 + OH- Kb = ?

Calculate Kb & compare to Ka

Kb = 1.0 x 10-14 4.7 x 10-11

Kb = 2.13 x 10-4

2.13 x 10-4

Kb > Ka Therefore

solution is BASIC