20–5. Day 1– Equilibrium Dissociation for Water, Acids and
Bases
a) Acid Dissociation
i) Recall, only weak acids will form an equilibrium.
H2S(aq) + H2O(l) qwwe HS–(aq) + H3O+(aq)
ii) Keq=Ka(cid) = [HS–][H3O+]
[H2S]
= 9.1 x 10–8
iii) Ka is called the
Acid Ionization Constant
iv) The greater the value of Ka,
the stronger the acid
v) Example: Write the reaction and
the Ka expression of H2SO3 with water
H2SO3(aq)+H2O(l)qeHSO3–(aq)+H3O+(aq)
Ka = [HSO3–][H3O+] = 1.5 x 10–2
[H2SO3]
b) Base Dissociation
i) Recall, only weak bases will form an equilibrium.
CO3–2(aq) + H2O(l) qwwe HCO3–(aq) + OH–(aq)
ii) Keq= Kb(ase) = [HCO3–][OH–] =
[CO3–2]
= 1.8 x 10–4 (not on the table, but can get it
from table.. see below)
iii) Kb is called the
Base Ionization Constant
iv) The greater the value of Kb,
the stronger the base.
c) Finding Kb
i) Theory
There is a relationship between Ka and Kb
Acid dissociation:
H2S(aq) + H2O(l)
qwwe HS–(aq) + H3O+(aq)
Ka = [HS–][H3O+] = 9.1 x 10–8
[H2S]
Conjugate Base dissociation:
HS–(aq) + H2O(l) qwwe H2S(aq) + OH–(aq)
Kb = [H2S][OH–] = 1.1 x 10–7 (not on the table)
[HS–]
Ka x Kb = [HS–][H3O+] x [H2S][OH–] =
[H2S]
[HS–]
[H3O+] [OH–] = Kw
Ka x Kb =Kw or Kb = Kw
Ka
ii) Procedure:
Look on Base side of Table
Of Relative Strengths to find
your base species
Find the Ka for the reverse
reaction.(at that line)
Calculate Kb from
Ka x Kb = Kw or better
***** Kb = Kw /Ka (of the conjugate acid)
iii) Example: Find the Kb for SO4–2
On the table we find:
HSO4– qwweH+ + SO4–2
Ka = 1.2 x 10–2
Kb SO42- = Kw / Ka (HSO4 -)
= 1.0 x 10–14
1.2 x 10–2
= 8.3 x 10–13
iv) Amphiprotic anions: be careful!
Example : Find Kb of HPO42–
H2PO4– qwweH+ + HPO4–2
Ka = 6.2 x 10–8
Kb (HPO42–) = Kw / Ka(H2PO4–)
= 1.0 x 10–14
6.2 x 10–8
= 1.6 x 10–7
Do questions: #29–30 page 127; #31–34 page 128; #35–37 page 130