The equilibrium constant expression is called the weak base dissociation constant, Kb,
and has the form:
The same reasoning that was used in dealing with weak acids is also true here: [HB+] =
[OH−]; [HB] ≈Minitially; the numerator can be represented as [OH−]^2 ; and knowing the ini-
tial molarity and Kbof the weak base, the [OH−] can easily be calculated. And if the initial
molarity and [OH−] are known, Kbcan be calculated.
For example, a 0.500 M solution of ammonia has a pH of 11.48. What is the Kbof
ammonia?
pH =11.48
[H+] = 10 −11.48
[H+] =3.3 × 10 −^12 M
Kw=[H+][OH−] =1.0 × 10 −^14
[OH−] =3.0 × 10 −^3 M
0.500 − xxx
[OH−] =[NH 4
+
] =3.0 × 10 −^3 M
[NH 3 ] =0.500 − 3.0 × 10 −^3 =0.497 M
The Kaand Kbof conjugate acid–base pairs are related through the Kwexpression:
Ka×Kb=Kw
This equation shows an inverse relationship between Kaand Kbfor any conjugate
acid–base pair.
This relationship may be used in problems such as: Determine the pH of a solution
made by adding 0.400 mol of strontium acetate to sufficient water to produce 2.000 L of
solution.
Solution:
The initial molarity is 0.400 mol/2.000 L =0.200 M.
When a salt is added to water dissolution will occur:
Sr (C H O 232 )( 22 →Sr2+aq) + 2C H O aq) 23 −(
Kb=
×
=×
−
(. ) −
(. )
.
30 10
0 497
18 10
32
5
Kb^4
NH+ OH
NH
=
[][]−
[] 3
NH 32 ++H ONH+− 4 OH
Kb
HB OH
HB]
+
=
[][ ]−
[
Equilibrium 221