5 Steps to a 5 AP Chemistry 2019

222 ❯ STEP 4. Review the Knowledge You Need to Score High

The equilibrium constant expression is called the weak base dissociation constant, Kb,

and has the form:

=

+ −

K

[HB][OH]

b [HB]

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

initial molarity and Kb of the weak base, the [OH-] can easily be calculated. And if the

initial molarity and [OH-] are known, Kb can be calculated.

For example, a 0.500 M solution of ammonia has a pH of 11.48. What is the Kb of

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

NHH 32 ++ON HO+− 4 H

0.500 - x x x

K

[NH][OH]

b [NH]

4

3

=

+−

[OH-] = [NH 4 +] = 3.0 × 10 -^3 M

[NH 3 ] = 0.500 - 3.0 × 10 -^3 = 0.497 M

=

×

=×

−

K (3.0 10 ) −

(0.497)

b 1.8^10

32

5

The Ka and Kb of conjugate acid–base pairs are related through the Kw expression:

Ka × Kb = Kw

This equation shows an inverse relationship between Ka and Kb for 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 23 HO 22 )S→+r(^2 +−aq)2CH 23 O( 2 aq)