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Hydrolysis of Bases

>> Parts of this equation/concept include:

A -	Strong Bases
B -	Weak Bases

Bases form OH^– rather than H₃O⁺. They can do this in two ways. If the hydroxide ion is combined with a metal as a salt, the reaction is the dissociation of the salt into its component ions. If it is a nitrogen base, the hydroxide is created by removing an H⁺ from water (leaving OH^–). The H⁺ then bonds to the lone pair of the nitrogen.

The strong bases are the group I hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) and the lower group II hydroxides [Ca(OH)₂, Sr(OH)₂, Ba(OH)₂]. Consequently, the reaction is the stoichiometric production of ions.

>> Example 1

What is the pOH and pH of 0.0902 M KOH?

Solution:

KOH K⁺ + OH^–

0.0902 mol KOH L

1 mol OH– 1 mol KOH

=

0.0902 mol OH–/L

pOH = –log[0.0902] = 1.045

pH = 14.00 – 1.045 = 12.95

>> Example 2

What is the pOH and pH of 0.15 M Ba(OH)₂?

Solution:

Ba(OH)₂ Ba²⁺ + 2 OH^–

0.15 mol Ba(OH)2 L

2 mol OH– 1 mol Ba(OH)2

=

0.30 M OH–

pOH = –log[0.30] = 0.52

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Weak bases can also be hydroxides salts in equilibrium with their ions. This type of equilibrium is addressed in the section on insoluble salts. The nitrogen bases react with water in an equilibrium reaction, accepting an H⁺ from the water. The equilibrium constant for these reactions is K_b.

The general formula for the equilibrium reaction is

B + H₂O BH⁺ + OH^–

>> Example 3

What is the pOH and pH of 0.17 M NH₃?

Solution:

NH₃ + H₂O NH₄⁺ + OH^–

Kb

=

[NH4+][OH–] [NH3]

=

1.8 x 10–5

Using the ICE table,

	NH3	NH4+	OH–
Initial	0.17	0	0
Change	–x	+x	+x
Equilibrium	0.17 – x	x	x

Using these values in the K_b equation

1.8 x 10–5

=

(x)(x) (0.17 – x)

Assuming x is small,

1.8 x 10–5

=

x2 (0.17)

1.7 x 10^–3 = x

Checking the assumption, 0.17 – 0.0017 = 0.1683 = 0.17, it is okay,

x = [OH^–] = 1.7 x 10^–3

pOH = 2.77

pH = 11.23

>> Example 4

What is the pOH and pH of a 0.35 M C₆H₅NH₂ solution?

Solution:

C₆H₅NH₂ + H₂O C₆H₅NH₃⁺ + OH^–

Kb

=

[C6H5NH3+][OH–] [C6H5NH2]

=

4.3 x 10–10

Using the ICE table,

	C6H5NH2	C6H5NH3+	OH–
Initial	0.35	0	0
Change	–x	+x	+x
Equilibrium	0.35 – x	x	x

4.3 x 10–10

=

(x)(x) (0.35 – x)

Assuming x is small,

4.3 x 10–10

=

x2 (0.35)

1.2 x 10^–5 = x = [OH^–]

Check the assumption, 0.35 – 1.2 x 10^–5 = 0.35; it's good.

pOH = 4.91

pH = 9.09

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