16.3-16.4: Base Ionization Equilibria-Acid/Base Properties of Salt Solutions

Base-Ionization Equilibria:

Equilibria involving weak bases are treated similarly to equilibria involving weak acids.

Example:
Ammonia ionizes in water as follows:

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH^- (aq)

In the general form, the weak base is replaced with B (remember acids were replaced with HA) as follows:

B (aq) + H₂O (l) ⇌ HB⁺ (aq) + OH^- (aq)

Base-Ionization Constant = K_b (yet another version of K_c)



Example:

In a solution of 2.35 M NH₃, what is the pH and pOH?

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH^- (aq)



K_b = (x²) / (2.35 - x) = 1.8 x 10^-5

Approximation holds, so:

x = 0.00650 = [OH^-]
pOH = -log(0.00650) = 2.19
pH = 14-pOH = 11.81

Acid-Base Properties of Salt Solutions:

A salt is the product of a neutralization (acid/base) reaction.

Parent Acid + Parent Base → Salt + Water

Example:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)
NaCl parents are HCl and NaOH (the cation is Na⁺ and the anion is Cl^-. Add OH^- to the cation and H⁺ to the anion to get the parents).

Example:

Salt: KC₂H₃O₂
Cation: K⁺
Anion: C₂H₃O₂^-
Parents: KOH, HC₂H₃O₂

Example:

Salt: MgBr₂
Cation: Mg²⁺
Anion: Br^-
Parents: Mg(OH)₂, 2HBr

Example:

Salt: LiCN
Cation: Li⁺
Anion: CN^-
Parents: LiOH, HCN

Consider a solution of sodium cyanide: NaCN



The second reaction can also be seen as a base ionization, and the K_b can easily be written from it.

The sodium ion is unreactive with water, but the cyanide ion reacts to produce HCN and OH^-. From the BrØnsted-Lowry point of view, CN^- acts as a base because it accepts a proton from H₂O. And since OH^- is a product, the solution will have a basic pH. This reaction is the hydrolysis of CN^-.

Hydrolysis: the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

Consider a solution with ammonium ion:

NH₄⁺ (aq) + H₂O (l) ⇌ NH₃ (aq) + H₃O⁺ (aq)

The ammonium ion acts as an acid, donating a proton to H₂O and forming NH₃. Since H₃O⁺ is a product, the solution will be acidic.

This reaction can also be seen as an acid ionization, and the K_a can easily be written from it.

Is the salt acidic, basic, or neutral? (How does the pH change?)

1. A salt of a strong base and a strong acid will be neutral, pH 7. The salt dissociates into ions in water that then react with water to form strong acids and bases, and we know from before that strong acids and bases dissociate 100% in water, so there is no net reaction and pH doesn't change.

2. A salt of a strong base and a weak acid will be basic, pH above 7.

3. A salt of a weak base and a strong acid will be acidic, pH below 7.

4. A salt of a weak base and weak acid is not as easy to predict. Both ions hydrolyze and the acidity or basicity depends on the relative strengths of the two ions. Must compare the K_a of the cation with the K_b of the anion. If K_a is larger, the solution is acidic. If K_b is larger, the solution is basic.

Example:
Salt: NH₄F

NH₄F → NH₄⁺ + F^-

NH₄⁺ + H₂O ⇌ H₃O⁺ + NH₃
What is the K_a for NH₃? (A hint that we need to find K_a and not K_b is H₃O⁺ is a product)
NH₃ is not on the list of K_a, but it is on the list of K_b in the text. And we know that:
K_w = 1 x 10^-14 = K_a x K_b, so: K_a = K_w / K_b

For NH₃, K_a = (1 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10

F^- + H₂O ⇌ OH^- + HF
What is the K_b for HF? (Need to find the K_b because OH^- is a product)
HF is not on the list of K_b in the text, but it is on the list of K_a.
K_b = K_w / K_a

For HF, K_b = (1 x 10^-14) / (6.8 x 10^-4) = 1.47 x 10^-11

K_a is greater than K_b here, so solution will be acidic.