CHM 1046

CHM 1046
General Chemistry II
Dr. Michael Blaber

Acid-Base Equilibria

Weak Bases

Weak bases in water react to release a hydroxide (OH^-) ion and their conjugate acid:

Weak Base(aq) + H₂O(l) ó Conjugate Acid(aq) + OH^-(aq)

A common weak base is ammonia (NH₃). Its conjugate acid is the ammonium ion (NH₄⁺):

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

Since the H₂O is the liquid state, and the concentration effectively doesn't change:

K_b is the base-dissociation constant (analogous to the K_a dissociation constant for acids)

K_b always refers to the equilibrium in which a base reacts with H₂O to form the conjugate acid and OH^-

The structures of weak bases contain one or more lone pairs of electrons

A lone pair is necessary to form a bond with an H⁺ ion (producing the conjugate acid and releasing OH^- from H₂O)
The lone pairs are typically located on Nitrogen atoms in the base

Why Nitrogen? Recall that for weak acids, the atom that the hydrogen is bonded to should have a high electronegativity value that promotes "taking" the shared electrons in the bond with hydrogen. Thus, a hydrogen that is covalently bonded to an oxygen (a highly electronegative element) will loose its electron when ionized (the hydrogen leaves as a proton, H⁺). In the base structures above we see Nitrogen as the atom bonded to hydrogen. Nitrogen has a lower electronegativity compared to oxygen (but still slightly higher than hydrogen, however). Therefore, any hydrogen that bonds to the nitrogen will exhibit less tendency to ionize and leave as H⁺. In other words, such compounds tend to accept H⁺ ions, but don't tend to release them - a characteristic of a base. Elements with electronegativity values lower than hydrogen are all metals. Metal hydrides release hydrogen as the H^- ion (the Hydrogen keeps the electrons in the bond). Hydride will react with water to produce OH^- and H₂(g).

Other weak bases are actually the anion forms of weak acids (i.e. the conjugate bases of weak acids)

(note: lone electron pairs to bond with H⁺ are located on O atoms in this case)
Types of Weak Bases

How can you tell from the formula whether a compound will behave as a weak base?

There are two general categories of weak bases:

1. Neutral compounds containing a non-bonding pair of electrons that can form a bond with a proton (i.e. can function as an acceptor of a proton)

Most of this type of base contains nitrogen (and it's the lone pair on the N that accepts the proton)
These include ammonia (NH₃) and related compounds called amines

An amine is like ammonia, exept one or more of the N-H bonds are replaced by a N-C bond. For example, the compound methylamine has the structure:

Amines can extract a proton from H₂O by forming a covalent bond with the lone pair of electrons on the Nitrogen atom

2. The second category includes anions of weak acids

The compound sodium hypochlorite (NaClO) dissociates to yield Na⁺ and ClO^- ions

In acid/base neutralization reactions, the Na⁺ ion would be a spectator ion

However, the ClO^- ion is the conjugate base of HClO (hypochlorous acid), a weak acid
Therefore, the ClO^- ion can potentially extract a proton from H₂O:

ClO^-(aq) + H₂O(l) ó HClO(aq) + OH^-(aq)

This is the definition of a base, and therefore ClO^- is a weak base in water

Pyridine is a weak base with a K_b = 1.7 x 10^-9. What is the pH of a 0.05M solution of pyridine in H₂O?

C₅H₅N(aq) + H₂O(l) ó C₅H₅NH⁺(aq) + OH^-(aq)

K_b = [C₅H₅NH⁺] [OH^-] / [C₅H₅N] = 1.7 x 10^-9

The initial concentration of C₅H₅N = 0.05M

The initial concentration of OH^-(aq) is 1 x 10^-7M for pure H₂O. However, this may be quite small compared to the concentration produced by the ionization of the weak base, so we will ignore it.

The initial concentration of C₅H₅NH⁺ = 0M

If X M = conc of C₅H₅N that ionizes, then from the stoichiometry, X M of OH^- and X M of C₅H₅NH⁺ are produced. Therefore:

K_b = 1.7 x 10^-9 = [C₅H₅NH⁺] [OH^-] / [C₅H₅N] = X² / (0.05 - X)

8.5 x 10^-11 - 1.7 x 10^-9X = X²

X² + 1.7 x 10^-9X - 8.5 x 10^-11 = 0

This is a quadratic:

a = 1, b = 1.7 x 10^-9, c = -8.5 x 10^-11

X = 0.92 x 10^-5M or

X = -0.92 x 10^-5M

X can't be negative, therefore [OH^-] = 0.922 x 10^-5M

pOH = -log(0.922 x 10^-5) = 5.04

Recall pH = 14.0 - pOH

pH = 14.0 - 5.04

pH = 8.96

This is a basic solution (pH > 7), as expected.

Did we have to use the quadratic? Let's assume that the amount that dissociates (X) is small in comparison to the initial concentration (0.05M)

K_b = 1.7 x 10^-9 = [C₅H₅NH⁺] [OH^-] / [C₅H₅N] = X² / (0.05 - X) » X² / (0.05 )

1.7 x 10^-9 * 0.05 = X²

X = 0.922 x 10^-5

This is the same answer from the quadratic. Also, the starting concentration of weak base divided by the value of the K_b is > 100

0.05 / 1.7 x 10-9 = 29,411,764 which is >100

Thus, the shortcut to the quadratic that we saw for weak acid calculations can also be used in weak base calculations

A solution of ammonia (NH₃) has a pH of 9.2. The K_b of ammonia is 1.8 x 10^-5. What is the molar concentration of the weak base?

pOH = 14.0 - pH

pOH = 14.0 - 9.2 = 4.8

4.8= -log([OH^-])

[OH^-] = 1.58 x 10^-5 M

The balanced equation of the ionization is:

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

Therefore, the stoichiometry is such that for each mole of OH^- produced, 1 mole of NH₃ was ionized, and 1 mole of NH₄⁺ conjugate ion was also produced.

If X = starting concentration of ammonia:

At equilibrium we have

[NH₃] = (X - 1.58 x 10^-5)

[OH^-] = 1.58 x 10^-5

[NH₄⁺] = 1.58 x 10^-5

K_b = 1.8 x 10^-5 = (1.58 x 10^-5)² / (X - 1.58 x 10^-5)

1.8 x 10^-5 = 2.51 x 10^-10 / (X - 1.58 x 10^-5)

X - 1.58 x 10^-5= 2.51 x 10^-10 / 1.8 x 10^-5

X - 1.58 x 10^-5= 1.40 x 10^-5

3.0 x 10^-5 M