Question

Part A. Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows:

C5H5N+H2O⇌C5H5NH++OH−

The pKb of pyridine is 8.75. What is the pH of a 0.140 M solution of pyridine?

Part B. Benzoic acid is a weak acid that has antimicrobial properties. Its sodium salt, sodium benzoate, is a preservative found in foods, medications, and personal hygiene products. Benzoic acid dissociates in water:

C6H5COOH⇌C6H5COO−+H+

The pKa of this reaction is 4.2. In a 0.69 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

PART A:

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH^-

K_b = [C₅H₅NH⁺] [OH^-] / [C₅H₅N]
pKb = - log Kb

Or, Kb= 10^^-pKb
Or, Kb = 10^-8.75
Or, Kb = 1.78 x 10^-9

C₅H₅N + H₂O ⇌ C₅H₅NH⁺ + OH^-
initial concentration: 0.14 0 0
Let x = change in [C₅H₅N]

At equilibrium:   0.14 –x x x
Putting the values in the expression for Kb,
1.78 x 10^-9 = x² / (0.14 –x)

neglecting x in 0.14–x we get,

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

We know,
[H+][OH-] = 1.0 x 10^-14
Or, [H+] = 1.0 x 10^-14 / 1.58 x 10^-5 = 6.34 x 10^^-10
Or, pH = - log[H+] = - log (6.34 x 10^-10) = -(-9.2)
Or, pH = 9.2

So the pH is 9.2

PART B:
C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺

Ka = [C₅H₅COO-] [H+] / [C₅H₅COOH]
pKa = - log Ka

Or, Ka= 10^^-pKa
Or, Ka = 10^^-4.2
Or, Ka = 6.31 x 10^-5

C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺

initial concentrations: 0.69 0 0

Equilibrium concentrations 0.69-x x x Let x = change in C₆H₅COOH

Putting the values in the expression for Ka,
6.31x 10^-5 = x² / (0.69 –x)

neglecting x in 0.69–x we get,

x = 6.6 x 10^-3 = [H+]
Percent Ionization = ([H+] / 0.69 )x100 = 0.956