PART A. Pyridine is a weak base that is used in the manufacture of pesticides an

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+H2OC5H5NH++OH

The pKb of pyridine is 8.75. What is the pH of a 0.220 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:

C6H5COOHC6H5COO+H+

The pKa of this reaction is 4.2. In a 0.62 M solution of benzoic acid, what percentage of the molecules are ionized?

Explanation / Answer

PART A

C5H5N + H2O C5H5NH+ + OH

Kb = [C5H5NH+] [OH-] / [ C5H5N]

[C5H5NH+] = [OH-]

So  [OH-]2 = Kb* [ C5H5N]

[OH-] = {Kb* [ C5H5N]}1/2

pKb = -log (Kb)

Kb = 10-pKb

Kb = 10-8.75

Kb = 1.778 x 10-9

[OH-] = [ 1.778 x 10-9 * 0.220] 1/2

[OH-] = 1.98 x 10-5 M

pOH = - log (1.98 x 10-5)

pOH = 4.7

So pH = 14 - 4.7 => 9.3

PART B

C6H5COOH C6H5COO + H+

Ka = [C6H5COO] [H+] / [ C6H5COOH]

[C6H5COO] = [H+]

So  [H+]2 = Ka* [C6H5COOH]

[H+] = {Ka* [C6H5COOH]}1/2

pKa = -log (Ka)

Ka = 10-pKa

Ka = 10-4.2

Ka = 6.31 x 10-5

[H+] = [6.31 x 10-5 * 0.62] 1/2

[H+] = 6.25 x 10-3 M

pH = - log (6.25 x 10-3)

pH = 2.2

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