Question

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 0.405 Msolution of pyridine? (Assume that the temperature is 25 ∘C.)

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 ionizes in water:

C6H5COOH⇌C6H5COO−+H+

The pKa of this reaction is 4.2. In a 0.51 M solution of benzoic acid, what percentage of the molecules are ionized?

Answer 1

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Pyridine dissociate as

C₅H₅N + H₂O C₅H₅NH⁺ + OH^-

K_b = [C₅H₅NH⁺] [OH^-] / [C₅H₅N]

pyridine is monobesic base therefore

[C₅H₅NH⁺] = [OH^-] = x

K_b = [x][x] /  [C₅H₅N]

K_b =[x]² /  [C₅H₅N]

[x]² = K_b  [C₅H₅N]

K_b = 10^-pKb = 10^-8.75 = 1.778 10^-9

[x]² = 1.778 10^-9 0.405 = 7.2 10^-10

[x] = 2.68 10^-5

[C₅H₅NH⁺] = [OH^-] = x = 2.68 10^-5

pOH = -log[OH^-] = -log(2.68 10^-5) = 4.57

pH = 14 - pOH = 14 - 4.57 = 9.43

pH of pyridine = 9.43

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benzoic acid dissociates as

C₆H₅COOH + H₂O C₆H₅COO^- + H₃O⁺

K_a = [C₆H₅COO^- ][H₃O⁺] / [C₆H₅COOH]

but  [C₆H₅COO^- ] = [H₃O⁺] = x

K_a = [x][x] / [C₆H₅COOH]

Ka = 10^-pKa = 10^-4.2 = 6.3 10^-5

Substitute the value in equation

6.3 10^-5 = [x]²/ 0.51

[x]² = 6.3 10^-5 0.51 = 3.22 10^-5

[x] = 0.00567 M

Concentration of H₃O⁺ = 0.00567 M

percentage ionization = [H₃O⁺] 100 / [C₆H₅COOH]

% ionization of benzoic acid = 0.00567 100 / 0.51 = 1.11 %

% ionization of benzoic acid  = 1.11%