Question

A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionization of the acid? 19% 0.42% 42% 0.19%

Answer 1

Answer:

0.42%

Explanation:

∵ pH = - log[H⁺].

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

∵ [H⁺] = √Ka.C

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

∵ Ka = α²C.

Where, α is the degree of dissociation.

∴ α = √(Ka/C) = √(8.065 x 10⁻⁶/0.45) = 4.234 x 10⁻³.

∴ percentage ionization of the acid = α x 100 = (4.233 x 10⁻³)(100) = 0.4233% ≅ 0.42%.

Answer 2

The answer is 0.42% just did it ;)