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sleet_krkn [62]
3 years ago
11

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 ionizati

on of the acid? 19% 0.42% 42% 0.19%
Chemistry
2 answers:
aleksandr82 [10.1K]3 years ago
5 0

The answer is 0.42% just did it ;)

aleksandr82 [10.1K]3 years ago
4 0

Answer:

0.42%

Explanation:

<em>∵ pH = - log[H⁺].</em>

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

<em>∵ [H⁺] = √Ka.C</em>

∴ [H⁺]² = Ka.C

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

<em>∵ Ka = α²C.</em>

Where, α is the degree of dissociation.

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

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

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