Calculate the percent ionization of 1.55 m aqueous acetic acid solution. for acetic acid, ka=1.8⋅10−5.

Answer is: percent ionization of aqueous acetic acid solution is 0.341%.
c(CH₃COOH) = 1.55 M.
Ka(CH₃COOH) = 1.8·10⁻⁵.
Chemical reaction: CH₃COOH ⇄ CH₃COO⁻ + H⁺.
[CH₃COO⁻] = [H⁺] = x.
[CH₃COOH] = 1.55 M - x.
Ka = [CH₃COO⁻] · [H⁺] / [CH₃COOH].
1.8·10⁻⁵ = x² / (1.55 M - x).
Solve quadratic equation: x = 0.0053 M.
percent of ionization:
α = 0,0053 M ÷ 1.55 M · 100% = 0,341%

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