The percent ionization solution of acetic acid at 25°C is 0.3 %. The answer is B.
At 25°C, the dissociation constant (Ka) for acetic acid is 1.75 × 10⁻⁵. The ionization reaction for acetic acid
CH₃COOH → CH₃COO⁻ + H⁺
initial concentration 2 - -
changing concentration - x + x + x
equilibrium concentration 2-x x x
[tex]Ka \:=\: \frac{[CH_3COO^-] \: [H^+]}{[CH_3COOH]}[/tex]
[tex]Ka \:=\: \frac{x \: x}{2-x}[/tex]
[tex]Ka \:=\: \frac{x^2}{2-x}[/tex]
Ka = x² ÷ (2-x)
1.75 × 10⁻⁵ = x² ÷ (2-x)
The magnitude of x is so small that x can be negligible.
1.75 × 10⁻⁵ = x² ÷ 2
1.75 × 10⁻⁵ × 2 = x²
x² = 3.50 × 10⁻⁵
[tex]x = \sqrt{3.50 \times 10^{- 5}}[/tex]
x = 5.92 × 10⁻³ M
The percentage ionization = (x ÷ 2) × 100%
= (5.92 × 10⁻³ ÷ 2) × 100%
= 2.96 × 10⁻³ × 100%
= 0.296%
= 0.3%
B
Learn more about the dissociation constant here: https://brainly.com/question/22668939
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