Answer:
1.34 × 10⁻⁵
Step-by-step explanation:
Let's consider the acid dissociation of propanoic acid.
CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)
We can find the acid dissociation constant (Ka) using an ICE chart.
CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)
I 0.65 0 0
C -x +x +x
E 0.65 - x x x
We know that [CH₃CH₂COO⁻] = [H₃O⁺] = x = 2.95 × 10⁻³ M
[CH₃CH₂COOH] = 0.65 - 2.95 × 10⁻³ = 0.64705 M
Ka = [CH₃CH₂COO⁻].[H₃O⁺]/[CH₃CH₂COOH]
Ka = (2.95 × 10⁻³)²/0.64705
Ka = 1.34 × 10⁻⁵