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Determine the Ka value of propanoic acid if it is found that 2.95x10-3 M of each conjugate exists when a 0.65 M solution is made.

User Neoteknic
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1 Answer

5 votes

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

User Tom Hunter
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