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.

1.34 × 10⁻⁵

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

gets less

ammonium chloride & tallow