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Caproic acid, C5H11COOH, occurs naturally in coconut and palm oil. It is a weak monoprotic acid, with Ka=1.3x10-5. A certain aqueous solution of caproic acid has a pH of 2.94. How much acid was dissolved to make 100 mL of this solution?

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Answer 1

C₅H₁₁COOH <==> H⁺ + C₅H₁₁COO^- ... dissociation of caproic acid when in water

Ka = 1.3x10^-5 = [H⁺][C₅H₁₁COO^-] / [C₅H₁₁COOH]

From the pH of 2.94, we can find the [H+] and hence the [C₅H₁₁COO^-], since they are the same.

pH = - log [H+]

[H⁺] = 10^-2.94

[H⁺] = [C₅H₁₁COO^-] = 1.15x10^-3 M

C₅H₁₁COOH <==> H⁺ + C₅H₁₁COO^-

X.............................0...............0...........Initial

-Z...........................+Z.............+Z.........Change

X-Z..........................Z...............Z..........Equilibrium

[H⁺] = [C₅H₁₁COO^-] = 1.15x10^-3 M = Z

Ka = 1.3x10^-5 = [H⁺][C₅H₁₁COO^-] / [C₅H₁₁COOH]

1.3x10^-5 = (1.15x10^-3 M)² / [C₅H₁₁COOH]

[C₅H₁₁COOH] = (1.15x10^-3)² / 1.5x10^-5

[C₅H₁₁COOH] = 8.82x10^-2 M = concentration @ equilibrium

Initial concentration would be 8.82x10^-2 M + 1.15x10^-3 M (see ICE table) = 8.94x10^-2 M initial concentration

In 100 ml (0.1 L) you would have 8.94x10^-2 mol/L x 0.1 L = 8.94x10^-3 mols caproic acid in 100 mls

If you want the grams in 100 mls, simply multiply mols by molar mass of caproic acid.