A certain carboxylic acid has a K_a of 2.7 × 10^-5. What is the pH of a 0.10 M solution of this acid?

Question

Quizplus · Accepted Answer

K_a = [H+][A-]/[HA] For any weak acid, the reaction can be represented as follows: HA + H2O ⇌ H3O+ + A- Therefore, the equilibrium constant (K_a) can also be written as: K_a = [H3O+][A-]/[HA] At equilibrium, the concentration of the acid (HA) that has not dissociated is (0.10 - [A-]). Therefore, we can rewrite the equation as: K_a = [H3O+][A-]/(0.10 - [A-]) Assuming x moles of acid have dissociated, we can write: [H3O+] = x mol/L [A-] = x mol/L Substituting these values, we get: 2.7 × 10^-5 = (x)(x)/(0.10 - x) Solving for x, we get x = 1.35 × 10^-3 Therefore, [H3O+] = [A-] = 1.35 × 10^-3 pH = -log[H3O+] = 2.78

A Certain Carboxylic Acid Has a Ka of 2

A Certain Carboxylic Acid Has a K_a of 2