A 0.10 M solution of a weak acid, HA, is found to be 1.50% ionized. Calculate K_a for this acid.

Question

Quizplus · Accepted Answer

Given that the acid is 1.50% ionized, this means that 1.50% of the initial concentration of the acid (0.10 M) dissociates into its ions. Therefore, the concentration of $H^+$ and $A^-$ ions produced is $0.10 \times \frac{1.50}{100} = 0.0015 \, M$. The equilibrium expression for the dissociation of a weak acid $HA$ is $K_a = \frac{[H^+][A^-]}{[HA]}$. Since the acid is only partially dissociated, the concentration of $HA$ at equilibrium is approximately the initial concentration minus the concentration that dissociated, which is $0.10 - 0.0015 = 0.0985 \, M$, but for simplicity in weak acid calculations, we can approximate the denominator as the initial concentration of the acid because the change is very small. Therefore, $K_a = \frac{(0.0015)(0.0015)}{0.10} = 2.25 \times 10^{-5}$, which rounds to $2.3 \times 10^{-5}$, matching choice C.

A 010 M Solution of a Weak Acid, HA, Is Found