What is the acid dissociation constant, Ka of the 0.150 mol dm–3 solution of weak acid HA with pH of 2.34?

First, It would be useful to write the equation for the dissociation of the weak acid HA, which is HA <--> H⁺+ A^-. Then, write the Ka expression of the weak acid HA, which is [H⁺][A^-]/[HA]. We know that [HA] = 0.150 mol dm^-3 as the concentration is given in the question. We also know the pH of HA is 2.34, we can find [H⁺] as pH = -log₁₀[H⁺] = 2.34. To arrange this equation, [H⁺] = 10^-pH= 10^-2.34 = 4.57 x 10^-3mol dm^-3. As this is a weak acid, it means that HA is only weakly dissociated, so only a very small amount of HA is dissociated into H⁺ and A^- ions. Therefore, we can make the assumption that [H⁺]=[A^-] = 4.57 x 10^-3mol dm^-3. Now, we have found all the concentrations we need, [H⁺], [A^-] and [HA], we can substitute these values into the Ka expression, so Ka = [H⁺][A^-]/[HA] = (4.57 x 10^-3)² / 0.150 = 1.39 x 10^-4 mol dm^-3 (3 s.f.).