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.).