The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the stems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0356 m solution of oxalic acid?
First we have to find Ka1 and Ka2 pKa1 = - log Ka1 so Ka1 = 0.059 pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵ Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid. Oxalic acid is H₂C₂O₄ H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻ 0.0356 M 0 0 0.0356 - x x x Ka1 = [tex] \frac{[H^+][HC2O4^-]}{[H2C2O4]} [/tex] = x² / 0.0356 - x x = 0.025 M pH = - log [H⁺] = - log (0.025) = 1.6
