The triprotic acid H3A has ionization constants of Ka1 = 5.4 Solution would choo

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would choose 2.02 as the answer and here is how you figure that out. As each proton is lost from the triprotic acid, the molecule becomes more electron rich, and less electron withdrawing. Therefore, the loss of each proton strengthens the acid-H bond and increases the pK a of the species. This trend is evident in the pK a data given. In general, it is true that K a1 > K a2 > K a3, and so on, for polyprotic acids. As you may have guessed, calculating the pH of a polyprotic acid solution is not as simple as it is for monoprotic acids. In fact, it is quite a messy problem. However, that mess can be quickly cleaned up by making the assumption that only the strongest acid (i.e. only the first dissociation) has a significant effect on the pH. Making that assumption turns the problem into one much easier to solve--calculating the pH of a weak acid solution. Weak acids do not completely dissociate in aqueous solution but are in equilibrium with their dissociated forms. Therefore, we must apply what we know about equilibria to solve these types of problems. Lets solve your problem looking at just the first pK a. To do this, we first write down the equilibrium involved and the expression for the equilibrium constant: Ka = [H+][H2A-] / [H3A] =x^2 / (.1-x) = 1x10^-3 Next, you should compile a table of values for the concentration of all species involved in the equilibrium. We already know that the initial concentration, [ ]o, of acetic acid is 0.10 M and that the initial concentration of H+ is 10-7 (since the solvent is neutral water). Even though there is an initial concentration of H+ in solution, it is so small compared to the amount produced by the acid that it is usually ignored. From the stoichiometry of the reaction, one mole of H+ and one mole of H2A- are produced for each mole of H3A dissociated. Therefore, if we denote the amount of H3A dissociated as x, the final concentrations of H+ and H2A- are both x, and the final concentration of H3A in solution is 0.10 M - x. This data is summarized in the equation above. You should note that the above equation is a quadratic equation in x and therefore requires use of the quadratic equation to solve for x. If, however, we can make the assumption that [H3A]o - x = [H3A]o, the equation becomes much easier to solve. We can do this if H3A is a weak enough acid that it dissociates very little, and the change in [H3A] is negligible. We can make the approximation [HA]o - x = [HA]o so long as x is less than 5% of the initial concentration of HA. X will be greater than 5% of [HA]o with stronger weak acids at low concentrations. Consider these guidelines when you decide whether or not to make the approximation: you can simplify the quadratic equation if the solution is more concentrated than 0.01 M and the pK a of the acid is greater than 3. So we have x^2/.1 = .001, x^2 = .0001, x = .01 Now that we know H+ =.01 we can take the negative log to get pH: -log(.01) = 2. Were a little bit off because of our oversimplification, but the answer should be 2.02, your first option.

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