# Tough question series – 2012 CHEM 13 NEWS Exam

(This is a reprint from the March 2013 issue of Chem 13 News, pages 6-7.)

The ionization of H3PO4 to PO43- involves three separate ionization reactions. The corresponding ionization constants are Ka1, Ka2 and Ka3, respectively. In 1 mol L-1 H3PO4 (aq), the equilibrium concentration, in moles per litre, of one of the species in solution is approximately equal to Ka2. Which species has an equilibrium concentration approximately equal to 6.2×10-8 mol L-1 For H3PO4; Ka1 = 6.9×10-3Ka2 = 6.2×10-8Ka3 = 4.8×10-13

A    H+

B    H3PO

C   H2PO4-

D   HPO42-

E    PO43-

The ionization of H3PO4 involves the following steps, each of which involves the transfer of a proton (H+) from a weak acid to water.

An exact solution to this equilibrium problem (i.e. finding the equilibrium concentrations of H3PO4, H2PO4-, HPO42- and PO43- when exactly one mole of H3PO4 is dissolved in water to make exactly one litre of solution) is a complicated problem. However, by assuming that the first ionization reaction occurs to the greatest extent, an approximate solution is readily obtained. In addition, we can also readily verify whether or not the assumption is valid.

If we assume that the first ionization of H3PO4 produces only a small amount of the singly-ionized form, H2PO4-, then the subsequent ionizations are not going to be very important.  Focusing only on the first ionization, and ignoring for the moment the subsequent ionizations (and the self-ionization of water), we can write

[H2PO4-]eq  ≈ [H3O+]eq and 1 mol L-1 ≈ [H3PO4]eq + [H2PO4-]eq

The equilibrium concentrations of H2PO4- and H3O+ are approximately equal because the first ionization produces equal amounts of these two species.  The equilibrium concentrations of H3PO4 and H2PO42- add up to 1 mol L-1 because, if we ignore the second and third ionization steps, the dissolved H3PO4 exists as either H3PO4 or H2PO4- at equilibrium.  We can use these two results to write the following expressions.

The second expression reveals the answer to the question that was originally asked:  HPO42- (D), the doubly-ionized form of H3PO4, has an equilibrium concentration approximately equal to 6.2×10-8 mol L-1. The equilibrium concentration of HPO42- is relatively small, evidence supporting the assumption that neither the second nor the third ionization step occurs to an appreciable extent.

It is instructive to assess the validity of the assumption in a more quantitative manner. To do this, we can rearrange the first expression and then solve it using the quadratic formula to obtain the result [H2PO4-]eq ≈ 0.080 mol L-1.  Thus, we see that of the H2PO4- generated by the first ionization step, only 100×(6.2×10-8/ 0.080) = 7.8×10-5 percent ionizes to HPO42-

Finally, by substituting
[H2PO4-] = 0.080 mol L-1 into the third expression, and rearranging, we obtain [PO43-] ≈ Ka2Ka3 / 0.080 = 3.7×10-19 mol L-1.  That is, of the HPO42- generated by the second ionization, only 100×(3.7×10-19/6.2×10-8) = 6.0×10-10 percent ionizes to PO43-.  Clearly, neglecting the second and third ionization steps was justified.

The results of this analysis suggest that for weak polyprotic acids, the first ionization step determines the pH.  Therefore, the calculation of pH is typically no more complicated than it is for a weak monoprotic acid. The reason is that, for a weak polyprotic acid, the first ionization produces only a small amount of the singly-ionized form. Thus, the amounts available for subsequent ionizations are typically too small to make a significant contribution to the total amount of H3O+ in solution.