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

The subject of this article is question #39 from the 2012 CHEM 13 NEWS Exam. Only 22% of students answered the question correctly and 52% of students did not answer it all. The question is reproduced below.

* The ionization of H_{3}PO_{4} to PO_{4}^{3-} involves three separate ionization reactions. The corresponding ionization constants are K_{a1}, K_{a2} and K_{a3}, respectively. In 1 mol L^{-1 }H_{3}PO_{4} (aq), the equilibrium concentration, in moles per litre, of one of the species in solution is approximately equal to K_{a2}. Which species has an equilibrium concentration approximately equal to 6.2×10^{-8} mol L^{-1}? * For H

_{3}PO

_{4; }

*K*

_{a1}= 6.9×10

^{-3};

*K*

_{a2}= 6.2×10

^{-8};

*K*

_{a3}= 4.8×10

^{-13}

**A** H^{+}

**B** H_{3}PO_{4 }

**C** H_{2}PO_{4}^{-}

**D** HPO_{4}^{2- }

**E** PO_{4}^{3-}

The ionization of H_{3}PO_{4} 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 H_{3}PO_{4}, H_{2}PO_{4}^{-}, HPO_{4}^{2- }and PO_{4}^{3-} when exactly one mole of H_{3}PO_{4} 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 H_{3}PO_{4} produces only a small amount of the singly-ionized form, H_{2}PO_{4}^{-}, 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

[H_{2}PO_{4}^{-}]_{eq} ≈ [H_{3}O^{+}]_{eq} and 1 mol L^{-1} ≈ [H_{3}PO_{4}]_{eq} + [H_{2}PO_{4}^{-}]_{eq}

The equilibrium concentrations of H_{2}PO_{4}^{-} and H_{3}O^{+} are approximately equal because the first ionization produces equal amounts of these two species. The equilibrium concentrations of H_{3}PO_{4} and H_{2}PO_{4}^{2-} add up to 1 mol L^{-1} because, if we ignore the second and third ionization steps, the dissolved H_{3}PO_{4} exists as either H_{3}PO_{4} or H_{2}PO_{4}^{-} 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: HPO_{4}^{2-} (D), the doubly-ionized form of H_{3}PO_{4}, has an equilibrium concentration approximately equal to 6.2×10^{-8} mol L^{-1}. The equilibrium concentration of HPO_{4}^{2-} 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 [H_{2}PO_{4}^{-}]eq ≈ 0.080 mol L^{-1}. Thus, we see that of the H_{2}PO_{4}^{-} generated by the first ionization step, only 100×(6.2×10^{-8}/ 0.080) = 7.8×10^{-5} percent ionizes to HPO_{4}^{2-}.

Finally, by substituting

[H_{2}PO_{4}^{-}] = 0.080 mol L^{-1} into the third expression, and rearranging, we obtain [PO_{4}^{3-}] ≈ *K*_{a2}*K*_{a3} / 0.080 = 3.7×10^{-19} mol L^{-1}. That is, of the HPO_{4}^{2-} generated by the second ionization, only 100×(3.7×10^{-19}/6.2×10^{-8}) = 6.0×10^{-10} percent ionizes to PO_{4}^{3-}. Clearly, neglecting the second and third ionization steps was justified.

The results of this analysis suggest that for **weak*** polyprotic acids*,

**. 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 H**

*the first ionization step determines the pH*_{3}O

^{+}in solution.