Question

When phosphoric acid is titrated with asolution, only two stoichiometric points are seen. Why?

Short answer

The reason for observation of two stoichiometric points during phosphoric acid titrated with NaOH solution can be discussed.

Step-by-step solution

Dissociation of H3PO4

The formula for phosphoric acid is , which means donation of three protons and make it triprotic acid. The dissociation can be explained as:

$$\begin{gathered} {\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{P}{{\mathrm{O}}_4}^{-}\left(\mathrm{aq}\right) {\mathrm{K}}_{\mathrm{a}1}=7.5\times {10}^{-3} \\ {\mathrm{H}}_2\mathrm{P}{{\mathrm{O}}_4}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+\mathrm{HP}{{\mathrm{O}}_4}^{2-}\left(\mathrm{aq}\right) {\mathrm{K}}_{\mathrm{a}2}=6.2\times {10}^{-8} \\ \mathrm{HP}{{\mathrm{O}}_4}^{2-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+\mathrm{P}{{\mathrm{O}}_4}^{3-}\left(\mathrm{aq}\right) {\mathrm{K}}_{\mathrm{a}3}=4.8\times {10}^{-13} \end{gathered}$$

Stoichiometric points during titration of phosphoric acid

In the reaction of phosphoric acid with strong base (), there is an observation of two breaks at stoichiometric points at K_a1and K_a2. The values of and are large enough compared tovalue. These large values show the stoichiometric point breaks. Due to the small value of K_a3 no stoichiometric break point is seen. Therefore, two stoichiometric points during phosphoric acid titration with NaOH can be seen.