Question

If the pH of a phosphoric acid (H3PO4) solution is adjusted to 6.50, what is the most abundant species and which is the second most abundant species? For H3PO4 Ka1 = 7.11 x 10-3, Ka2 = 6.34 x 10-8, and Ka3 = 4.22 x 10-13.

Answer 1

Explanation:

For the given values of $K_{a}$ we will have the values of $pK$ as follows.

As,        $pK_{a} = -log K_{a}$

Therefore,

     $pK_{a1}$ = 2.15,     $pK_{a2}$ = 7.20

      $pK_{a3}$ = 12.38

Now, at pH 6.50

      $H_{3}PO_{4} \rightarrow H_{2}PO^{-}_{4} + H^{+}$;   $K_{a1}$

At pH = 2.15;  $H_{3}PO_{4} = H_{2}PO^{-}_{4}$

       $H_{2}PO^{-}_{4} \rightarrow HPO^{2-}_{4} + H^{+}$;  $K_{a2}$

At pH 7.20;  $H_{2}PO^{-}_{4} = HPO^{2-}_{4}$

        $HPO^{2-}_{4} \rightarrow PO^{3-}_{4} + H^{+}$;     $K_{a3}$

Hence, we can conclude that most abundant species is $H_{2}PO^{-}_{4}$ and the second most abundant species is $HPO^{2-}_{4}$.