Question

Farmers who raise cotton once used arsenic acid, H₃AsO₄, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with Ka₁ = 2.5 × 10⁻⁴, Ka₂ = 5.6 × 10⁻⁸, and Ka₃ = 3 × 10⁻¹³. What is the pH of a 0.500 M solution of arsenic acid?

Answer 1

Explanation:

The reaction equation will be as follows.

      $H_{3}AsO_{4} \rightleftharpoons H_{2}AsO^{-}_{4} + H^{+}$

Hence, the expression for $K_{a}$ is as follows.

            $K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}$

Let us assume that the concentration of both $[H_{2}AsO^{-}_{4}]$ and $[H^{+}]$ is x.

           $2.5 \times 10^{-4} = \frac{x \times x}{0.5}$

                          x = 0.01118034

This means that the concentration of $[H^{+}]$ is 0.01118034.

Since, we know that the relation between pH and concentration of hydrogen ions is as follows.

              pH = $-log [H^{+}]$

                    = $-log (0.01118034)$

                    = 1.958

Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.