Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with K1 = 2.5 × 10–4, K2 = 5.6 × 10–8, and K3 = 3 × 10–13. What is the pH of a 0.500 M solution of arsenic acid?

pH = 1.95

Explanation:

For polyprotic acids, it is generally assumed that all H⁺ comes from the 1st ionization step. The amount of H⁺ delivered into solution for the 2nd and 3rd ionization steps are in the order of 10⁻⁴M and 10⁻⁶M respectively and provide very little change in pH from the quantity delivered in the 1st ionization step.

Therefore... the [H⁺] concentraion and pH are computed as follows...

[H⁺] = √Ka₁[H₃AsO₄] = √(2.5 x 10⁻⁴)(0.500) M = 0.1118M

pH = -log[H⁺] = -log(0.01118) = 1.95

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