Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5.5 \times\) \(10^{-3}, K_{\mathrm{a}_{2}}=1.7 \times 10^{-7}\), and \(K_{\mathrm{a}_{3}}=5.1 \times 10^{-12}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}{ }^{3-}\right]\) in a \(0.20-M\) arsenic acid solution.

The dissociation of each proton from arsenic acid can be described by three separate chemical reactions: 1) \[H_{3}AsO_{4} \rightleftharpoons H^{+} + H_{2}AsO_{4}^{-}\] 2) \[H_{2}AsO_{4}^{-} \rightleftharpoons H^{+} + HAsO_{4}^{2-}\] 3) \[HAsO_{4}^{2-} \rightleftharpoons H^{+} + AsO_{4}^{3-}\] We have been given the dissociation constant for each of these reactions: \[K_{a1} = 5.5 \times 10^{-3}\] \[K_{a2} = 1.7 \times 10^{-7}\] \[K_{a3} = 5.1 \times 10^{-12}\]

Since \(K_{a1} \gg K_{a2} \gg K_{a3}\), most of the H⁺ ions will come from the dissociation of the first proton (Reaction 1). Therefore, we treat H₃AsO₄ as a weak monoprotic acid and find the concentration of H⁺ ions using the formula: \[[H^+] = \sqrt{K_{a1} \times [\text{H}_{3}\text{AsO}_{4}]}\] Given, the concentration of arsenic acid is 0.20 M. Thus, \[[H^{+}] = \sqrt{(5.5 \times 10^{-3}) \times 0.20} \approx 0.033 \text{ M}\]

With the concentration of H⁺ ions, we can now find the concentrations of the other species in the solution. For the first dissociation reaction: \[\frac{[H^+][H_{2}AsO_{4}^{-}]}{[H_{3}AsO_{4}]}=K_{a1}\] \[[H_{2}AsO_{4}^{-}] = \frac{K_{a1} [H_{3}AsO_{4}]}{[H^+]}\] \[[H_{2}AsO_{4}^{-}] = \frac{(5.5\times10^{-3})(0.20)}{0.033}\approx0.033 \text{ M}\] For the second dissociation reaction: \[\frac{[H^+][HAsO_{4}^{2-}]}{[H_{2}AsO_{4}^{-}]}=K_{a2}\] \[[HAsO_{4}^{2-}] = \frac{K_{a2} [H_{2}AsO_{4}^{-}]}{[H^+]}\] \[[HAsO_{4}^{2-}] = \frac{(1.7\times10^{-7})(0.033)}{0.033}\approx1.7 \times 10^{-7} \text{ M}\] For the third dissociation reaction: \[\frac{[H^+][AsO_{4}^{3-}]}{[HAsO_{4}^{2-}]}=K_{a3}\] \[[AsO_{4}^{3-}] = \frac{K_{a3} [HAsO_{4}^{2-}]}{[H^+]}\] \[[AsO_{4}^{3-}] = \frac{(5.1\times10^{-12})(1.7\times10^{-7})}{0.033}\approx 2.6 \times 10^{-12} \text{ M}\] Finally, we find the concentration of OH⁻ ions to satisfy the equation \([H^{+}][OH^{-}] = K_{w}\), where \(K_w = 1.0 \times 10^{-14}\) is the ion product of water. \[[OH^{-}] = \frac{K_{w}}{[H^{+}]}\] \[[OH^{-}] = \frac{1.0 \times 10^{-14}}{0.033} \approx 3.03 \times 10^{-14} \text{ M}\]

The concentrations of all species in the 0.20 M arsenic acid solution are: \[[H^{+}] \approx 0.033 \text{ M}\] \[[OH^{-}] \approx 3.03 \times 10^{-14} \text{ M}\] \[[H_{3}AsO_{4}] \approx 0.20 - 0.033 \approx 0.167 \text{ M}\] \[[H_{2}AsO_{4}^{-}] \approx 0.033 \text{ M}\] \[[HAsO_{4}^{2-}] \approx 1.7 \times 10^{-7} \text{ M}\] \[[AsO_{4}^{3-}]\approx 2.6 \times 10^{-12} \text{ M}\]