Question

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.

Short answer

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

Step-by-step solution

Write the Dissociation Equations for each proton

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: \[K_{a1} = 5.5 \times 10^{-3}\] \[K_{a2} = 1.7 \times 10^{-7}\] \[K_{a3} = 5.1 \times 10^{-12}\]

Calculate the concentration of H⁺ ions

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, \[[H^{+}] = \sqrt{(5.5 \times 10^{-3}) \times 0.20} \approx 0.033 \text{ M}\]

Calculate the concentrations of the other species

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. \[[OH^{-}] = \frac{K_{w}}{[H^{+}]}\] \[[OH^{-}] = \frac{1.0 \times 10^{-14}}{0.033} \approx 3.03 \times 10^{-14} \text{ M}\]

Results

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

Key concepts

Acid Dissociation Constant

The acid dissociation constant, known as \(K_a\), measures an acid's strength by showing how well it dissociates into ions in water. For triprotic acids like arsenic acid \((\text{H}_3\text{AsO}_4)\), there are three dissociation constants: \(K_{a1}\), \(K_{a2}\), and \(K_{a3}\). These correspond to the ionization of each proton. With values of \(K_{a1} = 5.5 \times 10^{-3}\), \(K_{a2} = 1.7 \times 10^{-7}\), and \(K_{a3} = 5.1 \times 10^{-12}\), we see that the first proton dissociates much more readily than the second or third. This means that the majority of \(\text{H}^+\) ions in a solution of arsenic acid come from the first dissociation.

In practice, understanding \(K_a\) values helps predict the concentrations of various species in solution. It tells us which species will prevail at certain stages of ionization. For strong acids, \(K_a\) is large, indicating almost complete ionization, while for weak acids, it's small, indicating little ionization.

For students, knowing how to use \(K_a\) to calculate \(\text{H}^+\) concentration prepares you to solve real-world problems involving chemical reactions in solution.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, maintaining constant concentrations of reactants and products over time. In the context of arsenic acid dissociation, each step reaches equilibrium, described by the dissociation constants \(K_{a1}\), \(K_{a2}\), and \(K_{a3}\). These constants reveal how far the equilibrium lies toward the products.

When \(K_a1 \gg K_a2 \gg K_a3\), the first equilibrium is more favored, indicating more dissociation of the first proton compared to the others. At equilibrium, the concentration of each species remains stable, allowing us to solve for unknown concentrations.

To find these concentrations, we can use the expressions derived from each step's equilibrium state. For example, in the first step of arsenic acid's dissociation, the equation \(\frac{[\text{H}^+][\text{H}_2\text{AsO}_4^-]}{[\text{H}_3\text{AsO}_4]} = K_{a1}\) is used to find unknown concentrations at equilibrium. Understanding equilibrium is crucial for predicting how different conditions impact a reaction, helping illustrate the dynamic balance in chemical systems.

Acid-Base Equilibria

Acid-base equilibria involve acids donating protons and bases accepting them, establishing a balance of \, \(\text{H}^+\) \, and \, \(\text{OH}^-\) \, ions in solution. In solutions of triprotic acids like arsenic acid, this equilibrium involves multiple dissociations. Each dissociation reaches a distinct equilibrium described by its own constant—\(K_{a1}\), \(K_{a2}\), and \(K_{a3}\).

To analyze such systems, it's essential to consider each equilibrium step. Typically, the initial ionization (first \(K_a\)) contributes most to the \, \([\text{H}^+]\) \, concentration. We simplify complex problems by treating only the first dissociation as significant for estimating \, \([\text{H}^+]\). \, Further dissociations have minor impacts due to much smaller \(K_a\) values.

Working with acid-base equilibria also involves understanding the self-ionization of water, where \, \([\text{H}^+][\text{OH}^-] = K_w = 1.0 \times 10^{-14}\). This relationship helps us calculate hydroxide ion concentration \, \([\text{OH}^-]\) \, when needed, providing a full picture of the acid's effect on the solution. Mastery of acid-base equilibria is foundational for understanding reactions in biology, environmental science, and industrial processes.