Question

Write the ionization equation and the \(K_{\mathrm{a}}\) for each of the following acids. (a) \(\mathrm{AsH}_{4}{ }^{+}\) (b) \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

Short answer

Question: Write the ionization equations and the \(K_{\mathrm{a}}\) expressions for each of the following acids: (a) \(\mathrm{AsH}_{4}{ }^{+}\), (b) \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}\), and (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}\). Answer: (a) Ionization equation: \(\mathrm{AsH}_{4}{ }^{+}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{AsH}_{3}(aq)\) \(K_{\mathrm{a}}\) expression: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{AsH}_{3}]}{[\mathrm{AsH}_{4}{ }^{+}]}\) (b) Ionization equation: \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}(aq)\) \(K_{\mathrm{a}}\) expression: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}]}{[\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}]}\) (c) Ionization equation: \(\mathrm{H}_{2} \mathrm{SO}_{3}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{3}^{-}(aq)\) \(K_{\mathrm{a}}\) expression: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{SO}_{3}]}\)

Step-by-step solution

Write the ionization equation for each acid

We need to show how each acid dissociates in water, which involves the acid losing a proton (\(\mathrm{H}^{+}\)) and the remaining structure becoming a conjugate base. (a) For \(\mathrm{AsH}_{4}{ }^{+}\), the ionization equation is: \(\mathrm{AsH}_{4}{ }^{+}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{AsH}_{3}(aq)\) (b) For \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}\), the ionization equation is: \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}(aq)\) (c) For \(\mathrm{H}_{2} \mathrm{SO}_{3}\), the ionization equation is: \(\mathrm{H}_{2} \mathrm{SO}_{3}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{3}^{-}(aq)\)

Write the \(K_{\mathrm{a}}\) expression for each acid

The \(K_{\mathrm{a}}\) expression for each acid describes the equilibrium constant for the ionization reaction, with the concentrations of the products (ions) in the numerator and the concentration of the acid in the denominator. (a) For \(\mathrm{AsH}_{4}{ }^{+}\), the \(K_{\mathrm{a}}\) expression is: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{AsH}_{3}]}{[\mathrm{AsH}_{4}{ }^{+}]}\) (b) For \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}\), the \(K_{\mathrm{a}}\) expression is: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}]}{[\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}]}\) (c) For \(\mathrm{H}_{2} \mathrm{SO}_{3}\), the \(K_{\mathrm{a}}\) expression is: \(K_{\mathrm{a}} = \dfrac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{SO}_{3}]}\)

Key concepts

Acid Dissociation Constant

Understanding the strength of an acid depends greatly on its acid dissociation constant, denoted as \(K_{\mathrm{a}}\). This constant gives us an insight into the extent to which an acid can donate protons (or \(\mathrm{H}^{+}\)) ions when dissolved in water.

The \(K_{\mathrm{a}}\) is derived from the equilibrium constant for the acid ionization reaction equation, which shows the balance between the undissociated acid and the ions formed when an acid ionizes in solution. For each of the given acids, the ionization equation is written, followed by setting up the \(K_{\mathrm{a}}\) expression, which takes the form: \[K_{\mathrm{a}} = \frac{{[\mathrm{H}^{+}] [\text{{conjugate base}}]}}{{[\text{{acid}}]}}\]
Where the square brackets denote the molar concentrations of the reactants and products at equilibrium. Higher \(K_{\mathrm{a}}\) values indicate a stronger acid because it signifies a greater concentration of ions and, consequently, a greater degree of ionization.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, meaning there is no net change in the concentrations of reactants and products over time. It's a dynamic balance, not a static one, as reactions constantly occur in both directions.

In the context of acid ionization, equilibrium is represented by the double arrow in the ionization equation, signifying that the acid can dissociate to form \(\mathrm{H}^{+}\) and its conjugate base, and these can also recombine to form the original acid.

Importance of Equilibrium in Acid-Base Reactions

By examining chemical equilibrium, we can predict the direction in which a reaction will shift when conditions are changed (Le Châtelier's Principle). This is essential for understanding buffer solutions in biological and chemical systems, as well as in many industrial processes.

Conjugate Base

When an acid donates a proton during the ionization process, it forms its conjugate base. This conjugate base is crucial because it has the potential to accept a proton and revert back to its acid form. Therefore, it plays an important role in the acid-base equilibrium.

The nature of the conjugate base affects the acid's strength. A weak acid has a strong conjugate base that holds on to the \(\mathrm{H}^{+}\) weakly, making it more likely to release the proton, while a strong acid has a weak conjugate base that is unlikely to re-accept a \(\mathrm{H}^{+}\). As we see in the examples:

• For \(\mathrm{AsH}_{4}{ }^{+}\), the conjugate base is \(\mathrm{AsH}_{3}\).
• For \(\mathrm{H}_{2} \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{-}\), the conjugate base is \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}\).
• And for \(\mathrm{H}_{2} \mathrm{SO}_{3}\), the conjugate base is \(\mathrm{HSO}_{3}^{-}\).

Understanding the interplay between acids and their conjugate bases is essential to grasp the behavior of solutions and the mechanisms of acid-base reactions.