Question

Write out the stepwise \(K_{\mathrm{a}}\) reactions for the diprotic acid \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

Short answer

The stepwise Ka reactions for the diprotic acid H2SO3 are: 1. First ionization reaction: \( H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \) with Ka1 = \( \frac{[H^+][HSO_3^-]}{[H_2SO_3]} \) 2. Second ionization reaction: \( HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \) with Ka2 = \( \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]} \)

Step-by-step solution

Identify the first ionization reaction of H2SO3

The first ionization reaction occurs when H2SO3 loses one proton (H+). The chemical equation for the first ionization reaction is: \( H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \)

Write out the Ka expression for the first ionization

The acid dissociation constant, Ka, measures the strength of an acid in solution. It is the equilibrium constant for the reaction of an acid with water. For the first ionization reaction, the Ka expression is: \( K_{a1} = \frac{[H^+][HSO_3^-]}{[H_2SO_3]} \)

Identify the second ionization reaction of H2SO3

After the first ionization reaction, HSO3- can also lose a proton (H+), resulting in the second ionization reaction. The chemical equation for the second ionization reaction is: \( HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \)

Write out the Ka expression for the second ionization

Now, we need to write the Ka expression for the second ionization reaction: \( K_{a2} = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]} \) So, the stepwise Ka reactions for the diprotic acid H2SO3 are: 1. First ionization reaction: \( H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \) with Ka1 = \( \frac{[H^+][HSO_3^-]}{[H_2SO_3]} \) 2. Second ionization reaction: \( HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \) with Ka2 = \( \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]} \)

Key concepts

Acid Dissociation Constant (Ka)

The acid dissociation constant, abbreviated as Ka, is a crucial concept in acid-base chemistry. It serves as a quantitative measure of an acid's strength—that is, its ability to donate protons (H+) to the solution. The higher the value of Ka, the more strongly the acid dissociates, which corresponds to a stronger acid. Conversely, a lower Ka value indicates a weaker acid since it implies less dissociation.

To mathematically express the strength of an acid, especially a diprotic acid that can release two protons, we consider the ionization reactions in a stepwise manner. Each step has its own Ka value, representing the acid's strength at that particular stage of ionization.

First Ionization Reaction

The first ionization reaction of a diprotic acid involves the release of the initial proton (H+). In the example of sulfuric acid (H₂SO₃), this is represented by the chemical equation:

\( H_2SO_3 \rightleftharpoons H^+ + HSO_3^- \).

At this juncture, only one proton has been released, which makes the remaining ion, HSO₃^-, the bisulfite ion. Understanding this step is fundamental to grasping how diprotic acids behave in an aqueous solution, as the acid goes through a sequential loss of hydrogen ions, each with its equilibrium reaction affecting overall acidity.

Second Ionization Reaction

Following the first ionization, the second ionization reaction considers the bisulfite ion (HSO₃^-) acting as an acid itself and losing a proton. The equation is commonly written in the form:

\( HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \).

The species SO₃^2- formed is the sulfite ion. Since the first proton is more easily donated, the first ionization constant (Ka1) is generally larger than the second ionization constant (Ka2). This trend suggests that the first dissociation step is more favorable, while the second ionization is less so, depicting a common characteristic of polyprotic acids.

Equilibrium Constant

The concept of the equilibrium constant plays a pivotal role when discussing ionization reactions in acids. It applies not only to diprotic acids but to all reversible chemical reactions that reach a state of dynamic equilibrium. The equilibrium constant, which for acid dissociation is expressed as Ka, represents the ratio of the concentrations of the products to the reactants at equilibrium, raised to the power of their stoichiometric coefficients.

The equilibrium expressions for the sequential ionizations of a diprotic acid illustrate the idea that as one moves through the ionization steps, the strength of the acid (its tendency to lose a proton) decreases. This is explained by the Le Chatelier's Principle where a system at equilibrium, when disturbed, will adjust to minimize that disturbance. Therefore, as the concentration of hydrogen ions increases with the first ionization, it becomes less likely for the second ionization to proceed, reflecting in a smaller Ka value for the second ionization.