Question

\(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}\) functions as a monobasic acid in \(\mathrm{H}_{2} \mathrm{SO}_{4}\). (a) Write an equation to show what happens when \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}\) dissolves in \(\mathrm{H}_{2} \mathrm{SO}_{4}\). (b) Assess the strength of \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}\) as an acid given that the ionization constant is \(1.4 \times 10^{-2}\)

Short answer

(a) \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \rightleftharpoons \mathrm{HSO}_{3}^{+} + \mathrm{HSO}_{4}^{-} \). (b) \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) is a strong acid.

Step-by-step solution

Dissolution in Sulfuric Acid

To understand what happens when \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) (disulfuric acid) dissolves in sulfuric acid, we need to depict the equilibrium reaction. \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) dissociates to form \( \mathrm{HSO}_{3}^{+} \) and \( \mathrm{HSO}_{4}^{-} \). The chemical equation for this process is: \[ \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \rightleftharpoons \mathrm{HSO}_{3}^{+} + \mathrm{HSO}_{4}^{-}. \]

Interpret the Ionization Constant

The ionization constant, \( K_a \), of a substance tells us about its strength as an acid. A larger \( K_a \) value indicates a stronger acid because it signifies a higher degree of ionization. Here, the ionization constant of \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) is given as \( 1.4 \times 10^{-2} \). This relatively large value indicates that \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) is a strong acid, especially when compared to weak acids that typically have much smaller \( K_a \) values, often less than \( 10^{-5} \).

Key concepts

Acid Dissociation

Acid dissociation is the process by which an acid releases protons into a solution. When disulfuric acid (\( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \)) dissolves in sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)), it undergoes dissociation to form ions. This is an essential concept in chemistry, as it helps to understand how acids behave in different environments.

• In the dissolution process of \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) in \( \mathrm{H}_{2} \mathrm{SO}_{4} \), the acid breaks down into two parts: \( \mathrm{HSO}_{3}^{+} \) and \( \mathrm{HSO}_{4}^{-} \).
• The appearance of these ions marks the acid’s dissociation in the solution.

This dissociation is represented by the equilibrium reaction equation: \[ \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \rightleftharpoons \mathrm{HSO}_{3}^{+} + \mathrm{HSO}_{4}^{-}\] The double arrow indicates that the reaction can go back and forth, showing that it can reach an equilibrium state under certain conditions.

Acid Strength

Acid strength is gauged by how well an acid can donate protons. A stronger acid dissociates more completely in solution, resulting in more free ions.

• The strength of an acid is often discussed in terms of its ability to ionize in a solvent, such as water or sulfuric acid.
• For \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \), a relatively high ionization constant \( (K_a) \) of \( 1.4 \times 10^{-2} \) suggests it is quite strong compared to other acids with much smaller \( K_a \) values.

This means \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) more readily loses a proton, making it potent in protonating other substances or contributing to reactions.

Ionization Constant

The ionization constant, represented as \( K_a \), gives us quantitative information about acid dissociation.

• It tells us the extent to which an acid can dissociate into its ions in a particular solvent.
• A larger \( K_a \) value indicates the acid is more willing to dissociate and release protons, thus being a stronger acid.

For \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \), with a \( K_a \) of \( 1.4 \times 10^{-2} \), it shows that this acid ionizes significantly, supporting its strong acid characteristics. Understanding \( K_a \) is crucial as it allows chemists to predict how substances will behave in various chemical reactions, particularly in determining equilibrium states.

Equilibrium Reaction

An equilibrium reaction is a state in which the rates of the forward and reverse reactions are equal, leading to no net change in concentrations of reactants and products. This concept is crucial in understanding reversible reactions like acid dissociation in solutions.

• The dissociation of \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \) is a perfect example, where the forward reaction produces ions, and the reverse reaction allows these ions to recombine into the acid.
• In equilibrium, although the reactants and products are constantly being converted into each other, their overall concentrations remain stable over time.

Visualizing this through the equation \( \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7} \rightleftharpoons \mathrm{HSO}_{3}^{+} + \mathrm{HSO}_{4}^{- }\) helps to realize that both the acid and its ions coexist in the solution, dynamically interacting without losing balance. This understanding aids in predicting how changes in conditions like concentration, pressure, or temperature can shift this equilibrium.