Problem 50 - Chemistry: The Central Science in SI Units, Global Edition

Question

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{HSO}_{4}^{-}\), (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\).

Step-by-Step Solution

Answer

(a) Dissociation of \(\mathrm{HSO}_{4}^{-}\) with \(\mathrm{H}^{+}(a q)\) as a product and its \(K_a\) expression: \(\mathrm{HSO}_{4}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{SO}_{4}^{2-}(a q)\) \(K_a = \dfrac{[\mathrm{H}^{+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{HSO}_{4}^{-}]}\) (b) Dissociation of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) with \(\mathrm{H}^{+}(a q)\) as a product and its \(K_a\) expression: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}^{-}(a q) + \mathrm{H}^{+}(a q)\) \(K_a = \dfrac{[\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}]}\)

Step 1: (a) Dissociation of \(\mathrm{HSO}_{4}^{-}\) with \(\mathrm{H}^{+}(a q)\) as a product

\(\mathrm{HSO}_{4}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{SO}_{4}^{2-}(a q)\)

Step 2: (a) Dissociation of \(\mathrm{HSO}_{4}^{-}\) with hydronium ion

\(\mathrm{HSO}_{4}^{-}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(a q) + \mathrm{SO}_{4}^{2-}(a q)\)

Step 3: (a) Ka expression for \(\mathrm{HSO}_{4}^{-}\)

\(K_a = \dfrac{[\mathrm{H}^{+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{HSO}_{4}^{-}]}\)

Step 4: (b) Dissociation of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) with \(\mathrm{H}^{+}(a q)\) as a product

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}^{-}(a q) + \mathrm{H}^{+}(a q)\)

Step 5: (b) Dissociation of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) with hydronium ion

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}(l) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}^{-}(a q) + \mathrm{H}_{3}\mathrm{O}^{+}(a q)\)

Step 6: (b) Ka expression for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

\(K_a = \dfrac{[\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{O}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}]}\)

Key Concepts

Chemical Equations

When discussing chemical equations, especially those related to acid dissociation, it's important to understand how a chemical equation represents the reaction occurring in solution. A chemical equation provides a symbolic representation of a chemical reaction, illustrating the substances that react (reactants) and the substances produced (products). In the dissociation of acids, such as

• he hydrogen sulfate ion: \( \mathrm{HSO}_4^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_4^{2-}(aq) \),
• or phenol: \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}(l) \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}(aq) + \mathrm{H}^{+}(aq) \),

the chemical equation shows the clarity with which the acid (on the left) dissociates into its ions (on the right). For these reactions, two different notations may be used: one showing the production of the hydrogen ion (\(\mathrm{H}^{+}\)) and another showing the production of the hydronium ion (\(\mathrm{H}_3\mathrm{O}^{+}\)), which results from the interaction of hydrogen ions with water molecules. This clarifies how acids behave in water.

Equilibrium Constant

In chemistry, the equilibrium constant, denoted as \( K_a \) for acid dissociation reactions, helps us understand the extent to which an acid dissociates in an aqueous solution. The equation for the equilibrium constant is derived from the concentrations of the products and reactants at equilibrium. For example:

• In the dissociation of \(\mathrm{HSO}_4^{-}\), the \(K_a\) expression is \(K_a = \frac{[\mathrm{H}^{+}][\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^{-}]}\).
• Similarly, for phenol or \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\), it's \(K_a = \frac{[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}]}\).

This constant provides insight into the strength of an acid in solution. A larger \( K_a \) value suggests a stronger acid, indicating greater ionization in water. Understanding \( K_a \) helps chemists predict the behavior of acids in different conditions, such as changes in temperature or concentration, and their influence on reaction rates and dynamics.

Aqueous Solutions

Aqueous solutions are solutions where water acts as the solvent. They are crucial in understanding acid behavior because water not only facilitates the dissolution of compounds but also participates actively in acid dissociation reactions. In aqueous solutions, acids like the hydrogen sulfate ion \( \mathrm{HSO}_4^{-} \) or phenol \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \), dissociate to release ions. This dissociation can either produce hydrogen ions (\( \mathrm{H}^{+} \)) or form interaction complexes like the hydronium ion (\( \mathrm{H}_3\mathrm{O}^{+} \)).

• Hydrogen ions, when formed, imply a simple proton presence in the solution.
• Hydronium ions denote that the proton has bonded with a water molecule, forming \( \mathrm{H}_3\mathrm{O}^{+} \) which is a more accurate representation of what's happening in solution.

The interaction between acids and water is essential for understanding the nature and behavior of the resulting ions in aqueous environments, dictating the acid's strength and influence on the solution's properties like pH.