Question

Write the chemical equation and the \(K_{a}\) expression for the ionization 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) \(\mathrm{HBrO}_{2}\), (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\).

Short answer

For the ionization of the acids: (a) \(\mathrm{HBrO}_{2}\): Chemical equation with \(\mathrm{H}^{+}(a q)\): \(\mathrm{HBrO}_{2}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\) \(K_{a}\) expression: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\) With hydronium ion: \(\mathrm{HBrO}_{2}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\): Chemical equation with \(\mathrm{H}^{+}(a q)\): \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\) \(K_{a}\) expression: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\) With hydronium ion: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\)

Step-by-step solution

(a) Ionization of \(\mathrm{HBrO}_{2}\) with \(\mathrm{H}^{+}(a q)\)

The ionization of \(\mathrm{HBrO}_{2}\) in water is represented by the following chemical equation: \(\mathrm{HBrO}_{2}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\)

(a) \(K_{a}\) expression for \(\mathrm{HBrO}_{2}\)

The \(K_{a}\) expression for the ionization of \(\mathrm{HBrO}_{2}\) is given by: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\)

(a) Ionization of \(\mathrm{HBrO}_{2}\) with hydronium ion

The same ionization reaction with the hydronium ion instead of \(\mathrm{H}^{+}\) can be written as: \(\mathrm{HBrO}_{2}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\)

(b) Ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) with \(\mathrm{H}^{+}(a q)\)

The ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) in water is represented by the following chemical equation: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\)

(b) \(K_{a}\) expression for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\)

The \(K_{a}\) expression for the ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) is given by: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\)

(b) Ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) with hydronium ion

The same ionization reaction with the hydronium ion instead of \(\mathrm{H}^{+}\) can be written as: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\)

Key concepts

ionization equation

When an acid ionizes in an aqueous solution, it releases hydrogen ions (H⁺) or hydronium ions (\( ext{H}_3 ext{O}^+\)) into the solution. This process is often represented by an ionization equation. Consider the ionization of \( ext{HBrO}_2\): it can be depicted in two ways, using either \( ext{H}^+\) ions or \( ext{H}_3 ext{O}^+\) ions as the product.

• The equation with \( ext{H}^+\) is: \( ext{HBrO}_2(aq) \rightleftharpoons ext{H}^+(aq) + ext{BrO}_2^-(aq)\). In this equation, the \( ext{H}^+\) ion is shown directly as being released from \( ext{HBrO}_2\).
• Alternatively, water can play a role in the ionization, forming hydronium ions. The equation becomes: \( ext{HBrO}_2(aq) + ext{H}_2 ext{O}(l) \rightleftharpoons ext{H}_3 ext{O}^+(aq) + ext{BrO}_2^-(aq)\). Here, the \( ext{H}_3 ext{O}^+\) ion represents the same process, just including water in the ionization.

This distinction is important since \( ext{H}_3 ext{O}^+\) more accurately describes the proton's association with a water molecule in reality. Thus, both forms are correct, but the hydronium form is more commonly used in aqueous solutions.

acid dissociation constant

The acid dissociation constant, denoted as \(K_a\), measures the strength of an acid in a solution. For any acid ionization, \(K_a\) is expressed as the concentration ratio of the products to the reactants. Let's consider the ionization of \( ext{HBrO}_2\) as an example.In the ionization equation \( ext{HBrO}_2(aq) \rightleftharpoons ext{H}^+(aq) + ext{BrO}_2^-(aq)\), the \(K_a\) expression is:\[K_a = \frac{[\text{H}^+][\text{BrO}_2^-]}{[\text{HBrO}_2]}\]

• The products \([\text{H}^+]\) and \([\text{BrO}_2^-]\) are in the numerator.
• The reactant \([\text{HBrO}_2]\) is in the denominator.

A larger \(K_a\) value indicates a stronger acid, as it means more acid molecules are dissociated into ions, producing a higher concentration of \(\text{H}^+\). Conversely, a smaller \(K_a\) signifies a weaker acid. Recognizing \(K_a\) helps in comparing acid strengths and predicting the behavior of acids in different chemical reactions.

hydronium ion

Hydronium ions, represented as \( ext{H}_3 ext{O}^+\), form when a hydrogen ion reacts with a water molecule. This reaction is essential in aqueous solutions where acids are present. Unlike the isolated hydrogen ion, which is simply a proton, the hydronium ion is more stable in liquid water due to its association with water.

• The reaction: \( ext{H}^+(aq) + ext{H}_2 ext{O}(l) \rightarrow ext{H}_3 ext{O}^+(aq)\)
• This process highlights water's role as a solvent that stabilizes ions.

Understanding the hydronium ion's function helps clarify acid-base chemistry in aqueous solutions.The presence of \( ext{H}_3 ext{O}^+\) reflects an aqueous system's acidity, playing a role in defining pH levels. The higher the concentration of \( ext{H}_3 ext{O}^+\), the more acidic the solution. In acid ionization equations, showing hydronium ions captures the reality of solution chemistry more accurately than simply using \( ext{H}^+\).

proton transfer reaction

Proton transfer reactions are fundamental in acid-base chemistry, involving the movement of a proton (\( ext{H}^+\)) from one molecule to another. These reactions underpin the concept that acids donate protons, while bases accept them.

• In the ionization of \( ext{HBrO}_2\), for example, a proton is transferred from \( ext{HBrO}_2\) to a water molecule, forming \( ext{H}_3 ext{O}^+\).
• Such reactions are vital for understanding how acids and bases interact in different media.

In practice, the proton doesn't exist freely in solution. Instead, it is transferred to a nearby water molecule, forming hydronium ions. This transfer is often represented in equilibrium's forward and backward reactions, indicating the dynamic nature of the process.The Lewis and Brønsted–Lowry acid-base theories provide frameworks for these reactions, giving a broad understanding:

• Brønsted–Lowry theory defines acids as proton donors and bases as proton acceptors.
• Lewis theory highlights the pair of electrons in forming bonds during these transfer reactions.

Thus, proton transfer reactions form the core of many chemical processes and help explain the changes in pH that occur in solutions.