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

(a) For \(\mathrm{HBrO}_{2}\): Ionization reaction: \(\mathrm{HBrO}_{2}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\) and \(\mathrm{HBrO}_{2}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\) \(K_{a}\) expression: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\) and \(K_{a} = \frac{[\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\) (b) For \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\): Ionization reaction: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\) \(K_{a}\) expression: \(K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\) and \(K_{a} = \frac{[\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\)

Step-by-step solution

Write the formula for the ionization of \(\mathrm{HBrO}_{2}\) as a reaction with \(\mathrm{H}^{+}(a q)\) as a product

To write the formula, we need to show how the acid (\(\mathrm{HBrO}_{2}\)) dissociates into its ions in aqueous solution. The formula for this reaction is: \[\mathrm{HBrO}_{2}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\]

Write the \(K_{a}\) expression for the ionization of \(\mathrm{HBrO}_{2}\) with \(\mathrm{H}^{+}(a q)\) as a product

Now, we write the \(K_{a}\) expression for the above equilibrium reaction. The expression should include the products' concentration divided by the reactant's concentration. The formula for the \(K_{a}\) expression is: \[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\]

Write the formula for the ionization of \(\mathrm{HBrO}_{2}\) as a reaction with the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) as a product

Now, let's rewrite the ionization reaction using the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) as a product. The formula for this reaction is: \[\mathrm{HBrO}_{2}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q) + \mathrm{BrO}_{2}^{-}(a q)\]

Write the \(K_{a}\) expression for the ionization of \(\mathrm{HBrO}_{2}\) with the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) as a product

Finally, write the \(K_{a}\) expression for this equilibrium reaction. The expression should include product's concentrations divided by reactant's concentration. The formula for the \(K_{a}\) expression is: \[K_{a} = \frac{[\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\] (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\)

Write the formula for the ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) as a reaction with \(\mathrm{H}^{+}(a q)\) as a product

To write the formula, we need to show how the acid (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\)) dissociates into its ions in aqueous solution. The formula for this reaction is: \[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\]

Write the \(K_{a}\) expression for the ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) with \(\mathrm{H}^{+}(a q)\) as a product

Now, write the \(K_{a}\) expression for the above equilibrium reaction. The expression should include the products' concentration divided by the reactant's concentration. The formula for the \(K_{a}\) expression is: \[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\]

Write the formula for the ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) as a reaction with the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) as a product

Now, let's rewrite the ionization reaction using the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) as a product. The formula for this reaction is: \[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(a q) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q) + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)\]

Write the \(K_{a}\) expression for the ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) with the hydronium ion (\(\mathrm{H}_{3} \mathrm{O}^{+}\)) as a product

Finally, write the \(K_{a}\) expression for this equilibrium reaction. The expression should include product's concentrations divided by reactant's concentration. The formula for the \(K_{a}\) expression is: \[K_{a} = \frac{[\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\]

Key concepts

Ionization Reaction

Ionization reactions occur when an acid dissociates into ions in a solution. This is an important concept because it's how we understand acids in water. In simple terms, an acid breaks apart, losing a hydrogen ion (or proton) and forming its corresponding anion.
For example, in the ionization of \(\mathrm{HBrO}_{2}\), the acid breaks down as follows:\(\mathrm{HBrO}_{2}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{BrO}_{2}^{-}(aq)\).

• \(\mathrm{H}^{+}(aq)\) is the hydrogen ion, indicating the acidic property.
• \(\mathrm{BrO}_{2}^{-}(aq)\) is the resultant anion, in this case, a bromite ion.

The ionization reaction also involves the solvent, often water, especially when hydronium ions \((\mathrm{H}_{3} \mathrm{O}^{+}(aq))\) are produced instead of free hydrogen ions. Many find it helpful to picture water molecules capturing the released \(\mathrm{H}^{+}\) to form the \(\mathrm{H}_{3} \mathrm{O}^{+}\) ion, a more accurate representation of the acid's interaction in water.

Equilibrium Constant

The equilibrium constant, \(K_a\), gives us a measure of the strength of an acid in solution. It's a way to quantify how far an ionization reaction proceeds. Mathematically, \(K_a\) is expressed using the concentrations of the products and reactants at equilibrium.
For the ionization reaction of \(\mathrm{HBrO}_{2}\), the equilibrium expression is written as:\[K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\]

• The higher the \(K_a\) value, the stronger the acid, meaning it dissociates more in solution.
• A low \(K_a\) indicates a weak acid, with less dissociation.

Whenever hydronium ions are formed, the expression modifies slightly:\[K_{a} = \frac{[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{BrO}_{2}^{-}]}{[\mathrm{HBrO}_{2}]}\]This slight change mirrors the involvement of water molecules in the reaction, which are often considered constant in dilute solutions.

Aqueous Solution

An aqueous solution is one where water serves as the solvent. It’s critical for dissolving acids, which then ionize. Water's polar nature facilitates this process by surrounding and stabilizing ions as they form.

• When an acid is placed in water, it dissociates into hydrogen ions \(\mathrm{H}^{+}(aq)\) or \(\mathrm{H}_{3}\mathrm{O}^{+}(aq)\) and its corresponding anion.
• The role of water is vital because it stabilizes these ions, allowing the reaction to reach equilibrium.

For both \(\mathrm{HBrO}_{2}(aq)\) and \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH}(aq)\), the ions formed are part of the aqueous solution ensemble. Remember, under typical lab conditions, we assume all ions are in the aqueous phase unless stated otherwise.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They display reactants transforming into products. In the context of acids in water, they show ionization steps.
A chemical equation for an acid like \(\mathrm{HBrO}_{2}\) will be:\[\mathrm{HBrO}_{2}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{BrO}_{2}^{-}(aq)\]

• The arrow \(\rightleftharpoons\) signals that this is a reversible reaction, typical for weak acids.
• In another view with hydronium ions: \(\mathrm{HBrO}_{2}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{O}^{+}(aq) + \mathrm{BrO}_{2}^{-}(aq)\)

This notation makes balancing, understanding, and predicting reactions more transparent. Formulas are the key to tracking particle transformations through different phases of reactions.