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) Ionization of \(\mathrm{HBrO}_{2}\): 1) With \(\mathrm{H}^{+}\): \(\mathrm{HBrO}_{2}(aq) \rightleftharpoons \mathrm{BrO}_{2}^{-}(aq) + \mathrm{H}^{+}(aq)\) 2) With hydronium ion: \(\mathrm{HBrO}_{2}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{BrO}_{2}^{-}(aq) + \mathrm{H}_{3} \mathrm{O}^{+}(aq)\) \(K_{a}\) expression: \(K_{a} = \frac{[\mathrm{BrO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{HBrO}_{2}]}\) (b) Ionization of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\): 1) With \(\mathrm{H}^{+}\): \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(aq) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(aq) + \mathrm{H}^{+}(aq)\) 2) With hydronium ion: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(aq) + \mathrm{H}_{3} \mathrm{O}^{+}(aq)\) \(K_{a}\) expression: \(K_{a} = \frac{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}]}\)

Step-by-step solution

Write the chemical equation with H+ as the product

For the first case, consider \(\mathrm{HBrO}_{2}\) in aqueous solution and the dissociation reaction as follows: \[ \mathrm{HBrO}_{2}(aq) \rightleftharpoons \mathrm{BrO}_{2}^{-}(aq) + \mathrm{H}^{+}(aq) \]

Write the chemical equation with the hydronium ion

Now, let's write the dissociation equation with the hydronium ion as a product: \[ \mathrm{HBrO}_{2}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{BrO}_{2}^{-}(aq) + \mathrm{H}_{3} \mathrm{O}^{+}(aq) \]

Write the Ka expression

The \(K_{a}\) expression for the ionization of \(\mathrm{HBrO}_{2}\) can be written as: \[ K_{a} = \frac{[\mathrm{BrO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{HBrO}_{2}]} \] (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\)

Write the chemical equation with H+ as the product

For the first case, consider \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) in aqueous solution and the dissociation reaction as follows: \[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(aq) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(aq) + \mathrm{H}^{+}(aq) \]

Write the chemical equation with the hydronium ion

Now, let's write the dissociation equation with the hydronium ion as a product: \[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}(aq) + \mathrm{H}_{3} \mathrm{O}^{+}(aq) \]

Write the Ka expression

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

Key concepts

Chemical Equilibrium

When a weak acid dissolves in water, it doesn't fully ionize into its constituent ions. Instead, a dynamic equilibrium is established between the non-ionized acid and the ions it forms. This is a state where the rate at which the acid molecules lose protons (deprotonate) to form ions is equal to the rate at which the ions gain protons (reprotonate) to form the acid molecules. At equilibrium, the concentration of reactants and products remains constant over time.

Understanding this concept is crucial in predicting how the system will react to changes in conditions, which is governed by Le Chatelier's Principle. Chemical equilibrium is a fundamental concept in chemical reactions, including acid-base chemistry, implying that acid dissociation reactions are reversible, and that equilibrium can be described quantitatively using an equilibrium constant.

Ka Expression

The acid ionization constant, known as Ka, is a quantitative measure of the strength of an acid in solution. It is expressed as the equilibrium constant for the dissociation of the acid to produce the hydronium ion (H3O+) and the conjugate base of the acid. The general form of the Ka expression is:

\[K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}\]

where [HA] is the concentration of the non-ionized acid, [A-] is the concentration of the conjugate base, and [H+] is the concentration of hydrogen ions. For weak acids, the Ka value is small, indicating a lesser degree of ionization. Chemical equilibrium for weak acid ionization can thus be represented by a Ka expression, which allows us to predict the behavior and properties of the solutions they form.

Weak Acid Dissociation

Weak acids, unlike strong acids, do not completely dissociate in water. Instead, they exist as a mixture of undissociated acid molecules and the ions into which they can dissociate. The degree of dissociation of weak acids is represented by the acid dissociation constant (Ka).

For example, with the weak acid HBrO2 in water, the dissociation can be represented as follows:

\[HBrO_{2}(aq) \rightleftharpoons BrO_{2}^{-}(aq) + H^{+}(aq)\]

The position of the equilibrium can provide insight into the acid strength; a larger equilibrium constant (Ka) indicates a stronger acid. The dissociation process of weak acids is an essential factor in buffer solutions, affecting the pH stability.

Hydronium Ion Concentration

In an aqueous solution, the presence of hydrogen ions, H+, is often represented by the more accurate form of hydronium ions, H3O+, because hydrogen ions are highly reactive and tend to associate with water molecules. The concentration of hydronium ions in a solution is a determining factor for its acidity and is measured by the pH scale. The pH of the solution is inversely related to the hydronium ion concentration; a lower pH indicates a higher concentration of H3O+ ions.

Through the dissociation of weak acids, like C2H5COOH, the concentration of hydronium ions is established as seen in this equilibrium reaction:

\[C_{2}H_{5}COOH(aq) + H_{2}O(l) \rightleftharpoons C_{2}H_{5}COO^{-}(aq) + H_{3}O^{+}(aq)\]

Determining the hydronium ion concentration in solution is essential for calculating pH and understanding the acid's impact on chemical reactions and biological systems.