Question

Write the dissociation reaction and the corresponding \(K_{\mathrm{a}}\) equilibrium expression for each of the following acids in water. a. \(\mathrm{HCN}\) b. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Short answer

The dissociation reactions and corresponding Ka expressions for the given acids are: a. HCN: Reaction: \[ \mathrm{HCN} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H_3 O^+} + \mathrm{CN^-} \] Ka Expression: \[ K_{\mathrm{a}} = \frac{[\mathrm{H_3 O^+}][\mathrm{CN^-}]}{[\mathrm{HCN}]} \] b. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\): Reaction: \[ \mathrm{HOC}_{6} \mathrm{H}_{5} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H_3 O^+} + \mathrm{OC}_{6} \mathrm{H}_{5}^- \] Ka Expression: \[ K_{\mathrm{a}} = \frac{[\mathrm{H_3 O^+}][\mathrm{OC}_{6} \mathrm{H}_{5}^-]}{[\mathrm{HOC}_{6} \mathrm{H}_{5}]} \] c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\): Reaction: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H_3 O^+} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_2 \] Ka Expression: \[ K_{\mathrm{a}} = \frac{[\mathrm{H_3 O^+}][\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_2]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}]} \]

Step-by-step solution

Dissociation Reaction and Ka Expression for HCN

a. HCN Dissociation Reaction: \[ \mathrm{HCN} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H_3 O^+} + \mathrm{CN^-} \] Ka Expression: \[ K_{\mathrm{a}} = \frac{[\mathrm{H_3 O^+}][\mathrm{CN^-}]}{[\mathrm{HCN}]} \]

Dissociation Reaction and Ka Expression for HOC6H5

b. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) Dissociation Reaction: \[ \mathrm{HOC}_{6} \mathrm{H}_{5} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H_3 O^+} + \mathrm{OC}_{6} \mathrm{H}_{5}^- \] Ka Expression: \[ K_{\mathrm{a}} = \frac{[\mathrm{H_3 O^+}][\mathrm{OC}_{6} \mathrm{H}_{5}^-]}{[\mathrm{HOC}_{6} \mathrm{H}_{5}]} \]

Dissociation Reaction and Ka Expression for C6H5NH3+

c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) Dissociation Reaction: \[ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} + \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H_3 O^+} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_2 \] Ka Expression: \[ K_{\mathrm{a}} = \frac{[\mathrm{H_3 O^+}][\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_2]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}]} \]

Key concepts

Equilibrium Expression

In chemistry, understanding equilibrium expressions is essential for analyzing how reactions proceed and reach a state of balance. An equilibrium expression refers to the mathematical equation that represents the concentrations of reactants and products in a chemical reaction when equilibrium is achieved.
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This balance maintains constant concentrations of reactants and products, although they are not necessarily equal.

• The equilibrium expression shows the ratio of concentrations of products raised to their stoichiometric coefficients to those of reactants.
• It provides a quantifiable measure of the position of equilibrium, indicating whether the forward or reverse reaction is favored.

For instance, if we take the generic equation \(aA + bB \rightleftharpoons cC + dD\), the equilibrium expression, \(K\), is given by:\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]
This expression assists in determining how different conditions can affect the equilibrium state in chemical reactions.

Dissociation Reaction

A dissociation reaction is a type of chemical reaction where compounds split into smaller components, such as ions. For acids, this involves breaking apart to form hydrogen ions (\(H^+\)) and the corresponding anions, contributing to the acidic behavior in solutions.
When an acid such as HCN dissociates in water, it interacts with water molecules, splitting into \(H_3O^+\) (hydronium ion) and \(CN^-\) ions. This dissociation is crucial in defining the acid's strength and behavior in solutions.

• Dissociation reactions for acids are typically reversible, and the extent of dissociation varies with the nature of the acid.
• Weak acids partially dissociate, establishing an equilibrium where the reactants and products coexist.
• The dissociation equation helps visualize and calculate how effectively an acid releases its ions in a solution.

Here is an example of dissociation for a given acid, HCN:\[\text{HCN} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CN}^-\]
This equation is a fundamental tool for students to understand how acids behave when mixed with water.

Ka Expression

The acid dissociation constant, or \(K_a\), is a vital concept in understanding acid strength and behavior. It quantifies the extent of dissociation of an acid in water, allowing students to compare different acids’ tendencies to ionize.
The \(K_a\) expression follows a specific format, representing the ratio of the concentration of products (resulting from the acid dissociation) to that of the remaining undissociated acid.

• A higher \(K_a\) value indicates a stronger acid, which dissociates more completely in solution.
• Conversely, a lower \(K_a\) suggests a weaker acid, with minimal ionization.

For any given acid, the generic \(K_a\) expression can be represented as:\[K_a = \frac{[A^-][H_3O^+]}{[HA]}\]
This format is universal to all monoprotic acids, where \([HA]\) is the undissociated acid, and \([A^-]\) and \([H_3O^+]\) are the concentrations of the dissociated species. By understanding \(K_a\), you can gauge how different factors such as concentration, temperature, or presence of other substances influence the ionization and therefore the acidity of the solution. This understanding is crucial for predicting reactions and for practical applications in chemistry and related fields.