Problem 36 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

a. Dissociation reaction: \[ \mathrm{HCN} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{CN^-} \] \(K_\mathrm{a}\) expression: \[ K_\mathrm{a} = \frac{[\mathrm{H_3O^+}][\mathrm{CN^-}]}{[\mathrm{HCN}]} \] b. Dissociation reaction: \[ \mathrm{HOC_6 H_5} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{OC_6 H_5^-} \] \(K_\mathrm{a}\) expression: \[ K_\mathrm{a} = \frac{[\mathrm{H_3O^+}][\mathrm{OC_6 H_5^-}]}{[\mathrm{HOC_6 H_5}]} \] c. Dissociation reaction: \[ C_6 H_5 N H_3^+ + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{C_6 H_5 NH_2} \] \(K_\mathrm{a}\) expression: \[ K_\mathrm{a} = \frac{[\mathrm{H_3O^+}][\mathrm{C_6 H_5 NH_2}]}{[\mathrm{C_6 H_5 NH_3^+}]} \]

Step 1: Write the dissociation reaction for HCN in water

When HCN (hydrogen cyanide) dissociates in water, it donates a proton (H\(^+\)) to water and forms the cyanide ion (CN\(^-\)). The dissociation reaction is as follows: \[ \mathrm{HCN} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{CN^-} \]

Step 2: Write the \(K_\mathrm{a}\) equilibrium expression for HCN

The \(K_\mathrm{a}\) equilibrium expression for HCN is given by the concentration of the products divided by the concentration of the reactants. We need to exclude the concentration of water as it remains almost constant: \[ K_\mathrm{a} = \frac{[\mathrm{H_3O^+}][\mathrm{CN^-}]}{[\mathrm{HCN}]} \]

Step 3: Write the dissociation reaction for \(\mathrm{HOC}_6 \mathrm{H}_5\) in water

When \(\mathrm{HOC}_6 \mathrm{H}_5\) (phenol) dissociates in water, it donates a proton (H\(^+\)) to water and forms the phenoxide ion (\(\mathrm{OC}_6 \mathrm{H}_5^-\)). The dissociation reaction is as follows: \[ \mathrm{HOC_6 H_5} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{OC_6 H_5^-} \]

Step 4: Write the \(K_\mathrm{a}\) equilibrium expression for \(\mathrm{HOC}_6 \mathrm{H}_5\)

The \(K_\mathrm{a}\) equilibrium expression for \(\mathrm{HOC}_6 \mathrm{H}_5\) is given by the concentration of the products divided by the concentration of the reactants. We need to exclude the concentration of water as it remains almost constant: \[ K_\mathrm{a} = \frac{[\mathrm{H_3O^+}][\mathrm{OC_6 H_5^-}]}{[\mathrm{HOC_6 H_5}]} \]

Step 5: Write the dissociation reaction for \(C_6 H_5 N H_3^+\) in water

When \(C_6 H_5 N H_3^+\) (anilinium ion) dissociates in water, it donates a proton (H\(^+\)) to water and forms aniline (\(\mathrm{C_6 H_5 NH_2}\)). The dissociation reaction is as follows: \[ C_6 H_5 N H_3^+ + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{C_6 H_5 NH_2} \]

Step 6: Write the \(K_\mathrm{a}\) equilibrium expression for \(C_6 H_5 N H_3^+\)

The \(K_\mathrm{a}\) equilibrium expression for \(C_6 H_5 N H_3^+\) is given by the concentration of the products divided by the concentration of the reactants. We need to exclude the concentration of water as it remains almost constant: \[ K_\mathrm{a} = \frac{[\mathrm{H_3O^+}][\mathrm{C_6 H_5 NH_2}]}{[\mathrm{C_6 H_5 NH_3^+}]} \]

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is fundamental in exploring how acids behave in water. At equilibrium, the rate of the forward reaction, where the acid donates a proton to water, is equal to the rate of the reverse reaction, where the conjugate base accepts a proton. This does not mean the concentrations of products and reactants are the same but rather that they remain constant over time.

For instance, in the case of HCN in water, the equilibrium expression represents this delicate balance and is written as the product of the concentrations of the ions produced, hydronium (\textnormal{H}_3\textnormal{O}^{+}) and cyanide (\textnormal{CN}^{-}), divided by the concentration of the unchanged HCN, excluding water since its concentration is largely constant due to its excess amount. So for any weak acid, you can write:

\[ K_a = \frac{{[\textnormal{H}_3\textnormal{O}^{+}][\textnormal{A}^{-}]}}{{[\textnormal{HA}]}} \]

Here, \textnormal{HA} represents the acid and \textnormal{A}^{-} its conjugate base. This \textnormal{K}_a value is unique for each acid and gives insight into the acid's strength; lower values of \textnormal{K}_a indicate a weaker acid.

Dissociation Reaction

A dissociation reaction is a process in which a compound, such as an acid, breaks apart into its constituent ions. In acid-base chemistry, this reaction entails the release of hydrogen ions (\textnormal{H}^{+}) into the solution. When these hydrogen ions are released, the remaining part of the acid molecule forms a negatively charged ion known as the conjugate base.

Consider HCN's dissociation in water:

\[ \textnormal{HCN} + \textnormal{H}_2\textnormal{O} \rightleftharpoons \textnormal{H}_3\textnormal{O}^{+} + \textnormal{CN}^{-} \]

Here, HCN donates a hydrogen ion to water, forming hydronium (\textnormal{H}_3\textnormal{O}^{+}) and cyanide (\textnormal{CN}^{-}) ions, showcasing a classic example of a dissociation reaction in acid-base chemistry. This reversible process is crucial to understanding the dynamic nature of acids in solution and is described quantitatively by the acid dissociation constant (\textnormal{K}_a).

Acid-Base Chemistry

At the heart of acid-base chemistry is the transfer of protons from an acid to a base. Acids are proton donors, and bases are proton acceptors. In aqueous solutions, water often acts as the base, accepting protons from the acid to form hydronium ions (\textnormal{H}_3\textnormal{O}^{+}).

Each acid has a characteristic strength often expressed by its \textnormal{K}_a value, the acid dissociation constant, which quantifies its tendency to donate a proton in water. The dissociation of \textnormal{HOC}_6\textnormal{H}_5 (phenol), for example, can be seen as:

\[ \textnormal{HOC}_6\textnormal{H}_5 + \textnormal{H}_2\textnormal{O} \rightleftharpoons \textnormal{H}_3\textnormal{O}^{+} + \textnormal{OC}_6\textnormal{H}_5^{-} \]

Here, the produced \textnormal{OC}_6\textnormal{H}_5^{-} is the conjugate base of phenol. By studying these reactions and their equilibrium constants, students gain valuable insights into predicting the behavior of acids and bases in various chemical contexts.