Problem 67 - Chemistry The Molecular Science

Question

Hydrazoic acid, \(\mathrm{HN}_{3}\), is very explosive in its pure state but can be studied in aqueous solution. The acid is prepared by the reaction of hydrazine with nitrous acid. \(\mathrm{N}_{2} \mathrm{H}_{4}(\ell)+\mathrm{HNO}_{2}(\mathrm{aq}) \longrightarrow \mathrm{HN}_{3}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) (a) Determine the oxidation states of nitrogen in the compounds in this reaction. (b) What is the oxidizing agent in this reaction? (c) The \(K_{\mathrm{a}}\) of hydrazoic acid is \(1.0 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) of a \(0.010-\mathrm{M}\) solution of \(\mathrm{HN}_{3}\).

Step-by-Step Solution

Answer

Oxidation states: \(\begin{cases} \text{\(\mathrm{N}_2\mathrm{H}_4\): \(-2\)} \\ \text{\(\mathrm{HNO}_2\): \(+3\)} \\ \text{\(\mathrm{HN}_3\): \(-1/3\)} \end{cases}\); oxidizing agent: \(\mathrm{HNO}_2\); pH: 3.50.

Step 1: Determine Oxidation States

To determine the oxidation states of nitrogen in the given compounds, we first observe each molecule. For \( \mathrm{N}_{2}\mathrm{H}_4 \), we assign nitrogen an oxidation state of \(-2\) since hydrogen is \(+1\). In \( \mathrm{HNO}_{2} \), nitrogen typically has an oxidation state of \(+3\). In \( \mathrm{HN}_{3} \), the average oxidation state of nitrogen is \(-1/3\), since it is a resonance structure where one nitrogen is \(0\) and the others adjust to make the average. In water (\( \mathrm{H}_{2}\mathrm{O} \)), oxygen is \(-2\) and hydrogen is \(+1\).

Step 2: Identify the Oxidizing Agent

An oxidizing agent is the substance that gains electrons, thereby causing another substance to be oxidized. In the given reaction, \( \mathrm{HNO}_2 \) has nitrogen going from \(+3\) to \(-1/3\), indicating a gain of electrons, thus it is reduced. Therefore, \( \mathrm{HNO}_2 \) is the oxidizing agent.

Step 3: Calculate Hydrogen Ion Concentration

Given the \( K_{\mathrm{a}} = 1.0 \times 10^{-5} \) for \( \mathrm{HN}_3 \) and an initial concentration of \(0.010 \mathrm{M}\), we set up the expression for the acid dissociation constant: \[ K_a = \frac{[H^+][N_3^-]}{[HN_3]} = 1.0 \times 10^{-5} \]. Assuming \([H^+] = [N_3^-] = x\) and \([HN_3] = 0.010 - x \approx 0.010\), the expression simplifies to \[ K_a = \frac{x^2}{0.010} = 1.0 \times 10^{-5} \]. Solving for \(x\), \(x^2 = (1.0 \times 10^{-5}) \times (0.010)\), \(x = 3.16 \times 10^{-4}\).

Step 4: Calculate pH

The \([H^+]\) ion concentration found is \(x = 3.16 \times 10^{-4}\). The \( \mathrm{pH} \) is given by the negative logarithm of the \( [H^+] \), \( \mathrm{pH} = -\log_{10}(3.16 \times 10^{-4}) \). Calculating this gives \( \mathrm{pH} \approx 3.50 \).

Key Concepts

Oxidation States

Understanding oxidation states is key in analyzing chemical reactions, like the one involving hydrazoic acid formation. Oxidation states indicate how electrons are distributed in a compound and help in identifying which elements are oxidized or reduced. In the reaction of hydrazine (\( \mathrm{N}_{2}\mathrm{H}_4 \)) with nitrous acid (\( \mathrm{HNO}_{2} \)), we assign oxidation states to nitrogen based on bonding scenarios. In \( \mathrm{N}_{2}\mathrm{H}_4 \), nitrogen has an oxidation state of \(-2\), because each hydrogen contributes \(+1\), balancing the compound's charge. In \( \mathrm{HNO}_{2} \), nitrogen usually takes an oxidation state of \(+3\). For hydrazoic acid (\( \mathrm{HN}_{3} \)), the average nitrogen oxidation state is \(-1/3\) due to resonance, where different nitrogen atoms exhibit different states. In water (\( \mathrm{H}_{2}\mathrm{O} \)), oxygen is assigned \(-2\) and hydrogen \(+1\). Recognizing these states is crucial, as it lets us track electron movement in reactions.

Acid Dissociation Constant

The acid dissociation constant, \( K_a \), helps us understand the strength of an acid in solution. For hydrazoic acid (\( \mathrm{HN}_{3} \)), \( K_a = 1.0 \times 10^{-5} \) at 25°C, which indicates moderate dissociation into ions. A lower \( K_a \) value implies a weaker acid, meaning it doesn't fully dissociate in water. This constant is represented by the formula:\[ K_a = \frac{[\mathrm{H}^+][\mathrm{N}_3^-]}{[\mathrm{HN}_3]} \] This equation uses the concentrations of hydronium ions \([\mathrm{H}^+]\), azide ions \([\mathrm{N}_3^-]\), and undissociated hydrazoic acid \([\mathrm{HN}_3]\). Understanding and calculating \( K_a \) is essential to comprehend the acidic behavior and potential pH values of the solution.

pH Calculation

Calculating pH involves determining the hydronium ion concentration in a solution. When hydrazoic acid (\( \mathrm{HN}_3 \)) dissociates in water, it forms hydronium \([\mathrm{H}^+]\) and azide ions \( [\mathrm{N}_3^-] \). Given a 0.010 M solution of \( \mathrm{HN}_3 \), we use the established \( K_a \) to find \([\mathrm{H}^+]\). Assuming \([\mathrm{H}^+] = [\mathrm{N}_3^-] = x\) and noticing that \([\mathrm{HN}_3] \approx 0.010 - x \approx 0.010\), we plug these values into the \( K_a \) equation, solving for \( x \), which equates to the hydronium concentration \(3.16 \times 10^{-4}\). The pH is then calculated as the negative logarithm of the \([\mathrm{H}^+]\): \[ \mathrm{pH} = -\log_{10}(3.16 \times 10^{-4}) \] Resulting in a pH of roughly 3.50, indicating a weakly acidic solution.

Oxidizing Agent

An oxidizing agent in a chemical reaction is a substance that accepts electrons and undergoes a decrease in oxidation state, causing the other substance to be oxidized. In the hydrazoic acid formation reaction, nitrous acid (\( \mathrm{HNO}_2 \)) serves as the oxidizing agent. Here, nitrogen transitions from an oxidation state of \(+3\) in \( \mathrm{HNO}_2 \) to an average of \(-1/3\) in \( \mathrm{HN}_3 \), indicating a gain of electrons. By accepting these electrons, \( \mathrm{HNO}_2 \) facilitates the oxidation of hydrazine (\( \mathrm{N}_{2}\mathrm{H}_4 \)), which undergoes an increase in its nitrogen oxidation state. Identifying the oxidizing agent is crucial for understanding the electron flow and energy transformations in this chemical process.