Problem 63

Question

Calculate the percent ionization of hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) in solutions of each of the following concentrations \(\left(K_{a}\right.\) is given (b) \(0.100 M,(c) 0.0400 M\). in Appendix \(D\) ): (a) \(0.400 M\),

Step-by-Step Solution

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Answer
To calculate the percent ionization of hydrazoic acid (HN₃) in solutions with different concentrations, we follow these steps: 1. Write the acid ionization reaction: HN₃ (aq) ⇌ H⁺ (aq) + N₃⁻ (aq). 2. Write the Kₐ expression: \(Kₐ = \frac{[H^+][N_3^-]}{[HN_3]}\). 3. Create a table for initial and equilibrium concentrations with a being the initial concentration and x being the change in concentration. 4. Approximating x ≪ a, solve for x by using: \(Kₐ ≈ \frac{x^2}{a}\). 5. Calculate the percent ionization using: Percent Ionization = \(\frac{x}{a}\) × 100%. For each case, find x using the given Kₐ and initial concentration, then calculate the percent ionization. Analyze the results to understand how the ionization is affected by different initial concentrations.
1Step 1: Define the Acid Ionization Reaction
Hydrazoic acid, HN₃, when dissolved in water, ionizes into H⁺ and N₃⁻: HN₃ (aq) ⇌ H⁺ (aq) + N₃⁻ (aq)
2Step 2: Write the Kₐ Expression
The expression for Kₐ is given by: Kₐ = \(\frac{[H^+][N_3^-]}{[HN_3]}\) where [H⁺], [N₃⁻], and [HN₃] are the equilibrium concentrations of H⁺ ions, N₃⁻ ions, and hydrazoic acid, respectively.
3Step 3: Create the Table for the Initial and Equilibrium Concentrations
Now let's create the initial and equilibrium concentration table for each case: | | HN₃ | H⁺ | N₃⁻ | |-------|------|----|-----| | Initial | a | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | a-x | x | x | where a is the initial concentration of HN₃, and x is the change in concentration. In our cases: (a) a = 0.400 M (b) a = 0.100 M (c) a = 0.0400 M
4Step 4: Apply the Kₐ expression to find the Equilibrium Concentrations
Next, we will apply the following expression to find x: Kₐ = \(\frac{x (x)}{(a - x)}\) Since Kₐ is very small, we can approximate that x ≪ a, so the equation becomes: Kₐ ≈ \(\frac{x^2}{a}\) Now, we need to solve for x for each case after finding Kₐ in the Appendix D.
5Step 5: Calculate the Percent Ionization
Now, we will calculate the percent ionization for each case: Percent Ionization = \(\frac{x}{a}\) × 100% For each case, plug in the values obtained for x from Step 4 and the initial concentration values (a) from Step 3.
6Step 6: Interpret the Results
The percent ionization values for each case indicate how much the hydrazoic acid has ionized in the solution. As a weak acid, hydrazoic acid does not fully ionize, so the percent ionization will be less than 100%. By comparing the percent ionization values for each concentration, we can understand how the acid's ionization is affected by different initial concentrations.

Key Concepts

Hydrazoic AcidEquilibrium ConcentrationsPercent Ionization
Hydrazoic Acid
Hydrazoic acid, represented by the chemical formula \( \mathrm{HN}_3 \), is a weak acid, which means it does not completely dissociate in water. When hydrazoic acid is dissolved in water, it undergoes a reversible reaction. In this reaction, it partially ionizes to produce hydronium ions (H⁺) and azide ions (N₃⁻).
This dissociation is represented by the chemical equation:
\[ \mathrm{HN}_3 (aq) \rightleftharpoons \mathrm{H}^+ (aq) + \mathrm{N}_3^- (aq) \]
Understanding the nature of hydrazoic acid's ionization helps us calculate important parameters like its equilibrium concentrations and percent ionization. Since hydrazoic acid is a weak acid, it is essential to rely on its specific ionization constant \( K_a \) to predict its behavior in different solutions.
Equilibrium Concentrations
Equilibrium concentrations are crucial in predicting the behavior of weak acids like hydrazoic acid in solution. They tell us how much of each reactant and product is present once the system has reached equilibrium. For hydrazoic acid, this involves calculating the concentrations of \( \mathrm{HN}_3 \), \( \mathrm{H}^+ \), and \( \mathrm{N}_3^- \) ions in the solution.
When we start with an initial concentration \( a \) of hydrazoic acid, no ions are present initially (\( [\mathrm{H}^+] = 0, [\mathrm{N}_3^-] = 0 \)). As the acid starts ionizing, the concentrations of \( \mathrm{H}^+ \) and \( \mathrm{N}_3^- \) increase by \( x \), and the concentration of the undissociated \( \mathrm{HN}_3 \) decreases by the same amount \( x \).
This can be summarized in the equilibrium expression:
\[ K_a = \frac{[\mathrm{H}^+][\mathrm{N}_3^-]}{[\mathrm{HN}_3]} \]
For situations where the \( K_a \) value is quite small, it is a good approximation to assume \( x \ll a \). This simplifies the calculation by allowing us to ignore \( x \) in the denominator, leading to a straightforward quadratic equation to solve for \( x \).
Percent Ionization
The percent ionization of an acid is a measure of the extent to which it dissociates in solution. It is calculated by dividing the concentration of ionized acid by the initial concentration of the acid, then multiplying by 100%.
For hydrazoic acid, it is expressed as:
\[ \text{Percent Ionization} = \left(\frac{x}{a}\right) \times 100\% \]
where \( x \) represents the equilibrium concentration of \( \mathrm{H}^+ \) (or \( \mathrm{N}_3^- \)), and \( a \) is the initial concentration of \( \mathrm{HN}_3 \).
The percent ionization provides insight into the strength of the acid in solution. Typically, weak acids like hydrazoic acid have low percent ionizations due to the incomplete dissociation. As the concentration of hydrazoic acid decreases, the percent ionization tends to increase, illustrating how the fraction of dissociated molecules changes with concentration. Observing these changes helps predict the acid's behavior under different conditions.