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Chapter 16: Problem 63

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 in Appendix D): (a) $0.400 \mathrm{M}\(, (b) \)0.100 \mathrm{M}\(, (c) \)0.0400 \mathrm{M}$.

Short Answer

Expert verified
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.

Step by step solution

01

Define the Acid Ionization Reaction

Hydrazoic acid, HN₃, when dissolved in water, ionizes into H⁺ and N₃⁻: HN₃ (aq) ⇌ H⁺ (aq) + N₃⁻ (aq)
02

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.
03

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
04

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.
05

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.
06

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.

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Most popular questions from this chapter

Predict which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{K}^{+}\) or \(\mathrm{Cu}^{2+},(\mathrm{b}) \mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+},(\mathrm{c}) \mathrm{Al}^{3+}\) or \(\mathrm{Ga}^{3+} .\) Explain.

(a) What is the difference between the Arrhenius and the Bronsted-Lowry definitions of an acid? (b) \(\mathrm{NH}_{3}(g)\) and \(\mathrm{HCl}(g)\) react to form the ionic solid \(\mathrm{NH}_{4} \mathrm{Cl}(s) .\) Which substance is the Bronsted-Lowry acid in this reaction? Which is the Bronsted-Lowry base?

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Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic. $$ \begin{array}{llll} \hline\left[\mathrm{H}^{+}\right] & \mathrm{OH}^{-} \text {(aq) } && \mathrm{pH} & &\text { pOH }&& \text { Acidic or basic? } \\ \hline 7.5 \times 10^{-3} \mathrm{M} & & & & \\ & 3.6 \times 10^{-10} \mathrm{M} & & \\ & && 8.25 & & \\ & & &&& 5.70 & \end{array} $$
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