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Ch.16 - Acid-Base Equilibria
Chapter 16, Problem 63

Calculate the percent ionization of hydrazoic acid (HN3) in solutions of each of the following concentrations (Ka is given in Appendix D): (a) 0.400 M, (b) 0.100 M, (c) 0.0400 M.

Verified step by step guidance
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Step 1: Write the ionization equation for hydrazoic acid (HN3) in water: \[ \text{HN}_3 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{N}_3^- (aq) \].
Step 2: Set up the expression for the acid dissociation constant (Ka) using the concentrations at equilibrium: \[ K_a = \frac{[\text{H}^+][\text{N}_3^-]}{[\text{HN}_3]} \].
Step 3: Assume that the initial concentration of HN3 is \([\text{HN}_3]_0\) and that the change in concentration due to ionization is \(x\), so at equilibrium, \([\text{H}^+] = x\), \([\text{N}_3^-] = x\), and \([\text{HN}_3] = [\text{HN}_3]_0 - x\).
Step 4: Substitute the equilibrium concentrations into the Ka expression: \[ K_a = \frac{x^2}{[\text{HN}_3]_0 - x} \].
Step 5: Solve for \(x\) (the concentration of \(\text{H}^+\) ions) using the quadratic formula or by assuming \(x\) is small compared to \([\text{HN}_3]_0\), then calculate the percent ionization using \(\text{Percent Ionization} = \left( \frac{x}{[\text{HN}_3]_0} \right) \times 100\% \).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ionization and Acid Strength

Ionization refers to the process by which an acid donates protons (H+) to water, forming hydronium ions (H3O+). The strength of an acid is determined by its ability to ionize in solution, quantified by its acid dissociation constant (Ka). A higher Ka value indicates a stronger acid that ionizes more completely, while a lower Ka indicates a weaker acid with less ionization.
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Percent Ionization

Percent ionization is a measure of the extent to which an acid ionizes in solution, expressed as a percentage. It is calculated using the formula: (concentration of ionized acid / initial concentration of acid) × 100%. This value helps compare the strength of acids at different concentrations, as it often varies with dilution.
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Equilibrium Concentrations

In the context of acid ionization, equilibrium concentrations refer to the concentrations of reactants and products at equilibrium in a chemical reaction. For weak acids like hydrazoic acid, the equilibrium expression involves the concentrations of the ionized species and the undissociated acid, allowing for the calculation of percent ionization and the use of the Ka value to find the equilibrium state.
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Related Practice
Textbook Question

Determine the pH of each of the following solutions (Ka and Kb values are given in Appendix D): (c) 0.165 M hydroxylamine.

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Open Question
Saccharin, a sugar substitute, is a weak acid with pKa = 2.32 at 25 °C. It ionizes in aqueous solution as follows: HNC7H4SO31(aq) ⇌ H+(aq) + NC7H4SO3-(aq). What is the pH of a 0.10 M solution of this substance?
Open Question
The active ingredient in aspirin is acetylsalicylic acid 1HC9H7O42, a monoprotic acid with Ka = 3.3 * 10^-4 at 25 °C. What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, each containing 500 mg of acetylsalicylic acid, in 250 mL of water?
Textbook Question

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given in Appendix D): (a) 0.250 M (b) 0.0800 M (c) 0.0200 M

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Textbook Question

Citric acid, which is present in citrus fruits, is a triprotic acid (Table 16.3). (a) Calculate the pH of a 0.040 M solution of citric acid. (b) Did you have to make any approximations or assumptions in completing your calculations? (c) Is the concentration of citrate ion 1C6H5O7 3-2 equal to, less than, or greater than the H+ ion concentration?

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Open Question
Tartaric acid is found in many fruits, including grapes, and is partially responsible for the dry texture of certain wines. Calculate the pH and the tartrate ion C4H4O6²⁻ concentration for a 0.250 M solution of tartaric acid, for which the acid-dissociation constants are listed in Table 16.3. Did you have to make any approximations or assumptions in your calculation?