Textbook Question
What is the concentration of hydroxide ions 3OH-4 in a glass of wine with pH = 3.64? (LO 16.5, 16.6) (a) 2.3 * 10-4 M (b) 6.4 * 10-3 M (c) 6.8 * 10-9 M (d) 4.4 * 10-11 M
Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 9
Determine the following concentrations for a 0.40 M H2Se solution that has the stepwise dissociation constants of Ka1 = 1.3 * 10^-4 and Ka2 = 1.0 * 10^-11. (LO 16.10) (a) [H2Se] = 0.35, [HSe^-] = 5.0 * 10^-2, [H3O^+] = 3.0 * 10^-3, [Se^2-] = 1.3 * 10^-4 (b) [H2Se] = 0.39, [HSe^-] = 7.2 * 10^-3, [H3O^+] = 7.2 * 10^-3, [Se^2-] = 1.0 * 10^-11 (c) [H2Se] = 0.31, [HSe^-] = 9.0 * 10^-2, [H3O^+] = 9.0 * 10^-2, [Se^2-] = 1.0 * 10^-11 (d) [H2Se] = 0.40, [HSe^-] = 1.3 * 10^-4, [H3O^+] = 1.3 * 10^-4, [Se^2-] = 1.0 * 10^-11
Verified step by step guidance1
Step 1: Understand the dissociation process of H2Se. It dissociates in two steps: H2Se ⇌ HSe^- + H^+ (first dissociation) and HSe^- ⇌ Se^2- + H^+ (second dissociation).
Step 2: Use the first dissociation constant (Ka1 = 1.3 * 10^-4) to set up the equilibrium expression: Ka1 = [HSe^-][H^+]/[H2Se]. Assume initial [H2Se] = 0.40 M and solve for [HSe^-] and [H^+].
Step 3: Use the second dissociation constant (Ka2 = 1.0 * 10^-11) to set up the equilibrium expression: Ka2 = [Se^2-][H^+]/[HSe^-]. Use the [HSe^-] and [H^+] from the first dissociation to solve for [Se^2-].
Step 4: Compare the calculated concentrations of [H2Se], [HSe^-], [H^+], and [Se^2-] with the given options (a, b, c, d) to determine which set of concentrations matches the calculated values.
Step 5: Verify that the calculated concentrations satisfy both equilibrium expressions and the initial conditions, ensuring the correct choice is consistent with the dissociation constants and initial concentration.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Dissociation Constants (Ka)
Dissociation constants (Ka) quantify the strength of an acid in solution, indicating how well it donates protons (H+) to water. For polyprotic acids like H2Se, there are multiple dissociation steps, each with its own Ka value. The first dissociation constant (Ka1) is typically larger than the second (Ka2), reflecting the decreasing tendency to lose additional protons as the acid becomes more negatively charged.
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03:50
Characteristics of Ka and Kb
Equilibrium Concentrations
Equilibrium concentrations refer to the concentrations of reactants and products in a chemical reaction at equilibrium. In the context of acid dissociation, these concentrations can be determined using the initial concentration of the acid and the dissociation constants. The changes in concentration due to dissociation must be accounted for to find the equilibrium state, often using an ICE (Initial, Change, Equilibrium) table.
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02:35Thermal Equilibrium
Polyprotic Acids
Polyprotic acids can donate more than one proton per molecule, undergoing multiple dissociation steps. Each step has its own dissociation constant, and the overall behavior of the acid in solution is influenced by these successive dissociations. Understanding the sequential nature of these reactions is crucial for calculating the concentrations of the various species present in solution, such as H2Se, HSe^-, and Se^2-.
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03:25Polyprotic Buffers
Related Practice
Textbook Question
What is the pH of an aqueous solution of Ca1OH22 at 25.0 °C with a concentration of 6.3 * 10-5 M? (LO 16.7) (a) 4.20 (b) 10.10 (c) 11.36 (d) 9.80
Textbook Question
An acid solution with a concentration of 0.500 M has a pH = 3.21. What is the Ka of the acid? (LO 16.8) (a) 1.2 * 10-5 (b) 1.7 * 10-6 (c) 7.6 * 10-7 (d) 5.4 * 10-3
Textbook Question
Ammonia 1NH32 has base dissociation constant 1Kb2 of 1.8 * 10-5. What is the concentration of an aqueous ammonia solution that has a pH of 11.68? (LO 16.11) (a) 0.28 M (b) 3.6 M (c) 9.0 * 10-3 M (d) 1.3 M
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Textbook Question
Consider the reaction: SO2 + OH- S HSO3-. Which reaction scheme shows the correct use of the curved arrow notation representing thedonation of an electron pair and the correct labeling of the Lewis acid and Lewis base? (LO 16.14)(a) (b) (c) (d)