**Title: Understanding the Chemical Equilibria of HXO₄²⁻ in Aqueous Solution** **Introduction** The study of chemical equilibria is crucial in understanding the behavior of various ions in aqueous solutions. This tutorial explores the dissociation of HXO₄²⁻ in water and the resulting equilibria. **Equilibrium Reactions of HXO₄²⁻** **Scenario:** Solutions of HXO₄²⁻ (aq) can also be produced by dissolving Na₂HXO₄ (s) in water. When a 1.0 M solution of Na₂HXO₄ (aq) is prepared, two different reactions between HXO₄²⁻ (aq) and H₂O (l) can occur: 1. **First Equilibrium Reaction:** \[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{XO}_4^{3-} (\text{aq}) + \text{H}_3\text{O}^+ (\text{aq}) \quad K_a = 5 \times 10^{-13} \] 2. **Second Equilibrium Reaction:** \[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{H}_2\text{XO}_4 (\text{aq}) + \text{OH}^- (\text{aq}) \quad K_b = 2 \times 10^{-7} \] **Task e: Identifying Conjugate Acid and Base** Clearly indicate the conjugate acid and the conjugate base of HXO₄²⁻. i. **Conjugate Acid:** ii. **Conjugate Base:** **Task f: Equilibrium Calculations** At equilibrium, will [XO₄³⁻] be greater than, less than, or equal to [H₂XO₄⁻]? Justify your answer based on the magnitudes of \( K_a \) and \( K_b \). **Task g: Le Châtelier's Principle** When 0.020 mol of Na₃XO₄ (s) is added to the solution, [
**Title: Understanding the Chemical Equilibria of HXO₄²⁻ in Aqueous Solution** **Introduction** The study of chemical equilibria is crucial in understanding the behavior of various ions in aqueous solutions. This tutorial explores the dissociation of HXO₄²⁻ in water and the resulting equilibria. **Equilibrium Reactions of HXO₄²⁻** **Scenario:** Solutions of HXO₄²⁻ (aq) can also be produced by dissolving Na₂HXO₄ (s) in water. When a 1.0 M solution of Na₂HXO₄ (aq) is prepared, two different reactions between HXO₄²⁻ (aq) and H₂O (l) can occur: 1. **First Equilibrium Reaction:** \[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{XO}_4^{3-} (\text{aq}) + \text{H}_3\text{O}^+ (\text{aq}) \quad K_a = 5 \times 10^{-13} \] 2. **Second Equilibrium Reaction:** \[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{H}_2\text{XO}_4 (\text{aq}) + \text{OH}^- (\text{aq}) \quad K_b = 2 \times 10^{-7} \] **Task e: Identifying Conjugate Acid and Base** Clearly indicate the conjugate acid and the conjugate base of HXO₄²⁻. i. **Conjugate Acid:** ii. **Conjugate Base:** **Task f: Equilibrium Calculations** At equilibrium, will [XO₄³⁻] be greater than, less than, or equal to [H₂XO₄⁻]? Justify your answer based on the magnitudes of \( K_a \) and \( K_b \). **Task g: Le Châtelier's Principle** When 0.020 mol of Na₃XO₄ (s) is added to the solution, [
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Title: Understanding the Chemical Equilibria of HXO₄²⁻ in Aqueous Solution**
**Introduction**
The study of chemical equilibria is crucial in understanding the behavior of various ions in aqueous solutions. This tutorial explores the dissociation of HXO₄²⁻ in water and the resulting equilibria.
**Equilibrium Reactions of HXO₄²⁻**
**Scenario:**
Solutions of HXO₄²⁻ (aq) can also be produced by dissolving Na₂HXO₄ (s) in water. When a 1.0 M solution of Na₂HXO₄ (aq) is prepared, two different reactions between HXO₄²⁻ (aq) and H₂O (l) can occur:
1. **First Equilibrium Reaction:**
\[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{XO}_4^{3-} (\text{aq}) + \text{H}_3\text{O}^+ (\text{aq}) \quad K_a = 5 \times 10^{-13} \]
2. **Second Equilibrium Reaction:**
\[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{H}_2\text{XO}_4 (\text{aq}) + \text{OH}^- (\text{aq}) \quad K_b = 2 \times 10^{-7} \]
**Task e: Identifying Conjugate Acid and Base**
Clearly indicate the conjugate acid and the conjugate base of HXO₄²⁻.
i. **Conjugate Acid:**
ii. **Conjugate Base:**
**Task f: Equilibrium Calculations**
At equilibrium, will [XO₄³⁻] be greater than, less than, or equal to [H₂XO₄⁻]? Justify your answer based on the magnitudes of \( K_a \) and \( K_b \).
**Task g: Le Châtelier's Principle**
When 0.020 mol of Na₃XO₄ (s) is added to the solution, [](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc6d4b638-a19d-4307-8723-23f937d8c9b8%2Fcea7f874-2815-452c-8a8f-df94c30d8d02%2F2fa31ck.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Understanding the Chemical Equilibria of HXO₄²⁻ in Aqueous Solution**
**Introduction**
The study of chemical equilibria is crucial in understanding the behavior of various ions in aqueous solutions. This tutorial explores the dissociation of HXO₄²⁻ in water and the resulting equilibria.
**Equilibrium Reactions of HXO₄²⁻**
**Scenario:**
Solutions of HXO₄²⁻ (aq) can also be produced by dissolving Na₂HXO₄ (s) in water. When a 1.0 M solution of Na₂HXO₄ (aq) is prepared, two different reactions between HXO₄²⁻ (aq) and H₂O (l) can occur:
1. **First Equilibrium Reaction:**
\[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{XO}_4^{3-} (\text{aq}) + \text{H}_3\text{O}^+ (\text{aq}) \quad K_a = 5 \times 10^{-13} \]
2. **Second Equilibrium Reaction:**
\[ \text{HXO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \leftrightharpoons \text{H}_2\text{XO}_4 (\text{aq}) + \text{OH}^- (\text{aq}) \quad K_b = 2 \times 10^{-7} \]
**Task e: Identifying Conjugate Acid and Base**
Clearly indicate the conjugate acid and the conjugate base of HXO₄²⁻.
i. **Conjugate Acid:**
ii. **Conjugate Base:**
**Task f: Equilibrium Calculations**
At equilibrium, will [XO₄³⁻] be greater than, less than, or equal to [H₂XO₄⁻]? Justify your answer based on the magnitudes of \( K_a \) and \( K_b \).
**Task g: Le Châtelier's Principle**
When 0.020 mol of Na₃XO₄ (s) is added to the solution, [
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