A chemist has a bottle that contains 250 mL of a 0.787 M solution of aluminum chloride. What are the concentrations of chloride ions and of aluminum ions in the bottle?
A chemist has a bottle that contains 250 mL of a 0.787 M solution of aluminum chloride. What are the concentrations of chloride ions and of aluminum ions in the bottle?
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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![**Educational Content: Determining Ion Concentrations in a Solution**
**Problem Statement:**
A chemist has a bottle that contains 250 mL of a 0.787 M solution of aluminum chloride. What are the concentrations of chloride ions and of aluminum ions in the bottle?
**Options:**
- A. \([Cl^-] = 0.60 \, \text{mol/L}, \, [Al^{3+}] = 0.20 \, \text{mol/L}\)
- B. \([Cl^-] = 0.787 \, \text{mol/L}, \, [Al^{3+}] = 0.787 \, \text{mol/L}\)
- C. \([Cl^-] = 0.787 \, \text{mol/L}, \, [Al^{3+}] = 2.36 \, \text{mol/L}\)
- D. \([Cl^-] = 0.0031 \, \text{mol/L}, \, [Al^{3+}] = 0.0031 \, \text{mol/L}\)
- E. \([Cl^-] = 2.36 \, \text{mol/L}, \, [Al^{3+}] = 0.787 \, \text{mol/L}\)
**Analysis of Problem:**
Aluminum chloride, \(\text{AlCl}_3\), dissociates into one aluminum ion (\(Al^{3+}\)) and three chloride ions (\(Cl^-\)) in solution. For every mole of \(\text{AlCl}_3\) that dissociates, one mole of \(Al^{3+}\) ions and three moles of \(Cl^-\) ions are produced.
Given:
- The concentration of the \(\text{AlCl}_3\) solution is 0.787 M.
**Calculation:**
- The concentration of \(Al^{3+}\) ions is equal to the concentration of \(\text{AlCl}_3\), which is 0.787 M.
- The concentration of \(Cl^-\) ions is three times the concentration of \(\text{AlCl}_3\), which is \(3 \times 0.787 \, \text{M} = 2.361 \, \text{M}\).
**Correct Answer:**
- E. \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c289562-bb12-43d3-a3c9-e79f806a283f%2F4dbf7efd-e8ae-479d-aebf-8981bc9ef02c%2Fceoudiq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Determining Ion Concentrations in a Solution**
**Problem Statement:**
A chemist has a bottle that contains 250 mL of a 0.787 M solution of aluminum chloride. What are the concentrations of chloride ions and of aluminum ions in the bottle?
**Options:**
- A. \([Cl^-] = 0.60 \, \text{mol/L}, \, [Al^{3+}] = 0.20 \, \text{mol/L}\)
- B. \([Cl^-] = 0.787 \, \text{mol/L}, \, [Al^{3+}] = 0.787 \, \text{mol/L}\)
- C. \([Cl^-] = 0.787 \, \text{mol/L}, \, [Al^{3+}] = 2.36 \, \text{mol/L}\)
- D. \([Cl^-] = 0.0031 \, \text{mol/L}, \, [Al^{3+}] = 0.0031 \, \text{mol/L}\)
- E. \([Cl^-] = 2.36 \, \text{mol/L}, \, [Al^{3+}] = 0.787 \, \text{mol/L}\)
**Analysis of Problem:**
Aluminum chloride, \(\text{AlCl}_3\), dissociates into one aluminum ion (\(Al^{3+}\)) and three chloride ions (\(Cl^-\)) in solution. For every mole of \(\text{AlCl}_3\) that dissociates, one mole of \(Al^{3+}\) ions and three moles of \(Cl^-\) ions are produced.
Given:
- The concentration of the \(\text{AlCl}_3\) solution is 0.787 M.
**Calculation:**
- The concentration of \(Al^{3+}\) ions is equal to the concentration of \(\text{AlCl}_3\), which is 0.787 M.
- The concentration of \(Cl^-\) ions is three times the concentration of \(\text{AlCl}_3\), which is \(3 \times 0.787 \, \text{M} = 2.361 \, \text{M}\).
**Correct Answer:**
- E. \
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