What is the concentration of ions present when 0.35 m Ba(NO3)2 dissociates into ions in an aqueous solution? Enter your answer with 3 significant digits and no unit.
What is the concentration of ions present when 0.35 m Ba(NO3)2 dissociates into ions in an aqueous solution? Enter your answer with 3 significant digits and no unit.
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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![### Ion Concentration Calculation
**Problem Statement:**
What is the concentration of ions present when 0.35 m Ba(NO₃)₂ dissociates into ions in an aqueous solution?
**Instructions:**
Enter your answer with 3 significant digits and no unit.
**Answer Box:**
\[ \text{Answer:} \ \square\square\square \]
To calculate the concentration of ions, you must consider the dissociation equation and how many ions each formula unit produces when it dissolves. For barium nitrate (Ba(NO₃)₂), this dissociation can be represented as:
\[ \text{Ba(NO}_{3}\text{)}_{2} \rightarrow \text{Ba}^{2+} + 2\text{NO}_{3}^{-} \]
Given:
1. The initial molarity of \( Ba(NO₃)₂ \) is 0.35 m.
2. Each mole of \( Ba(NO₃)₂ \) produces 1 mole of \( Ba^{2+} \) ions and 2 moles of \( NO₃^- \) ions.
Let’s break it down:
- For \( Ba^{2+} \): 0.35 m
- For \( NO₃^- \): 2 × 0.35 m = 0.70 m
**Total ion concentration:**
\[ 0.35 \, \text{m (Ba}^{2+}\text{)} + 0.70 \, \text{m (2NO}_3^- \text{)} = 1.05 \, \text{m} \]
Thus, the concentration of ions in the solution is 1.05 (with 3 significant digits).
\[ \text{Answer: 1.05} \]
This comprehensive breakdown ensures a clear understanding and accurate calculation of the ion concentration.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc25f84cd-37fe-4eda-bb3a-0db4abed4038%2Fc06456fb-e97d-44a0-b252-5caa75725020%2Fyijqovn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Ion Concentration Calculation
**Problem Statement:**
What is the concentration of ions present when 0.35 m Ba(NO₃)₂ dissociates into ions in an aqueous solution?
**Instructions:**
Enter your answer with 3 significant digits and no unit.
**Answer Box:**
\[ \text{Answer:} \ \square\square\square \]
To calculate the concentration of ions, you must consider the dissociation equation and how many ions each formula unit produces when it dissolves. For barium nitrate (Ba(NO₃)₂), this dissociation can be represented as:
\[ \text{Ba(NO}_{3}\text{)}_{2} \rightarrow \text{Ba}^{2+} + 2\text{NO}_{3}^{-} \]
Given:
1. The initial molarity of \( Ba(NO₃)₂ \) is 0.35 m.
2. Each mole of \( Ba(NO₃)₂ \) produces 1 mole of \( Ba^{2+} \) ions and 2 moles of \( NO₃^- \) ions.
Let’s break it down:
- For \( Ba^{2+} \): 0.35 m
- For \( NO₃^- \): 2 × 0.35 m = 0.70 m
**Total ion concentration:**
\[ 0.35 \, \text{m (Ba}^{2+}\text{)} + 0.70 \, \text{m (2NO}_3^- \text{)} = 1.05 \, \text{m} \]
Thus, the concentration of ions in the solution is 1.05 (with 3 significant digits).
\[ \text{Answer: 1.05} \]
This comprehensive breakdown ensures a clear understanding and accurate calculation of the ion concentration.
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