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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![**Understanding the Molarity of Sodium Ions in a \( \text{Na}_2\text{CO}_3 \) Solution**
**Problem Statement:**
What is the molarity of \( \text{Na}^+ \) ions in a 0.20 M \( \text{Na}_2\text{CO}_3 \) solution?
**Chemical Reaction:**
\[ \text{Na}_2\text{CO}_3 (s) \rightarrow 2 \text{Na}^+ (aq) + \text{CO}_3^{2-} (aq) \]
### Explanation:
When sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) dissolves in water, it dissociates into sodium ions (\( \text{Na}^+ \)) and carbonate ions (\( \text{CO}_3^{2-} \)). From the balanced chemical equation, it is evident that one mole of \( \text{Na}_2\text{CO}_3 \) yields two moles of \( \text{Na}^+ \) ions.
Given that the molarity of the sodium carbonate solution is 0.20 M,
- Molarity of \( \text{Na}_2\text{CO}_3 \) = 0.20 M
Using the stoichiometry of the reaction:
- 1 mole of \( \text{Na}_2\text{CO}_3 \) produces 2 moles of \( \text{Na}^+ \)
Therefore, the molarity of \( \text{Na}^+ \) in the solution is:
\[ 0.20 \text{ M} \times 2 = 0.40 \text{ M} \]
So, the molarity of \( \text{Na}^+ \) ions in a 0.20 M \( \text{Na}_2\text{CO}_3 \) solution is 0.40 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15ddb6cd-6983-4657-9770-05e88534b1be%2Fcc21346d-1414-4a6c-ac95-f440299ef9e7%2Fx34c1.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding the Molarity of Sodium Ions in a \( \text{Na}_2\text{CO}_3 \) Solution**
**Problem Statement:**
What is the molarity of \( \text{Na}^+ \) ions in a 0.20 M \( \text{Na}_2\text{CO}_3 \) solution?
**Chemical Reaction:**
\[ \text{Na}_2\text{CO}_3 (s) \rightarrow 2 \text{Na}^+ (aq) + \text{CO}_3^{2-} (aq) \]
### Explanation:
When sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) dissolves in water, it dissociates into sodium ions (\( \text{Na}^+ \)) and carbonate ions (\( \text{CO}_3^{2-} \)). From the balanced chemical equation, it is evident that one mole of \( \text{Na}_2\text{CO}_3 \) yields two moles of \( \text{Na}^+ \) ions.
Given that the molarity of the sodium carbonate solution is 0.20 M,
- Molarity of \( \text{Na}_2\text{CO}_3 \) = 0.20 M
Using the stoichiometry of the reaction:
- 1 mole of \( \text{Na}_2\text{CO}_3 \) produces 2 moles of \( \text{Na}^+ \)
Therefore, the molarity of \( \text{Na}^+ \) in the solution is:
\[ 0.20 \text{ M} \times 2 = 0.40 \text{ M} \]
So, the molarity of \( \text{Na}^+ \) ions in a 0.20 M \( \text{Na}_2\text{CO}_3 \) solution is 0.40 M.
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