**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 MUsing 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.

Concentration of Na2CO3 = 0.20M Molarity of Na+=?

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What is the molarity of Na* ions in a 0.20 M Na2CO3 solution? N22CO3 (s) → 2 Na* (aq) + CO3 2- (aq.

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

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

**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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