What is the concentration of sodium ions in 0.285 M NażCO3? M

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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2 part question.

**Question:**

What is the concentration of sodium ions in 0.285 M \( \text{Na}_2\text{CO}_3 \)?

**Answer Box:** 

M

---

**Explanation:**

To solve this problem, we need to determine the concentration of sodium ions (\( \text{Na}^+ \)) in a solution of sodium carbonate (\( \text{Na}_2\text{CO}_3 \)). 

**Step-by-step Solution:**

- \( \text{Na}_2\text{CO}_3 \) dissociates in water to form two sodium ions and one carbonate ion:
  \[
  \text{Na}_2\text{CO}_3 \rightarrow 2 \text{Na}^+ + \text{CO}_3^{2-}
  \]

- Therefore, for every mole of \( \text{Na}_2\text{CO}_3 \), two moles of \( \text{Na}^+ \) are produced.

- Given the solution concentration is 0.285 M \( \text{Na}_2\text{CO}_3 \), the concentration of sodium ions is:
  \[
  2 \times 0.285 \, \text{M} = 0.570 \, \text{M}
  \]

**Conclusion:**

The concentration of sodium ions \( (\text{Na}^+) \) in the solution is 0.570 M.
Transcribed Image Text:**Question:** What is the concentration of sodium ions in 0.285 M \( \text{Na}_2\text{CO}_3 \)? **Answer Box:** M --- **Explanation:** To solve this problem, we need to determine the concentration of sodium ions (\( \text{Na}^+ \)) in a solution of sodium carbonate (\( \text{Na}_2\text{CO}_3 \)). **Step-by-step Solution:** - \( \text{Na}_2\text{CO}_3 \) dissociates in water to form two sodium ions and one carbonate ion: \[ \text{Na}_2\text{CO}_3 \rightarrow 2 \text{Na}^+ + \text{CO}_3^{2-} \] - Therefore, for every mole of \( \text{Na}_2\text{CO}_3 \), two moles of \( \text{Na}^+ \) are produced. - Given the solution concentration is 0.285 M \( \text{Na}_2\text{CO}_3 \), the concentration of sodium ions is: \[ 2 \times 0.285 \, \text{M} = 0.570 \, \text{M} \] **Conclusion:** The concentration of sodium ions \( (\text{Na}^+) \) in the solution is 0.570 M.
**Problem Statement:**

100.0 mL of a 0.710 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?

---

**Graph/Diagram Explanation:**

There is no graph or diagram, but there is a partially filled line with the label "M", indicating that the answer should be expressed in molarity (M).
Transcribed Image Text:**Problem Statement:** 100.0 mL of a 0.710 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution? --- **Graph/Diagram Explanation:** There is no graph or diagram, but there is a partially filled line with the label "M", indicating that the answer should be expressed in molarity (M).
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