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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![**Problem:**
If 5.0 mL of 0.00200 M Fe(NO₃)₃ is diluted to 50.0 mL, what is the concentration of Fe³⁺ in the diluted solution?
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**Solution Explanation:**
To find the concentration of Fe³⁺ after dilution, you can use the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the initial concentration (0.00200 M)
- \( V_1 \) is the initial volume (5.0 mL)
- \( C_2 \) is the final concentration
- \( V_2 \) is the final volume (50.0 mL)
Substituting the known values:
\[ (0.00200 \, \text{M}) (5.0 \, \text{mL}) = C_2 (50.0 \, \text{mL}) \]
\[ 0.010 \, \text{M} \cdot \text{mL} = C_2 \cdot 50.0 \, \text{mL} \]
\[ C_2 = \frac{0.010}{50.0} \]
\[ C_2 = 0.00020 \, \text{M} \]
Thus, the concentration of Fe³⁺ in the diluted solution is 0.00020 M.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b6b6912-3084-4759-aee0-bc0c3c3b661a%2F09615b64-bb3c-4519-8991-e443b8a418ae%2Fz2pn30m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
If 5.0 mL of 0.00200 M Fe(NO₃)₃ is diluted to 50.0 mL, what is the concentration of Fe³⁺ in the diluted solution?
---
**Solution Explanation:**
To find the concentration of Fe³⁺ after dilution, you can use the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the initial concentration (0.00200 M)
- \( V_1 \) is the initial volume (5.0 mL)
- \( C_2 \) is the final concentration
- \( V_2 \) is the final volume (50.0 mL)
Substituting the known values:
\[ (0.00200 \, \text{M}) (5.0 \, \text{mL}) = C_2 (50.0 \, \text{mL}) \]
\[ 0.010 \, \text{M} \cdot \text{mL} = C_2 \cdot 50.0 \, \text{mL} \]
\[ C_2 = \frac{0.010}{50.0} \]
\[ C_2 = 0.00020 \, \text{M} \]
Thus, the concentration of Fe³⁺ in the diluted solution is 0.00020 M.
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