Determine the volume (in mL) of water that needs to be added to 25.0 mL of a 0.250 M NaBr to produce a 0.0395 M solution. Assume the volumes are additive.
Determine the volume (in mL) of water that needs to be added to 25.0 mL of a 0.250 M NaBr to produce a 0.0395 M solution. Assume the volumes are additive.
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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![**Dilution Calculation Example**
**Problem Statement:**
Determine the volume (in mL) of water that needs to be added to 25.0 mL of a 0.250 M NaBr solution to produce a 0.0395 M solution. Assume the volumes are additive.
**Calculation Steps:**
1. Identify the initial concentration (\( C_i \)) and volume (\( V_i \)):
- Initial concentration (\( C_i \)): 0.250 M
- Initial volume (\( V_i \)): 25.0 mL
2. Identify the final concentration (\( C_f \)):
- Final concentration (\( C_f \)): 0.0395 M
3. Use the dilution formula: \( C_i \times V_i = C_f \times V_f \)
- Rearrange to find the final volume (\( V_f \)): \( V_f = \frac{C_i \times V_i}{C_f} \)
4. Substitute the known values:
\[
V_f = \frac{0.250 \, \text{M} \times 25.0 \, \text{mL}}{0.0395 \, \text{M}}
\]
5. Calculate \( V_f \):
\[
V_f = \frac{6.25}{0.0395} \approx 158.2 \, \text{mL}
\]
6. Determine the volume of water to add:
\[
\text{Volume of water to add} = V_f - V_i
\]
\[
\text{Volume of water to add} = 158.2 \, \text{mL} - 25.0 \, \text{mL} = 133.2 \, \text{mL}
\]
Therefore, 133.2 mL of water needs to be added to the 25.0 mL of 0.250 M NaBr solution to produce a 0.0395 M solution.
**Diagram:**
The diagram on an educational page would typically include a step-by-step visual representation of the dilution process, with measured volumes and concentration calculations depicted clearly. Since a visual diagram is not present in the text provided, make sure to follow each step methodically to ensure accurate understanding and replication of the calculation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b712500-b2ac-4dfb-be2f-1c8b439ce3a4%2F7e1decba-f44d-4bea-aaff-31a9fb56c4a6%2Fz1g3sj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Dilution Calculation Example**
**Problem Statement:**
Determine the volume (in mL) of water that needs to be added to 25.0 mL of a 0.250 M NaBr solution to produce a 0.0395 M solution. Assume the volumes are additive.
**Calculation Steps:**
1. Identify the initial concentration (\( C_i \)) and volume (\( V_i \)):
- Initial concentration (\( C_i \)): 0.250 M
- Initial volume (\( V_i \)): 25.0 mL
2. Identify the final concentration (\( C_f \)):
- Final concentration (\( C_f \)): 0.0395 M
3. Use the dilution formula: \( C_i \times V_i = C_f \times V_f \)
- Rearrange to find the final volume (\( V_f \)): \( V_f = \frac{C_i \times V_i}{C_f} \)
4. Substitute the known values:
\[
V_f = \frac{0.250 \, \text{M} \times 25.0 \, \text{mL}}{0.0395 \, \text{M}}
\]
5. Calculate \( V_f \):
\[
V_f = \frac{6.25}{0.0395} \approx 158.2 \, \text{mL}
\]
6. Determine the volume of water to add:
\[
\text{Volume of water to add} = V_f - V_i
\]
\[
\text{Volume of water to add} = 158.2 \, \text{mL} - 25.0 \, \text{mL} = 133.2 \, \text{mL}
\]
Therefore, 133.2 mL of water needs to be added to the 25.0 mL of 0.250 M NaBr solution to produce a 0.0395 M solution.
**Diagram:**
The diagram on an educational page would typically include a step-by-step visual representation of the dilution process, with measured volumes and concentration calculations depicted clearly. Since a visual diagram is not present in the text provided, make sure to follow each step methodically to ensure accurate understanding and replication of the calculation.
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