Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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.
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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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