Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter4: Chemical Reactions In Solution
Section: Chapter Questions
Problem 4.9QE: Addition of water to concentrated sulfuric acid is dangerous because it generates enough heat to...
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![**Question 10: Determining Acetic Acid Volume for Solution Preparation**
To solve the following problem, you will apply the concept of molarity and dilution:
**Problem Statement:**
What volume of 0.200 M acetic acid (in mL) is needed to prepare 500. mL of a 0.028 M acetic acid solution?
**Options:**
- A) 70.0 mL
- B) 0.014 mL
- C) 70 mL
- D) 140 mL
**Explanation:**
To find the correct volume, you can use the dilution formula, which is:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) = initial concentration (0.200 M)
- \( V_1 \) = volume of the initial solution you need to find
- \( C_2 \) = final concentration (0.028 M)
- \( V_2 \) = final volume (500 mL)
Rearranging the formula to solve for \( V_1 \):
\[ V_1 = \frac{C_2V_2}{C_1} \]
Substitute the given values:
\[ V_1 = \frac{0.028 \times 500}{0.200} \]
\[ V_1 = 70.0 \, \text{mL} \]
Thus, the correct answer is **70.0 mL**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f591c64-216a-4120-ac44-1b23dfa641af%2F8cd0fc0e-95d2-44e2-8bcb-7735031296f5%2Fuca6lsc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 10: Determining Acetic Acid Volume for Solution Preparation**
To solve the following problem, you will apply the concept of molarity and dilution:
**Problem Statement:**
What volume of 0.200 M acetic acid (in mL) is needed to prepare 500. mL of a 0.028 M acetic acid solution?
**Options:**
- A) 70.0 mL
- B) 0.014 mL
- C) 70 mL
- D) 140 mL
**Explanation:**
To find the correct volume, you can use the dilution formula, which is:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) = initial concentration (0.200 M)
- \( V_1 \) = volume of the initial solution you need to find
- \( C_2 \) = final concentration (0.028 M)
- \( V_2 \) = final volume (500 mL)
Rearranging the formula to solve for \( V_1 \):
\[ V_1 = \frac{C_2V_2}{C_1} \]
Substitute the given values:
\[ V_1 = \frac{0.028 \times 500}{0.200} \]
\[ V_1 = 70.0 \, \text{mL} \]
Thus, the correct answer is **70.0 mL**.
Expert Solution
Step 1
Concentration is inversely proportional to dilution. Hence on dilution concentration decreases
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