### Solution to Dilution Problem: Calculating Required Volume of Concentrated HCl #### Problem Statement:How many liters of 0.8 M HCl is needed to make 0.80 L of 0.17 M HCl? (give your answer to 3 decimal places)#### Detailed Explanation:To solve this problem, we need to use the dilution equation:\[ C_1V_1 = C_2V_2 \]Where:- \( C_1 \) is the concentration of the stock solution.- \( V_1 \) is the volume of the stock solution needed.- \( C_2 \) is the concentration of the diluted solution.- \( V_2 \) is the volume of the diluted solution.Given values:- \( C_1 = 0.8 \) M- \( C_2 = 0.17 \) M- \( V_2 = 0.80 \) LWe need to find \( V_1 \).\[ 0.8 \, \text{M} \times V_1 = 0.17 \, \text{M} \times 0.80 \, \text{L} \]\[ V_1 = \frac{0.17 \, \text{M} \times 0.80 \, \text{L}}{0.8 \, \text{M}} \]\[ V_1 = \frac{0.136 \, \text{M} \cdot \text{L}}{0.8 \, \text{M}} \]\[ V_1 = 0.170 \, \text{L} \]#### Answer:To prepare 0.80 L of a 0.17 M HCl solution, you need 0.170 L of 0.8 M HCl.

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How many liters of 0.8 M HCI is needed to make 0.80 L of 0.17 M HCI? (give your answer to 3 decimal places)

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

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Question

### Solution to Dilution Problem: Calculating Required Volume of Concentrated HCl

#### Problem Statement:
How many liters of 0.8 M HCl is needed to make 0.80 L of 0.17 M HCl? (give your answer to 3 decimal places)

#### Detailed Explanation:

To solve this problem, we need to use the dilution equation:
\[ C_1V_1 = C_2V_2 \]

Where:
- \( C_1 \) is the concentration of the stock solution.
- \( V_1 \) is the volume of the stock solution needed.
- \( C_2 \) is the concentration of the diluted solution.
- \( V_2 \) is the volume of the diluted solution.

Given values:
- \( C_1 = 0.8 \) M
- \( C_2 = 0.17 \) M
- \( V_2 = 0.80 \) L

We need to find \( V_1 \).

\[ 0.8 \, \text{M} \times V_1 = 0.17 \, \text{M} \times 0.80 \, \text{L} \]

\[ V_1 = \frac{0.17 \, \text{M} \times 0.80 \, \text{L}}{0.8 \, \text{M}} \]

\[ V_1 = \frac{0.136 \, \text{M} \cdot \text{L}}{0.8 \, \text{M}} \]

\[ V_1 = 0.170 \, \text{L} \]

#### Answer:
To prepare 0.80 L of a 0.17 M HCl solution, you need 0.170 L of 0.8 M HCl.

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