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What is the molarity of an HCI solution if 30.5 mL of 0.500 M NaOH is required to neutralize 0.0600 L of HCI during a titration? A M

What is the molarity of an HCI solution if 30.5 mL of 0.500 M NaOH is required to neutralize 0.0600 L of HCI during a titration? A M

Chemistry
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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**Titration Problem: Calculating Molarity of HCl Solution** **Problem Statement:** What is the molarity of an HCl solution if 30.5 mL of 0.500 M NaOH is required to neutralize 0.0600 L of HCl during a titration? **Solution:** To find the molarity of the HCl solution, use the formula derived from the concept of titration: \[ \text{Molarity of Acid (HCl)} = \frac{\text{Molarity of Base (NaOH) × Volume of Base}}{\text{Volume of Acid (HCl)}} \] **Given:** - Molarity of NaOH = 0.500 M - Volume of NaOH = 30.5 mL = 0.0305 L (convert mL to L) - Volume of HCl = 0.0600 L **Calculation:** \[ \text{Molarity of HCl} = \frac{0.500 \, \text{M} \times 0.0305 \, \text{L}}{0.0600 \, \text{L}} \] \[ \text{Molarity of HCl} = \frac{0.01525}{0.0600} \] \[ \text{Molarity of HCl} = 0.2542 \, \text{M} \] Thus, the molarity of the HCl solution is 0.2542 M.
Transcribed Image Text:**Titration Problem: Calculating Molarity of HCl Solution** **Problem Statement:** What is the molarity of an HCl solution if 30.5 mL of 0.500 M NaOH is required to neutralize 0.0600 L of HCl during a titration? **Solution:** To find the molarity of the HCl solution, use the formula derived from the concept of titration: \[ \text{Molarity of Acid (HCl)} = \frac{\text{Molarity of Base (NaOH) × Volume of Base}}{\text{Volume of Acid (HCl)}} \] **Given:** - Molarity of NaOH = 0.500 M - Volume of NaOH = 30.5 mL = 0.0305 L (convert mL to L) - Volume of HCl = 0.0600 L **Calculation:** \[ \text{Molarity of HCl} = \frac{0.500 \, \text{M} \times 0.0305 \, \text{L}}{0.0600 \, \text{L}} \] \[ \text{Molarity of HCl} = \frac{0.01525}{0.0600} \] \[ \text{Molarity of HCl} = 0.2542 \, \text{M} \] Thus, the molarity of the HCl solution is 0.2542 M.
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