22. A 25.00 mL of an unknown HCl solution requires titration with 22.50 mL of 0.2000 M NaOH to reach the equivalence point. What is the concentration of the unknown HCI solution? A) 0.1800 M C) 0.3600 M B) 0.0900 M D) 0.2222 M
22. A 25.00 mL of an unknown HCl solution requires titration with 22.50 mL of 0.2000 M NaOH to reach the equivalence point. What is the concentration of the unknown HCI solution? A) 0.1800 M C) 0.3600 M B) 0.0900 M D) 0.2222 M
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 Example**
**Question:**
A 25.00 mL sample of an unknown HCl solution requires titration with 22.50 mL of 0.2000 M NaOH to reach the equivalence point. What is the concentration of the unknown HCl solution?
**Options:**
A) 0.1800 M
B) 0.0900 M
C) 0.3600 M
D) 0.2222 M
**Correct Answer:**
A) 0.1800 M
**Explanation:**
To determine the concentration of the unknown HCl solution, use the formula derived from the titration process, where the moles of acid equal the moles of base at the equivalence point:
\[ \text{Molarity of HCl} \times \text{Volume of HCl} = \text{Molarity of NaOH} \times \text{Volume of NaOH} \]
Given that the molarity of NaOH is 0.2000 M and the volume used is 22.50 mL, calculate the moles of NaOH:
\[ \text{Moles of NaOH} = 0.2000 \, \text{M} \times 0.02250 \, \text{L} = 0.004500 \, \text{mol} \]
Since moles of HCl = moles of NaOH (because HCl and NaOH react in a 1:1 ratio), the moles of HCl = 0.004500 mol. Using the volume of HCl (25.00 mL or 0.02500 L), find its molarity:
\[ \text{Molarity of HCl} = \frac{0.004500 \, \text{mol}}{0.02500 \, \text{L}} = 0.1800 \, \text{M} \]
Thus, the concentration of the unknown HCl solution is 0.1800 M (Option A).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffddc283e-bfa7-4492-b564-1add28da6d82%2F49f3e558-e6a4-49f5-8fc6-d55b84c68070%2Fh82p963_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration Problem Example**
**Question:**
A 25.00 mL sample of an unknown HCl solution requires titration with 22.50 mL of 0.2000 M NaOH to reach the equivalence point. What is the concentration of the unknown HCl solution?
**Options:**
A) 0.1800 M
B) 0.0900 M
C) 0.3600 M
D) 0.2222 M
**Correct Answer:**
A) 0.1800 M
**Explanation:**
To determine the concentration of the unknown HCl solution, use the formula derived from the titration process, where the moles of acid equal the moles of base at the equivalence point:
\[ \text{Molarity of HCl} \times \text{Volume of HCl} = \text{Molarity of NaOH} \times \text{Volume of NaOH} \]
Given that the molarity of NaOH is 0.2000 M and the volume used is 22.50 mL, calculate the moles of NaOH:
\[ \text{Moles of NaOH} = 0.2000 \, \text{M} \times 0.02250 \, \text{L} = 0.004500 \, \text{mol} \]
Since moles of HCl = moles of NaOH (because HCl and NaOH react in a 1:1 ratio), the moles of HCl = 0.004500 mol. Using the volume of HCl (25.00 mL or 0.02500 L), find its molarity:
\[ \text{Molarity of HCl} = \frac{0.004500 \, \text{mol}}{0.02500 \, \text{L}} = 0.1800 \, \text{M} \]
Thus, the concentration of the unknown HCl solution is 0.1800 M (Option A).
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