By titration, it is found that 29.9 mL of 0.108 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution. [HCI] = M
By titration, it is found that 29.9 mL of 0.108 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution. [HCI] = 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
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![**Titration Problem: Calculating the Concentration of an HCl Solution**
By titration, it is found that 29.9 mL of 0.108 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.
\[[\text{HCl}] = \_\_\_\_\_\_\_\_ \text{M} \]
**Explanation:**
To find the concentration of the HCl solution, use the concept of molarity and the titration formula:
1. Begin with the balanced chemical equation for the reaction:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
Each mole of HCl reacts with one mole of NaOH.
2. Calculate moles of NaOH used:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity} = 0.0299 \, \text{L} \times 0.108 \, \text{M} = 0.0032292 \, \text{mol}
\]
3. Since the molar ratio of HCl to NaOH is 1:1, moles of HCl = moles of NaOH = 0.0032292 mol.
4. Calculate the molarity of the HCl solution:
\[
[\text{HCl}] = \frac{\text{Moles of HCl}}{\text{Volume of HCl (L)}} = \frac{0.0032292 \, \text{mol}}{0.025 \, \text{L}} = 0.129168 \, \text{M}
\]
Thus, the concentration of the HCl solution is approximately 0.129 M.
**Source:** McQuarrie, Rock, and Gallogly 4e – General Chemistry
**Publisher:** University Science Books](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffdc195bb-9009-47db-866e-65b6514806e8%2F7c892236-ba39-4512-9798-95b4a3a693e0%2Fnnzln7n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration Problem: Calculating the Concentration of an HCl Solution**
By titration, it is found that 29.9 mL of 0.108 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.
\[[\text{HCl}] = \_\_\_\_\_\_\_\_ \text{M} \]
**Explanation:**
To find the concentration of the HCl solution, use the concept of molarity and the titration formula:
1. Begin with the balanced chemical equation for the reaction:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
Each mole of HCl reacts with one mole of NaOH.
2. Calculate moles of NaOH used:
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity} = 0.0299 \, \text{L} \times 0.108 \, \text{M} = 0.0032292 \, \text{mol}
\]
3. Since the molar ratio of HCl to NaOH is 1:1, moles of HCl = moles of NaOH = 0.0032292 mol.
4. Calculate the molarity of the HCl solution:
\[
[\text{HCl}] = \frac{\text{Moles of HCl}}{\text{Volume of HCl (L)}} = \frac{0.0032292 \, \text{mol}}{0.025 \, \text{L}} = 0.129168 \, \text{M}
\]
Thus, the concentration of the HCl solution is approximately 0.129 M.
**Source:** McQuarrie, Rock, and Gallogly 4e – General Chemistry
**Publisher:** University Science Books
Expert Solution
Step 1
Given -
Volume of NaOH= 29.9 mL = V1
Concentration of NaOH= 0.1 08 M = M1
Volume of HCl =25.0 mL = V2
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