5. A student conducted a titration experiment. He/she used 12.0 mL of silver nitrate (titrant) solution to reach the endpoint. If the student used 25.4 mL of the chloride (analyte) for the titration and found the concentration of chloride ions to be 2.2 M, what is the mass (in grams) of silver nitrate titrant LaClic
5. A student conducted a titration experiment. He/she used 12.0 mL of silver nitrate (titrant) solution to reach the endpoint. If the student used 25.4 mL of the chloride (analyte) for the titration and found the concentration of chloride ions to be 2.2 M, what is the mass (in grams) of silver nitrate titrant LaClic
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Titration Experiment Calculation**
In this problem, a student conducted a titration experiment. The student used 12.0 mL of silver nitrate (the titrant solution) to reach the endpoint of the titration. In conducting this experiment, the student used 25.4 mL of the chloride analyte solution and determined that the concentration of chloride ions in the solution was 2.2 M. The task is to calculate the mass (in grams) of the silver nitrate titrant used in the experiment.
**Problem Statement:**
A student conducted a titration experiment. He/she used 12.0 mL of silver nitrate (titrant) solution to reach the endpoint. If the student used 25.4 mL of the chloride (analyte) for the titration and found the concentration of chloride ions to be 2.2 M, what is the mass (in grams) of silver nitrate titrant?
**Step-by-Step Solution:**
To find the mass of the silver nitrate titrant, follow these steps:
1. **Calculate the moles of chloride ions:**
- The concentration of chloride ions given is 2.2 M.
- The volume of the chloride solution used is 25.4 mL, which is equivalent to 0.0254 L (since 1 mL = 0.001 L).
Using the formula:
\[ \text{Moles of Chloride Ions} = \text{Molarity} \times \text{Volume (in liters)} \]
\[ \text{Moles of Chloride Ions} = 2.2 \, \text{M} \times 0.0254 \, \text{L} \]
\[ \text{Moles of Chloride Ions} = 0.05588 \, \text{mol} \]
2. **Determine the moles of silver nitrate:**
- The reaction between silver nitrate (AgNO₃) and chloride ions (Cl⁻) is as follows:
\[ \text{AgNO₃} + \text{Cl}^- \rightarrow \text{AgCl} + \text{NO₃}^- \]
In this reaction, 1 mole of AgNO₃ reacts with 1 mole of Cl⁻. Therefore, the moles of AgNO₃](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9feb948d-7f0e-4488-9fa7-4d9b218ef655%2Fca478151-59f4-45e3-8033-5dd41ef7d98c%2Fq38jll_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration Experiment Calculation**
In this problem, a student conducted a titration experiment. The student used 12.0 mL of silver nitrate (the titrant solution) to reach the endpoint of the titration. In conducting this experiment, the student used 25.4 mL of the chloride analyte solution and determined that the concentration of chloride ions in the solution was 2.2 M. The task is to calculate the mass (in grams) of the silver nitrate titrant used in the experiment.
**Problem Statement:**
A student conducted a titration experiment. He/she used 12.0 mL of silver nitrate (titrant) solution to reach the endpoint. If the student used 25.4 mL of the chloride (analyte) for the titration and found the concentration of chloride ions to be 2.2 M, what is the mass (in grams) of silver nitrate titrant?
**Step-by-Step Solution:**
To find the mass of the silver nitrate titrant, follow these steps:
1. **Calculate the moles of chloride ions:**
- The concentration of chloride ions given is 2.2 M.
- The volume of the chloride solution used is 25.4 mL, which is equivalent to 0.0254 L (since 1 mL = 0.001 L).
Using the formula:
\[ \text{Moles of Chloride Ions} = \text{Molarity} \times \text{Volume (in liters)} \]
\[ \text{Moles of Chloride Ions} = 2.2 \, \text{M} \times 0.0254 \, \text{L} \]
\[ \text{Moles of Chloride Ions} = 0.05588 \, \text{mol} \]
2. **Determine the moles of silver nitrate:**
- The reaction between silver nitrate (AgNO₃) and chloride ions (Cl⁻) is as follows:
\[ \text{AgNO₃} + \text{Cl}^- \rightarrow \text{AgCl} + \text{NO₃}^- \]
In this reaction, 1 mole of AgNO₃ reacts with 1 mole of Cl⁻. Therefore, the moles of AgNO₃
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