mL In the titration of 46.0 mL of 0.400 M HCOOH with 0.150 M 1 3 LIOH, how many mL of LIOH are required to reach the halfway point? 4 5 6. C 7 8 9
mL In the titration of 46.0 mL of 0.400 M HCOOH with 0.150 M 1 3 LIOH, how many mL of LIOH are required to reach the halfway point? 4 5 6. C 7 8 9
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 Calculation Example
**Problem Statement:**
In the titration of 46.0 mL of 0.400 M HCOOH (formic acid) with 0.150 M LiOH (lithium hydroxide), how many mL of LiOH are required to reach the halfway point?
**Steps to Solve:**
1. **Understanding the Problem**:
- We have a formic acid solution with a known concentration and volume.
- We need to find the volume of LiOH solution needed to reach the halfway point of the titration.
2. **Key Concepts**:
- The halfway point in titration of a weak acid with a strong base is when half of the acid has been neutralized.
- At the halfway point, the number of moles of base added equals half the number of moles of acid initially present.
3. **Calculations**:
- **Determine Initial Moles of HCOOH**:
\[
\text{Moles of HCOOH} = \text{Molarity} \times \text{Volume} \quad (\text{in Liters})
\]
\[
\text{Moles of HCOOH} = 0.400 \, \text{M} \times 46.0 \, \text{mL} \times \frac{1\, \text{L}}{1000\, \text{mL}} = 0.0184 \, \text{moles}
\]
- **Half of the Moles of HCOOH**:
\[
\text{Half} = \frac{0.0184 \, \text{moles}}{2} = 0.0092 \, \text{moles}
\]
- **Volume of 0.150 M LiOH to Provide 0.0092 moles**:
\[
\text{Volume} = \frac{\text{Moles}}{\text{Molarity}}
\]
\[
\text{Volume LiOH} = \frac{0.0092 \, \text{moles}}{0.150 \, \text{M}} = 0.0613 \, \text{L}
\]
\[
\text{Volume LiOH in mL}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabf3f71a-8f38-4387-af5c-ba72dda07b09%2Fc4e01f82-28bb-4170-98b1-193d22ed3111%2F5xtkppu_processed.png&w=3840&q=75)
Transcribed Image Text:### Titration Calculation Example
**Problem Statement:**
In the titration of 46.0 mL of 0.400 M HCOOH (formic acid) with 0.150 M LiOH (lithium hydroxide), how many mL of LiOH are required to reach the halfway point?
**Steps to Solve:**
1. **Understanding the Problem**:
- We have a formic acid solution with a known concentration and volume.
- We need to find the volume of LiOH solution needed to reach the halfway point of the titration.
2. **Key Concepts**:
- The halfway point in titration of a weak acid with a strong base is when half of the acid has been neutralized.
- At the halfway point, the number of moles of base added equals half the number of moles of acid initially present.
3. **Calculations**:
- **Determine Initial Moles of HCOOH**:
\[
\text{Moles of HCOOH} = \text{Molarity} \times \text{Volume} \quad (\text{in Liters})
\]
\[
\text{Moles of HCOOH} = 0.400 \, \text{M} \times 46.0 \, \text{mL} \times \frac{1\, \text{L}}{1000\, \text{mL}} = 0.0184 \, \text{moles}
\]
- **Half of the Moles of HCOOH**:
\[
\text{Half} = \frac{0.0184 \, \text{moles}}{2} = 0.0092 \, \text{moles}
\]
- **Volume of 0.150 M LiOH to Provide 0.0092 moles**:
\[
\text{Volume} = \frac{\text{Moles}}{\text{Molarity}}
\]
\[
\text{Volume LiOH} = \frac{0.0092 \, \text{moles}}{0.150 \, \text{M}} = 0.0613 \, \text{L}
\]
\[
\text{Volume LiOH in mL}
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