A chemist titrates 90.0 mL of a 0.7050M methylamine (CH3NH₂) solution with 0.5016M HBr solution at 25 °C. Calculate the pH at equivalence. The pK, of methylamine is 3.36. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added. pH = X
A chemist titrates 90.0 mL of a 0.7050M methylamine (CH3NH₂) solution with 0.5016M HBr solution at 25 °C. Calculate the pH at equivalence. The pK, of methylamine is 3.36. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added. pH = X
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...
Related questions
Question
![### Titration of Methylamine with HBr Solution
**Problem Statement:**
A chemist titrates 90.0 mL of a 0.7050 M methylamine (CH₃NH₂) solution with 0.5016 M HBr solution at 25 °C. Calculate the pH at equivalence. The pK_b of methylamine is 3.36.
- Round your answer to 2 decimal places.
**Note for Advanced Students:**
You may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.
**Equation:**
- Methylamine: \( \text{CH}_3\text{NH}_2 \)
- \( \text{pH} = \, \)
(There is a calculation box followed by buttons for actions such as confirming or resetting.)
---
**Detailed Explanation:**
In this titration problem, the aim is to determine the pH of the solution at the equivalence point. Methylamine (CH₃NH₂) is a weak base, and HBr is a strong acid. At the equivalence point, the moles of HBr added will be equal to the moles of CH₃NH₂ initially present.
1. **Determine moles of CH₃NH₂:**
- Initial moles of CH₃NH₂ = Volume \(\times\) Molarity = 90.0 mL \(\times\) 0.7050 M = 0.06345 moles
2. **Volume of HBr required:**
- Since the moles of HBr added will be equal to the moles of CH₃NH₂,
Moles of HBr = 0.06345 moles
\[
\text{Volume of HBr} = \frac{\text{Moles of HBr}}{\text{Molarity of HBr}} = \frac{0.06345 \, \text{moles}}{0.5016 \, \text{M}} = 0.1265 \, \text{L} = 126.5 \, \text{mL}
\]
3. **Calculate the new volume at equivalence:**
- Total volume = Volume of CH₃NH₂ + Volume of HBr = 90.0 mL + 126](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c14a699-de9b-4c1d-9356-4bb35f1041ad%2Feb69ec09-30ca-4027-8382-d0749711574a%2F2b48b3n_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Titration of Methylamine with HBr Solution
**Problem Statement:**
A chemist titrates 90.0 mL of a 0.7050 M methylamine (CH₃NH₂) solution with 0.5016 M HBr solution at 25 °C. Calculate the pH at equivalence. The pK_b of methylamine is 3.36.
- Round your answer to 2 decimal places.
**Note for Advanced Students:**
You may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.
**Equation:**
- Methylamine: \( \text{CH}_3\text{NH}_2 \)
- \( \text{pH} = \, \)
(There is a calculation box followed by buttons for actions such as confirming or resetting.)
---
**Detailed Explanation:**
In this titration problem, the aim is to determine the pH of the solution at the equivalence point. Methylamine (CH₃NH₂) is a weak base, and HBr is a strong acid. At the equivalence point, the moles of HBr added will be equal to the moles of CH₃NH₂ initially present.
1. **Determine moles of CH₃NH₂:**
- Initial moles of CH₃NH₂ = Volume \(\times\) Molarity = 90.0 mL \(\times\) 0.7050 M = 0.06345 moles
2. **Volume of HBr required:**
- Since the moles of HBr added will be equal to the moles of CH₃NH₂,
Moles of HBr = 0.06345 moles
\[
\text{Volume of HBr} = \frac{\text{Moles of HBr}}{\text{Molarity of HBr}} = \frac{0.06345 \, \text{moles}}{0.5016 \, \text{M}} = 0.1265 \, \text{L} = 126.5 \, \text{mL}
\]
3. **Calculate the new volume at equivalence:**
- Total volume = Volume of CH₃NH₂ + Volume of HBr = 90.0 mL + 126
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY