The diagram below represents a monoprotic acid, HA. If 14.75 mL 0.0200 M KOH are added to 32.47 mL 0.0100 M solution of the above acid, which is the final pH? O4.08 4.23 2.08 3.08
The diagram below represents a monoprotic acid, HA. If 14.75 mL 0.0200 M KOH are added to 32.47 mL 0.0100 M solution of the above acid, which is the final pH? O4.08 4.23 2.08 3.08
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Mark S. Cracolice, Ed Peters
Chapter9: Chemical Change
Section: Chapter Questions
Problem 79E
Related questions
Concept explainers
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
Question
**Description of Diagram:**
The diagram shows a beaker with a liquid containing two types of particles: larger red spheres and smaller blue spheres. The larger red spheres likely represent the undissociated form of the monoprotic acid (HA), while the smaller blue spheres represent either the hydrogen ions (H⁺) or the conjugate base (A⁻) formed upon dissociation.
**Problem Statement:**
If 14.75 mL of 0.0200 M KOH is added to 32.47 mL of 0.0100 M solution of the above acid, what is the final pH?
**Answer Choices:**
- 4.08
- 4.23
- 2.08
- 3.08
---
Here's a step-by-step explanation for finding the pH after the addition of KOH to the monoprotic acid solution:
1. Calculate the moles of KOH added:
\[
\text{Moles of KOH} = 14.75 \text{ mL} \times \left( \frac{0.0200 \text{ moles}}{1000 \text{ mL}} \right) = 0.000295 \text{ moles}
\]
2. Calculate the moles of the acid (HA) initially present:
\[
\text{Moles of HA} = 32.47 \text{ mL} \times \left( \frac{0.0100 \text{ moles}}{1000 \text{ mL}} \right) = 0.0003247 \text{ moles}
\]
3. Since KOH is a strong base, it will completely neutralize the acid. Determine the remaining moles of HA and the moles of A⁻ produced:
\[
\text{Moles of HA after reaction} = 0.0003247 \text{ moles} - 0.000295 \text{ moles} = 0.0000297 \text{ moles}
\]
\[
\text{Moles of A⁻ formed} = 0.000295 \text{ moles}
\]
4. Determine](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F277ae00b-d056-42ed-a729-386f4fca5973%2F02935c2b-281d-41b9-9b57-4db0a43ccd16%2Fto10p9b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 7
The diagram below represents a monoprotic acid, HA.
**Description of Diagram:**
The diagram shows a beaker with a liquid containing two types of particles: larger red spheres and smaller blue spheres. The larger red spheres likely represent the undissociated form of the monoprotic acid (HA), while the smaller blue spheres represent either the hydrogen ions (H⁺) or the conjugate base (A⁻) formed upon dissociation.
**Problem Statement:**
If 14.75 mL of 0.0200 M KOH is added to 32.47 mL of 0.0100 M solution of the above acid, what is the final pH?
**Answer Choices:**
- 4.08
- 4.23
- 2.08
- 3.08
---
Here's a step-by-step explanation for finding the pH after the addition of KOH to the monoprotic acid solution:
1. Calculate the moles of KOH added:
\[
\text{Moles of KOH} = 14.75 \text{ mL} \times \left( \frac{0.0200 \text{ moles}}{1000 \text{ mL}} \right) = 0.000295 \text{ moles}
\]
2. Calculate the moles of the acid (HA) initially present:
\[
\text{Moles of HA} = 32.47 \text{ mL} \times \left( \frac{0.0100 \text{ moles}}{1000 \text{ mL}} \right) = 0.0003247 \text{ moles}
\]
3. Since KOH is a strong base, it will completely neutralize the acid. Determine the remaining moles of HA and the moles of A⁻ produced:
\[
\text{Moles of HA after reaction} = 0.0003247 \text{ moles} - 0.000295 \text{ moles} = 0.0000297 \text{ moles}
\]
\[
\text{Moles of A⁻ formed} = 0.000295 \text{ moles}
\]
4. Determine
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 5 steps
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
Introductory Chemistry: An Active Learning Approa…
Chemistry
ISBN:
9781305079250
Author:
Mark S. Cracolice, Ed Peters
Publisher:
Cengage Learning
General Chemistry - Standalone book (MindTap Cour…
Chemistry
ISBN:
9781305580343
Author:
Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:
Cengage Learning
Introductory Chemistry: A Foundation
Chemistry
ISBN:
9781337399425
Author:
Steven S. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Introductory Chemistry: An Active Learning Approa…
Chemistry
ISBN:
9781305079250
Author:
Mark S. Cracolice, Ed Peters
Publisher:
Cengage Learning
General Chemistry - Standalone book (MindTap Cour…
Chemistry
ISBN:
9781305580343
Author:
Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:
Cengage Learning
Introductory Chemistry: A Foundation
Chemistry
ISBN:
9781337399425
Author:
Steven S. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:
9781938168390
Author:
Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:
OpenStax
World of Chemistry, 3rd edition
Chemistry
ISBN:
9781133109655
Author:
Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:
Brooks / Cole / Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:
9781337399074
Author:
John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:
Cengage Learning