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

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...
icon
Related questions
icon
Concept explainers
Question
### Question 7

The diagram below represents a monoprotic acid, HA.

![Diagram of monoprotic acid](link-to-image)

**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
Transcribed Image Text:### Question 7 The diagram below represents a monoprotic acid, HA. ![Diagram of monoprotic acid](link-to-image) **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
steps

Step by step

Solved in 5 steps

Blurred answer
Knowledge Booster
Ionic Equilibrium
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.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY