Mr. Smith titrated exactly one-half of a weak acid solution to the endpoint. He then combined it with the other half of the weak acid solution. The pH was measured at 5.75. What is the Ka for the acid? A Ka = [?] x 10[²] X
Mr. Smith titrated exactly one-half of a weak acid solution to the endpoint. He then combined it with the other half of the weak acid solution. The pH was measured at 5.75. What is the Ka for the acid? A Ka = [?] x 10[²] 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...
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![### Acid-Base Titration Analysis
**Problem Statement:**
Mr. Smith titrated exactly one-half of a weak acid solution to the endpoint. He then combined it with the other half of the weak acid solution. The pH was measured at 5.75.
**Question:**
What is the \(K_a\) for the acid?
### Calculation
To find the acid dissociation constant (\(K_a\)), use the given information and the formula for \(K_a\).
For a weak acid (HA) and its salt (A^-), at the point where the solution is made up of equal parts of HA and A^-, the pH equals the pKa.
Given:
- pH = 5.75
Since pH = pKa at this point,
\[ \text{pKa} = \text{pH} = 5.75 \]
Convert pKa to \(K_a\):
\[ \text{pKa} = -\log(K_a) \]
\[ 5.75 = -\log(K_a) \]
\[ K_a = 10^{-5.75} \]
### Input Fields:
- Coefficient (green): \[?\]
- Exponent (yellow): \[?\]
The solution to the \(K_a\) value:
\[ K_a = [1.78] \times 10^{-6} \]
Thus, the:
- **Coefficient (green)** is \(1.78\)
- **Exponent (yellow)** is \(-6\)
### Interactive Component:
Users can enter the values of the coefficient and exponent in the provided fields, then press the "Enter" button to verify their answers.
\[
K_a = 1.78 \times 10^{-6}
\]
**Note:**
The structure of this problem encourages understanding of the relationship between pH, pKa, and \(K_a\), and how to transition between these values using logarithmic functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4c65346-a6bd-47be-a891-a239e86f6fd1%2F05eb8447-b4eb-4329-b7cb-7cae2ca21dca%2Ft1qx1k7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Acid-Base Titration Analysis
**Problem Statement:**
Mr. Smith titrated exactly one-half of a weak acid solution to the endpoint. He then combined it with the other half of the weak acid solution. The pH was measured at 5.75.
**Question:**
What is the \(K_a\) for the acid?
### Calculation
To find the acid dissociation constant (\(K_a\)), use the given information and the formula for \(K_a\).
For a weak acid (HA) and its salt (A^-), at the point where the solution is made up of equal parts of HA and A^-, the pH equals the pKa.
Given:
- pH = 5.75
Since pH = pKa at this point,
\[ \text{pKa} = \text{pH} = 5.75 \]
Convert pKa to \(K_a\):
\[ \text{pKa} = -\log(K_a) \]
\[ 5.75 = -\log(K_a) \]
\[ K_a = 10^{-5.75} \]
### Input Fields:
- Coefficient (green): \[?\]
- Exponent (yellow): \[?\]
The solution to the \(K_a\) value:
\[ K_a = [1.78] \times 10^{-6} \]
Thus, the:
- **Coefficient (green)** is \(1.78\)
- **Exponent (yellow)** is \(-6\)
### Interactive Component:
Users can enter the values of the coefficient and exponent in the provided fields, then press the "Enter" button to verify their answers.
\[
K_a = 1.78 \times 10^{-6}
\]
**Note:**
The structure of this problem encourages understanding of the relationship between pH, pKa, and \(K_a\), and how to transition between these values using logarithmic functions.
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