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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What is the pH of the solution
![### Calculating pH from Hydroxide Ion Concentration
**Problem Statement**:
Given that the hydroxide ion concentration \([OH^-]\) is \(3.5 \times 10^{-5}\), what is the pH of the solution?
---
To solve this problem, we need to follow these steps:
1. **Calculate the pOH**:
Use the formula:
\[
\text{pOH} = -\log[OH^-]
\]
Plugging in the given concentration:
\[
\text{pOH} = -\log(3.5 \times 10^{-5})
\]
Using logarithm properties, this can be approximated as:
\[
\text{pOH} \approx -(\log(3.5) + \log(10^{-5}))
\]
Since \( \log(10^{-5}) = -5 \):
\[
\text{pOH} \approx -(\log(3.5) - 5)
\]
The value of \(\log(3.5)\) is approximately \(0.54\):
\[
\text{pOH} \approx -(0.54 - 5) = 5 - 0.54 = 4.46
\]
2. **Calculate the pH**:
Use the relationship between pH and pOH:
\[
\text{pH} + \text{pOH} = 14
\]
Substitute the pOH into the equation:
\[
\text{pH} = 14 - \text{pOH} = 14 - 4.46 = 9.54
\]
Thus, the pH of the solution is \(9.54\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11cb493f-74d8-445b-bfd0-ead085c73a2d%2F9df71cc0-492e-43c6-94f7-f0873cdc50d0%2Fejw9fqr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating pH from Hydroxide Ion Concentration
**Problem Statement**:
Given that the hydroxide ion concentration \([OH^-]\) is \(3.5 \times 10^{-5}\), what is the pH of the solution?
---
To solve this problem, we need to follow these steps:
1. **Calculate the pOH**:
Use the formula:
\[
\text{pOH} = -\log[OH^-]
\]
Plugging in the given concentration:
\[
\text{pOH} = -\log(3.5 \times 10^{-5})
\]
Using logarithm properties, this can be approximated as:
\[
\text{pOH} \approx -(\log(3.5) + \log(10^{-5}))
\]
Since \( \log(10^{-5}) = -5 \):
\[
\text{pOH} \approx -(\log(3.5) - 5)
\]
The value of \(\log(3.5)\) is approximately \(0.54\):
\[
\text{pOH} \approx -(0.54 - 5) = 5 - 0.54 = 4.46
\]
2. **Calculate the pH**:
Use the relationship between pH and pOH:
\[
\text{pH} + \text{pOH} = 14
\]
Substitute the pOH into the equation:
\[
\text{pH} = 14 - \text{pOH} = 14 - 4.46 = 9.54
\]
Thus, the pH of the solution is \(9.54\).
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