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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If a buffer solution is 0.110 M in a weak base (?b= 7.2×10^−5) and 0.460 M in its conjugate acid, what is the pH?
![### Calculating the pH of a Buffer Solution
**Problem:**
If a buffer solution is 0.110 M in a weak base (K_b = 7.2 × 10^-5) and 0.460 M in its conjugate acid, what is the pH?
**Solution:**
To find the pH of a buffer solution, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
First, we need to find \( \text{p}K_a \). Since we are given \( K_b \) for the weak base, we can convert it to \( K_a \) using the relationship:
\[ K_a \cdot K_b = K_w \]
where \( K_w = 1.0 \times 10^{-14} \).
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{7.2 \times 10^{-5}} \]
\[ K_a \approx 1.39 \times 10^{-10} \]
Now, we calculate \( \text{p}K_a \):
\[ \text{p}K_a = -\log(K_a) = -\log(1.39 \times 10^{-10}) \]
\[ \text{p}K_a \approx 9.86 \]
Next, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = 9.86 + \log\left(\frac{0.110}{0.460}\right) \]
Calculating the ratio inside the logarithm:
\[ \frac{0.110}{0.460} \approx 0.239 \]
Finally,
\[ \text{pH} = 9.86 + \log(0.239) \]
\[ \text{log}(0.239) \approx -0.622 \]
\[ \text{pH} = 9.86 - 0.622 \]
\[ \text{pH} \approx 9.24 \]
**pH = 9.24**
This calculated pH represents the value for the buffer solution containing 0.110 M of the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85650e2f-17ad-4aa4-9d56-acd675fbc675%2Fbe7105aa-ad29-4542-b69b-198fae389ce0%2Fnhf7aru_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the pH of a Buffer Solution
**Problem:**
If a buffer solution is 0.110 M in a weak base (K_b = 7.2 × 10^-5) and 0.460 M in its conjugate acid, what is the pH?
**Solution:**
To find the pH of a buffer solution, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
First, we need to find \( \text{p}K_a \). Since we are given \( K_b \) for the weak base, we can convert it to \( K_a \) using the relationship:
\[ K_a \cdot K_b = K_w \]
where \( K_w = 1.0 \times 10^{-14} \).
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{7.2 \times 10^{-5}} \]
\[ K_a \approx 1.39 \times 10^{-10} \]
Now, we calculate \( \text{p}K_a \):
\[ \text{p}K_a = -\log(K_a) = -\log(1.39 \times 10^{-10}) \]
\[ \text{p}K_a \approx 9.86 \]
Next, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = 9.86 + \log\left(\frac{0.110}{0.460}\right) \]
Calculating the ratio inside the logarithm:
\[ \frac{0.110}{0.460} \approx 0.239 \]
Finally,
\[ \text{pH} = 9.86 + \log(0.239) \]
\[ \text{log}(0.239) \approx -0.622 \]
\[ \text{pH} = 9.86 - 0.622 \]
\[ \text{pH} \approx 9.24 \]
**pH = 9.24**
This calculated pH represents the value for the buffer solution containing 0.110 M of the
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