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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![### Buffer Solution Calculation Example
A buffer solution made from HClO and KClO has a pH of 7.75. If \( pK_a \) for HClO is 7.46, what is the \(\frac{[\text{ClO}^-]}{[\text{HClO}]}\) in the buffer?
Using the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where:
- \( \text{pH} \) is the pH of the buffer solution.
- \( pK_a \) is the acid dissociation constant.
- \(\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\) is the ratio of the concentrations of the conjugate base (\(\text{A}^-\)) and the weak acid (\(\text{HA})).\)
Substituting the given values:
\[ 7.75 = 7.46 + \log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) \]
\[\log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) = 7.75 - 7.46 \]
\[\log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) = 0.29 \]
Converting the logarithmic value back to the ratio:
\[\frac{[\text{ClO}^-]}{[\text{HClO}]} = 10^{0.29} \]
Using a calculator, we find:
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} \approx 1.95 \]
Thus, the ratio \(\frac{[\text{ClO}^-]}{[\text{HClO}]}\) in the buffer is approximately 1.95, which should be entered into the provided box.
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} = \boxed{1.95} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa527ffdc-16df-4a1f-9953-96956a75d05e%2F9326771c-8a18-4529-910d-f320777d2efd%2F2xxkapu.png&w=3840&q=75)
Transcribed Image Text:### Buffer Solution Calculation Example
A buffer solution made from HClO and KClO has a pH of 7.75. If \( pK_a \) for HClO is 7.46, what is the \(\frac{[\text{ClO}^-]}{[\text{HClO}]}\) in the buffer?
Using the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where:
- \( \text{pH} \) is the pH of the buffer solution.
- \( pK_a \) is the acid dissociation constant.
- \(\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\) is the ratio of the concentrations of the conjugate base (\(\text{A}^-\)) and the weak acid (\(\text{HA})).\)
Substituting the given values:
\[ 7.75 = 7.46 + \log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) \]
\[\log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) = 7.75 - 7.46 \]
\[\log\left(\frac{[\text{ClO}^-]}{[\text{HClO}]}\right) = 0.29 \]
Converting the logarithmic value back to the ratio:
\[\frac{[\text{ClO}^-]}{[\text{HClO}]} = 10^{0.29} \]
Using a calculator, we find:
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} \approx 1.95 \]
Thus, the ratio \(\frac{[\text{ClO}^-]}{[\text{HClO}]}\) in the buffer is approximately 1.95, which should be entered into the provided box.
\[ \frac{[\text{ClO}^-]}{[\text{HClO}]} = \boxed{1.95} \]
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