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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![The text in the image is as follows:
**The pKa value for H₂S is 7.00. What mole ratio of NaHS to H₂S is needed to prepare a buffer with a pH of 7.31?**
\[ \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} = \]
### Explanation for Educational Website
This problem involves the preparation of a buffer solution using sodium hydrogen sulfide (NaHS) and hydrogen sulfide (H₂S). To determine the required mole ratio of NaHS to H₂S to achieve a specific pH, one can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Where:
- pH is the desired pH of the buffer solution.
- pKa is the acid dissociation constant of the weak acid (H₂S in this case).
- [Base] is the concentration of the base form (HS⁻).
- [Acid] is the concentration of the acid form (H₂S).
In this case:
- pH = 7.31
- pKa = 7.00
Using the Henderson-Hasselbalch equation, the calculation becomes:
\[ 7.31 = 7.00 + \log \left( \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \right) \]
Solving for the ratio \(\frac{[\text{HS}^-]}{[\text{H}_2\text{S}]}\):
\[ 7.31 - 7.00 = \log \left( \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \right) \]
\[ 0.31 = \log \left( \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \right) \]
\[ 10^{0.31} = \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \]
Using a calculator, we find:
\[ \frac{[\text{HS}^-]}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fab4efe3b-fd54-4aa0-8660-b031c889fb2d%2Ffe31c28d-9047-469a-9b8e-9bc015c7134d%2Fnk4r40r_processed.png&w=3840&q=75)
Transcribed Image Text:The text in the image is as follows:
**The pKa value for H₂S is 7.00. What mole ratio of NaHS to H₂S is needed to prepare a buffer with a pH of 7.31?**
\[ \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} = \]
### Explanation for Educational Website
This problem involves the preparation of a buffer solution using sodium hydrogen sulfide (NaHS) and hydrogen sulfide (H₂S). To determine the required mole ratio of NaHS to H₂S to achieve a specific pH, one can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Where:
- pH is the desired pH of the buffer solution.
- pKa is the acid dissociation constant of the weak acid (H₂S in this case).
- [Base] is the concentration of the base form (HS⁻).
- [Acid] is the concentration of the acid form (H₂S).
In this case:
- pH = 7.31
- pKa = 7.00
Using the Henderson-Hasselbalch equation, the calculation becomes:
\[ 7.31 = 7.00 + \log \left( \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \right) \]
Solving for the ratio \(\frac{[\text{HS}^-]}{[\text{H}_2\text{S}]}\):
\[ 7.31 - 7.00 = \log \left( \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \right) \]
\[ 0.31 = \log \left( \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \right) \]
\[ 10^{0.31} = \frac{[\text{HS}^-]}{[\text{H}_2\text{S}]} \]
Using a calculator, we find:
\[ \frac{[\text{HS}^-]}{
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