Since phosphoric acid is polyprotic, it can create several buffer regions. One buffer forms between H2PO4- and HPO42-. Given the conditions below, what is the pH of this buffer solution? = 6.23 x 10-8 Канарот 0.50 M H2PO4- + 0.50 M HPO42- pH of buffer=[?] pH of Buffer I pH of Buffer Enter
Since phosphoric acid is polyprotic, it can create several buffer regions. One buffer forms between H2PO4- and HPO42-. Given the conditions below, what is the pH of this buffer solution? = 6.23 x 10-8 Канарот 0.50 M H2PO4- + 0.50 M HPO42- pH of buffer=[?] pH of Buffer I pH of Buffer Enter
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![### Phosphoric Acid Buffer Regions
Phosphoric acid is a polyprotic acid, meaning it can dissociate multiple times, creating several buffer regions. One significant buffer system involves the equilibrium between the dihydrogen phosphate ion (H₂PO₄⁻) and the hydrogen phosphate ion (HPO₄²⁻). This buffer system can be used to maintain pH in specific ranges.
#### Buffer System Between H₂PO₄⁻ and HPO₄²⁻
Given the dissociation constant (Ka) and the concentrations of the acid and its conjugate base, we can calculate the pH of the buffer solution. For the given buffer system:
- Ka(H₂PO₄⁻) = 6.23 × 10⁻⁸
- Concentrations:
- 0.50 M H₂PO₄⁻
- 0.50 M HPO₄²⁻
#### pH Calculation
To find the pH of this buffer solution, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Given that the concentrations of the acid and base are equal:
\[ \text{pH} = \text{pKa} \]
To find pKa, we take the negative logarithm of Ka:
\[ \text{pKa} = -\log(6.23 \times 10⁻⁸) \]
Therefore, the pH of the buffer is simply the pKa value. Use the equation above for the calculation.
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#### Interactive Element
At the bottom of the image, there is an interactive element for users to input the calculated pH value. Users can type in their answer in the provided text box and then click the "Enter" button to submit their response, likely to check if it is correct.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e6a7604-eb1c-41a1-a623-753603ba28f3%2F23ee0c10-260f-4143-ae3f-7824ae5378a6%2Flxsw2w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Phosphoric Acid Buffer Regions
Phosphoric acid is a polyprotic acid, meaning it can dissociate multiple times, creating several buffer regions. One significant buffer system involves the equilibrium between the dihydrogen phosphate ion (H₂PO₄⁻) and the hydrogen phosphate ion (HPO₄²⁻). This buffer system can be used to maintain pH in specific ranges.
#### Buffer System Between H₂PO₄⁻ and HPO₄²⁻
Given the dissociation constant (Ka) and the concentrations of the acid and its conjugate base, we can calculate the pH of the buffer solution. For the given buffer system:
- Ka(H₂PO₄⁻) = 6.23 × 10⁻⁸
- Concentrations:
- 0.50 M H₂PO₄⁻
- 0.50 M HPO₄²⁻
#### pH Calculation
To find the pH of this buffer solution, we use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Given that the concentrations of the acid and base are equal:
\[ \text{pH} = \text{pKa} \]
To find pKa, we take the negative logarithm of Ka:
\[ \text{pKa} = -\log(6.23 \times 10⁻⁸) \]
Therefore, the pH of the buffer is simply the pKa value. Use the equation above for the calculation.
---
#### Interactive Element
At the bottom of the image, there is an interactive element for users to input the calculated pH value. Users can type in their answer in the provided text box and then click the "Enter" button to submit their response, likely to check if it is correct.
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