A buffer solution contains 0.447 M hypochlorous acid and 0.236 M potassium hypochlorite. If 0.0335 moles of sodium hydroxide are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding sodium hydroxide) pH = =
A buffer solution contains 0.447 M hypochlorous acid and 0.236 M potassium hypochlorite. If 0.0335 moles of sodium hydroxide are added to 250 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding sodium hydroxide) pH = =
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
Problem 57P
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![**Buffer Solution Problem**
A buffer solution contains 0.447 M hypochlorous acid and 0.236 M potassium hypochlorite.
If 0.0335 moles of sodium hydroxide are added to 250 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume does not change upon adding sodium hydroxide)
\[ \text{pH} = \_\_\_\_ \]
---
**Explanation:**
To solve this problem, you will need to understand how adding a strong base (sodium hydroxide) to a buffer affects its pH. You can use the Henderson-Hasselbalch equation which is given as:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{p}K_a\) is the negative base-10 logarithm of the acid dissociation constant (\(K_a\)) of the weak acid (hypochlorous acid in this case).
- \([\text{A}^-]\) is the concentration of the conjugate base (hypochlorite ion).
- \([\text{HA}]\) is the concentration of the weak acid (hypochlorous acid).
**Steps to Calculate:**
1. **Calculate the initial moles of Hypochlorous Acid (HA) and Hypochlorite Ion (A⁻)**:
- Initial moles of hypochlorous acid ([\(\text{HA}\)]): \(0.447 \, M \times 0.250 \, L = 0.11175 \, \text{moles}\)
- Initial moles of hypochlorite ([\(\text{A}^-\)]): \(0.236 \, M \times 0.250 \, L = 0.059 \, \text{moles}\)
2. **New concentrations after adding NaOH**:
- Since NaOH is a strong base, it will react completely with the weak acid (\(\text{HA}\)).
- Moles of NaOH added: \(0.0335 \, \text{moles}\)
- After reaction: New \([\text{HA}]\) = Initial \([\text{HA](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5baa0102-60f7-4f8d-865b-4da67ad08e15%2F2e41cc2c-9dbe-400f-9530-c25b5c3efbb1%2F4jpnr5m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Buffer Solution Problem**
A buffer solution contains 0.447 M hypochlorous acid and 0.236 M potassium hypochlorite.
If 0.0335 moles of sodium hydroxide are added to 250 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume does not change upon adding sodium hydroxide)
\[ \text{pH} = \_\_\_\_ \]
---
**Explanation:**
To solve this problem, you will need to understand how adding a strong base (sodium hydroxide) to a buffer affects its pH. You can use the Henderson-Hasselbalch equation which is given as:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{p}K_a\) is the negative base-10 logarithm of the acid dissociation constant (\(K_a\)) of the weak acid (hypochlorous acid in this case).
- \([\text{A}^-]\) is the concentration of the conjugate base (hypochlorite ion).
- \([\text{HA}]\) is the concentration of the weak acid (hypochlorous acid).
**Steps to Calculate:**
1. **Calculate the initial moles of Hypochlorous Acid (HA) and Hypochlorite Ion (A⁻)**:
- Initial moles of hypochlorous acid ([\(\text{HA}\)]): \(0.447 \, M \times 0.250 \, L = 0.11175 \, \text{moles}\)
- Initial moles of hypochlorite ([\(\text{A}^-\)]): \(0.236 \, M \times 0.250 \, L = 0.059 \, \text{moles}\)
2. **New concentrations after adding NaOH**:
- Since NaOH is a strong base, it will react completely with the weak acid (\(\text{HA}\)).
- Moles of NaOH added: \(0.0335 \, \text{moles}\)
- After reaction: New \([\text{HA}]\) = Initial \([\text{HA
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