Pre-lab question #9: Assuming that Ka is 1.85 x 105 for acetic acid, calculate the pH at the one-half the equivalence point for a titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH. 4.73 0.100 5.98 1.85 x 10-5
Pre-lab question #9: Assuming that Ka is 1.85 x 105 for acetic acid, calculate the pH at the one-half the equivalence point for a titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH. 4.73 0.100 5.98 1.85 x 10-5
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Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Pre-lab Question #9:
**Question:** Assuming that \(K_a\) is \(1.85 \times 10^{-5}\) for acetic acid, calculate the pH at the one-half equivalence point for a titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH.
**Answer Options:**
1. \( \bigcirc \) 4.73
2. \( \bigcirc \) 0.100
3. \( \bigcirc \) 5.98
4. \( \bigcirc \) \( 1.85 \times 10^{-5} \)
---
**Explanation:**
At the one-half equivalence point in a titration of a weak acid with a strong base, the pH of the solution is equal to the pKa of the acid. This is because the concentration of the acid is equal to the concentration of its conjugate base, and thus the Henderson-Hasselbalch equation simplifies to:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
\[ \text{pH} = \text{pKa} + \log (1) \]
\[ \text{pH} = \text{pKa} \]
Given \( K_a = 1.85 \times 10^{-5} \),
\[ \text{pKa} = -\log(1.85 \times 10^{-5}) \approx 4.73 \]
Thus, the correct answer is:
\[ \bigcirc \, 4.73 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcffe6fb8-a8b1-4738-951c-b41ccaeecf4e%2F7b1a7286-83bd-4d66-b7be-97cf6e9652aa%2Finr16zd_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Pre-lab Question #9:
**Question:** Assuming that \(K_a\) is \(1.85 \times 10^{-5}\) for acetic acid, calculate the pH at the one-half equivalence point for a titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH.
**Answer Options:**
1. \( \bigcirc \) 4.73
2. \( \bigcirc \) 0.100
3. \( \bigcirc \) 5.98
4. \( \bigcirc \) \( 1.85 \times 10^{-5} \)
---
**Explanation:**
At the one-half equivalence point in a titration of a weak acid with a strong base, the pH of the solution is equal to the pKa of the acid. This is because the concentration of the acid is equal to the concentration of its conjugate base, and thus the Henderson-Hasselbalch equation simplifies to:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
\[ \text{pH} = \text{pKa} + \log (1) \]
\[ \text{pH} = \text{pKa} \]
Given \( K_a = 1.85 \times 10^{-5} \),
\[ \text{pKa} = -\log(1.85 \times 10^{-5}) \approx 4.73 \]
Thus, the correct answer is:
\[ \bigcirc \, 4.73 \]
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