Volume vs pH 14 12 10 equivalence point 2 equivoler ce point - 10 15 20 25 30 Volume of base added (mL), PH= pka pka= 5.0 %3D from the more precise graph. 5. 00 Hd
Volume vs pH 14 12 10 equivalence point 2 equivoler ce point - 10 15 20 25 30 Volume of base added (mL), PH= pka pka= 5.0 %3D from the more precise graph. 5. 00 Hd
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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I dont know if I’m doing it right

Transcribed Image Text:### Titration Curve Analysis
#### Task
2. Show equivalence point, ½ equivalence point, and pKa on the more precise graph.
#### Graph: Volume vs pH
The graph displayed shows the relationship between the volume of base added (in mL) and the resulting pH of the solution. The x-axis represents the volume of base added, ranging from 0 to 30 mL, while the y-axis represents the pH values, ranging from 0 to 14.
**Important Points on the Graph:**
- **Equivalence Point**: The steepest part of the curve, indicating a rapid change in pH. It is marked on the graph where the pH suddenly rises from approximately 7 to 12 with additional base volume.
- **½ Equivalence Point**: This is marked on the graph at half the volume of the equivalence point, where the pH measures around 5.
- **pKa Determination**: At the ½ equivalence point, pH = pKa is asserted, and hence pKa is identified as 5.0.
#### Calculation
3. **Calculate \( K_a \) from the more precise graph**:
- Given that:
- \( \text{pKa} = 5.0 \)
- The equation used:
- \( K_a = 10^{-\text{pKa}} \)
- \( K_a = 10^{-5.0} \)
The result of this calculation is not provided in the image but should result in \( K_a \) being equal to \( 1.0 \times 10^{-5} \).
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