A 30.0-mL volume of 0.50 M CH3COOH (Ka = 1.8 × 10-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 °C. Express the pH numerically. • View Available Hint(s) ? pH = %3D
A 30.0-mL volume of 0.50 M CH3COOH (Ka = 1.8 × 10-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 °C. Express the pH numerically. • View Available Hint(s) ? pH = %3D
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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Transcribed Image Text:**Acetic Acid Titration Problem**
A 30.0-mL volume of 0.50 M CH₃COOH (Kₐ = 1.8 × 10⁻⁵) was titrated with 0.50 M NaOH. Calculate the pH after the addition of 30.0 mL of NaOH at 25 °C.
**Express the pH numerically.**
**Hint Section**
- A hint section is available for additional guidance during problem-solving.
**Graphical and Text Entry Tools**
- Tools for entering equations and text are provided, including options for inserting mathematical symbols and undo/redo functionality.
**Input Box**
- A field is provided to enter the calculated pH value.

Transcribed Image Text:### Titration Curve Explanation
This graph represents a titration curve, demonstrating the change in pH as a solution of 0.100 M NaOH is added to an acidic solution.
- **X-axis (Horizontal):** This axis measures the volume of 0.100 M NaOH added, in milliliters (mL), ranging from 0 to 80 mL.
- **Y-axis (Vertical):** This axis shows the pH level of the solution, ranging from 0 to 14.
- **Curve Description:** The curve starts at a low pH (acidic) and gradually increases as NaOH is added. A steep rise in pH is observed around the midpoint, indicating the rapid increase of pH as the acid is neutralized by the base.
- **Equivalence Point:** Marked by a blue dot, the equivalence point is where the moles of NaOH added are stoichiometrically equivalent to the moles of acid present in the solution. In this graph, the equivalence point occurs roughly at 40 mL of NaOH added, with a pH around 7, indicating the complete neutralization of the acid by the base.
This titration curve is typical of a strong acid-strong base titration, where the pH at the equivalence point is neutral.
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