**Calculating the pH at Equivalence of a Titration****Problem Statement:**A chemist titrates 160.0 mL of a 0.5839 M benzoic acid \((HC_7H_5O_2)\) solution with 0.4219 M KOH solution at 25°C. Calculate the pH at equivalence. The \(pK_a\) of benzoic acid is 4.20.Round your answer to 2 decimal places.**Note for Advanced Students:** You may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added.**Solution Input:**pH = \(\_\)[ ] **Options:**- Explanation- Check---This educational module presents a typical titration problem where a weak acid reacts with a strong base. At equivalence, all benzoic acid has reacted with potassium hydroxide, forming water and the conjugate base. Calculating the pH involves understanding the relationship between the \(pK_a\) of the acid and the concentration of its conjugate base in solution.

Given: Volume of benzoic acid = 160.0 mL Molarity of benzoic acid = 0.5839 M Molarity of KOH =…

Answered: A chemist titrates 160.0 mL of a 0.5839…

A chemist titrates 160.0 mL of a 0.5839 M benzoic acid (HC,H,CO₂) solution with 0.4219M KOH solution at 25 °C. Calculate the pH at equivalence. The pK, of benzoic acid is 4.20.

Chemistry

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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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Question

**Calculating the pH at Equivalence of a Titration**

**Problem Statement:**

A chemist titrates 160.0 mL of a 0.5839 M benzoic acid \((HC_7H_5O_2)\) solution with 0.4219 M KOH solution at 25°C. Calculate the pH at equivalence. The \(pK_a\) of benzoic acid is 4.20.

Round your answer to 2 decimal places.

**Note for Advanced Students:** You may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added.

**Solution Input:**

pH = \(\_\)

[ ]

**Options:**

- Explanation
- Check

---

This educational module presents a typical titration problem where a weak acid reacts with a strong base. At equivalence, all benzoic acid has reacted with potassium hydroxide, forming water and the conjugate base. Calculating the pH involves understanding the relationship between the \(pK_a\) of the acid and the concentration of its conjugate base in solution.

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