Try Again Your answer is incorrect. 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. 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.
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![**Calculating the pH at Equivalence of a Titration**
A chemist titrates 160.0 mL of a 0.5839 M benzoic acid \((\text{HC}_7\text{H}_5\text{O}_2)\) solution with a 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.*
**Calculated pH:**
pH = 9.22
[Explanation] [Recheck]
This setup involves the calculation of pH at the equivalence point during a titration where a weak acid (benzoic acid) reacts with a strong base (KOH). At equivalence, all the benzoic acid has reacted with KOH, forming the benzoate ion. The pH is then determined by the hydrolysis of this ion. The key data inputs include the initial volumes and concentrations, the \(pK_a\) of benzoic acid, and finally, the rounded calculation for pH based on these inputs.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f35f950-23fc-4555-bd77-58b8b6d22853%2F8fd523f1-a743-4019-83e6-062ef2b6f8da%2Ftaie54_processed.jpeg&w=3840&q=75)
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