E Calculating the pH at equivalence of a titration A chemist titrates 160.0 mL of a 0.4531 M butanoic acid (HC,H,CO₂) solution with 0.4662M KOH solution at 25 °C. Calculate the pH at equivalence. The PK, of butanoic acid is 4.82. 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. pH = X G
E Calculating the pH at equivalence of a titration A chemist titrates 160.0 mL of a 0.4531 M butanoic acid (HC,H,CO₂) solution with 0.4662M KOH solution at 25 °C. Calculate the pH at equivalence. The PK, of butanoic acid is 4.82. 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. pH = X G
Chemistry
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Chapter1: Chemical Foundations
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![**Calculating the pH at Equivalence of a Titration**
A chemist titrates 60.0 mL of a 0.4531 M butanoic acid \((\text{HC}_4\text{H}_7\text{O}_2)\) solution with 0.4662 M KOH solution at 25°C. Calculate the pH at equivalence. The \(pK_a\) of butanoic acid is 4.82.
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.*
Enter pH = [Input Box]
[Buttons: X (Clear) Check]
### Explanation
This problem involves a titration of a weak acid (butanoic acid) with a strong base (KOH). At the equivalence point, all the butanoic acid has been neutralized by the KOH. The solution will contain the conjugate base of butanoic acid, \(\text{C}_4\text{H}_7\text{O}_2^-\), which will affect the pH. The pH can be calculated using the formula:
\[ \text{pH} = \frac{1}{2}(\text{pK}_a + \text{pK}_w - \text{pK}_b) \]
Where \(\text{pK}_w\) at 25°C is 14. Because the solution contains only the conjugate base at equivalence, you can calculate the pH using its concentration and the \(K_a\) relationship:
\[ \text{pH} = 7 + \frac{1}{2}(\text{pK}_w - \text{pK}_a) \]
This is a simplified equation for calculating the pH at equivalence in a titration involving a weak acid and a strong base.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f35f950-23fc-4555-bd77-58b8b6d22853%2F9d4125c8-e19d-40bb-b577-4cb5d4a43c86%2F4hrcpyu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculating the pH at Equivalence of a Titration**
A chemist titrates 60.0 mL of a 0.4531 M butanoic acid \((\text{HC}_4\text{H}_7\text{O}_2)\) solution with 0.4662 M KOH solution at 25°C. Calculate the pH at equivalence. The \(pK_a\) of butanoic acid is 4.82.
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.*
Enter pH = [Input Box]
[Buttons: X (Clear) Check]
### Explanation
This problem involves a titration of a weak acid (butanoic acid) with a strong base (KOH). At the equivalence point, all the butanoic acid has been neutralized by the KOH. The solution will contain the conjugate base of butanoic acid, \(\text{C}_4\text{H}_7\text{O}_2^-\), which will affect the pH. The pH can be calculated using the formula:
\[ \text{pH} = \frac{1}{2}(\text{pK}_a + \text{pK}_w - \text{pK}_b) \]
Where \(\text{pK}_w\) at 25°C is 14. Because the solution contains only the conjugate base at equivalence, you can calculate the pH using its concentration and the \(K_a\) relationship:
\[ \text{pH} = 7 + \frac{1}{2}(\text{pK}_w - \text{pK}_a) \]
This is a simplified equation for calculating the pH at equivalence in a titration involving a weak acid and a strong base.
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