**Titration Calculation Example**Consider the titration of 15.75 mL of 0.549 M potassium propanoate, CH₃CH₂CO₂K, with 0.207 M HNO₃. What is the pH of the solution after 39.83 mL of the HNO₃ has been added? (pKa of CH₃CH₂CO₂H = 4.691)---*This problem explores the titration of a weak base, potassium propanoate, with a strong acid, nitric acid (HNO₃). The objective is to determine the pH of the solution after a specific volume of HNO₃ has been added.*1. **Understand the Reactants and Products**: - Potassium propanoate (CH₃CH₂CO₂K) dissociates in water into CH₃CH₂CO₂⁻ and K⁺. - HNO₃ is a strong acid that dissociates completely in water into H⁺ and NO₃⁻.2. **Initial Moles Calculation**: - Calculate the initial moles of CH₃CH₂CO₂K: \[ \text{Moles of CH₃CH₂CO₂K} = 15.75 \text{ mL} \times 0.549 \text{ M} = 8.65 \text{ mmol} \] - Calculate the moles of HNO₃ added: \[ \text{Moles of HNO₃} = 39.83 \text{ mL} \times 0.207 \text{ M} = 8.25 \text{ mmol} \]3. **Reaction Completion**: - This is a stoichiometric reaction between CH₃CH₂CO₂⁻ and H⁺: \[ \text{CH₃CH₂CO₂⁻} + \text{H⁺} \rightarrow \text{CH₃CH₂CO₂H} \] - After reaction, the remaining moles of CH₃CH₂CO₂⁻ = 8.65 mmol - 8.25 mmol = 0.40 mmol. - Moles of CH₃CH₂CO₂H formed = 8.25 mmol.4.

Given : Concentration of CH3CH2CO2K = 0.549 M Volume of CH3CH2CO2K solution = 15.75 mL = 0.01575 L…

Answered: Consider the titration of 15.75 mL of…

Consider the titration of 15.75 mL of 0.549 M potassium propanoate, CH3CH;CO,K, with 0.207 M HNO3. What is the pH of the solution after 39.83 mL of the HNO3 has been added? (pK, of CH3CH;COH = 4.691)

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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