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7. What volume of 0.120 M NaOH will neutralize half of a 53.8-mL solution of 0.07 M propionic acid, CH3CH₂COOH (a weak monoprotic acid)?

7. What volume of 0.120 M NaOH will neutralize half of a 53.8-mL solution of 0.07 M propionic acid, CH3CH₂COOH (a weak monoprotic acid)?

Chemistry
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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**Question 7: Neutralization Calculation** What volume of 0.120 M NaOH will neutralize half of a 53.8 mL solution of 0.07 M propionic acid, CH₃CH₂COOH (a weak monoprotic acid)? --- **Explanation:** This question involves a basic neutralization problem where you calculate the volume of a sodium hydroxide (NaOH) solution needed to neutralize a portion of a propionic acid solution. Here are the steps to solve it: 1. **Identify Moles of Acid:** - Calculate the moles of propionic acid in half of the given solution. - Given: 53.8 mL total, so half is 26.9 mL. - Molarity of propionic acid = 0.07 M. - Moles = Molarity × Volume (in liters). 2. **Balance the Reaction:** - The reaction between NaOH and CH₃CH₂COOH produces water and the sodium salt of the acid. - This is a 1:1 mole reaction because propionic acid is monoprotic. 3. **Calculate Volume of NaOH:** - Use the moles of acid to find the moles of NaOH required (1:1 ratio). - Use molarity of NaOH to convert moles into the volume needed (in liters). This setup allows students to understand the concept of stoichiometry in an acid-base neutralization context.
Transcribed Image Text:**Question 7: Neutralization Calculation** What volume of 0.120 M NaOH will neutralize half of a 53.8 mL solution of 0.07 M propionic acid, CH₃CH₂COOH (a weak monoprotic acid)? --- **Explanation:** This question involves a basic neutralization problem where you calculate the volume of a sodium hydroxide (NaOH) solution needed to neutralize a portion of a propionic acid solution. Here are the steps to solve it: 1. **Identify Moles of Acid:** - Calculate the moles of propionic acid in half of the given solution. - Given: 53.8 mL total, so half is 26.9 mL. - Molarity of propionic acid = 0.07 M. - Moles = Molarity × Volume (in liters). 2. **Balance the Reaction:** - The reaction between NaOH and CH₃CH₂COOH produces water and the sodium salt of the acid. - This is a 1:1 mole reaction because propionic acid is monoprotic. 3. **Calculate Volume of NaOH:** - Use the moles of acid to find the moles of NaOH required (1:1 ratio). - Use molarity of NaOH to convert moles into the volume needed (in liters). This setup allows students to understand the concept of stoichiometry in an acid-base neutralization context.
Expert Solution
Step 1

Given:

Volume of propionic acid (V1) is 53.8 mL2=26.9 mL

Molarity of propionic acid (M1) is 0.07 M.

Molarity of NaOH (M2) is 0.120 M. 

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