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E MasteringCh x P Chemical eg X P Chemical eg X P Chemical eg X P Chemical ec X P Chemical ec X A session.masteringchemistry.com/myct/itemView?assignmentProblemID3D144082727&offset%3Dnext P Chemical ec X P Chemical e > CHE154-H Gen Chem II Bronikowski S20

E MasteringCh x P Chemical eg X P Chemical eg X P Chemical eg X P Chemical ec X P Chemical ec X A session.masteringchemistry.com/myct/itemView?assignmentProblemID3D144082727&offset%3Dnext P Chemical ec X P Chemical e > CHE154-H Gen Chem II Bronikowski S20

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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E MasteringCh x P Chemical eg X P Chemical eg X P Chemical eg X P Chemical ec X P Chemical ec X A session.masteringchemistry.com/myct/itemView?assignmentProblemID3D144082727&offset%3Dnext P Chemical ec X P Chemical e > CHE154-H Gen Chem II Bronikowski S20 <CHE154 S20 Ch18 Sec4-6 Chapter 18 Algorithmic Question 60 Part A A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150M NaOH solution. What is the pH at the equivalence point? The K of benzoic acid is 6.5 x 10. 11.20 4.20 9.80 7.20 8.53 Submit Request Answer vide Feedback
Transcribed Image Text:E MasteringCh x P Chemical eg X P Chemical eg X P Chemical eg X P Chemical ec X P Chemical ec X A session.masteringchemistry.com/myct/itemView?assignmentProblemID3D144082727&offset%3Dnext P Chemical ec X P Chemical e > CHE154-H Gen Chem II Bronikowski S20 <CHE154 S20 Ch18 Sec4-6 Chapter 18 Algorithmic Question 60 Part A A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150M NaOH solution. What is the pH at the equivalence point? The K of benzoic acid is 6.5 x 10. 11.20 4.20 9.80 7.20 8.53 Submit Request Answer vide Feedback
Expert Solution
Step 1

Given,

Volume of benzoic acid = 25.0 mL = 0.025 L     (1 mL = 0.001 L)

Molarity of benzoic acid = 0.150 M = 0.150 mol/L

Molarity of NaOH = 0.150 M = 0.150 mol/L

Ka = 6.5 × 10-5

Step 2

Moles of benzoic acid can be calculated as :

Chemistry homework question answer, step 2, image 1

Step 3

The reaction of benzoic acid with strong base, NaOH can be written as :

 C6H5COOH (aq) + OH- (aq) → C6H5COO- (aq) + H2O (l)

At equivalence point, the moles of benzoic acid will be equal to the moles of NaOH. Thus, we can calculate the volume of NaOH used as :

Chemistry homework question answer, step 3, image 1

Step 4

At equivalence point, the concentration of C6H5COO- will be :

Chemistry homework question answer, step 4, image 1

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