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The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required 27.36 mL of 0.5106 M NaOH for complete reaction. Addition of 13.68 mL of 0.5106 M HCl to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH = 3.48. Determine the molar mass of acetylsalicylic acid and its K a value. State any assumptions you must make to reach your answer.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 95AE
Textbook Problem
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The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required 27.36 mL of 0.5106 M NaOH for complete reaction. Addition of 13.68 mL of 0.5106 M HCl to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH = 3.48. Determine the molar mass of acetylsalicylic acid and its Ka value. State any assumptions you must make to reach your answer.

Interpretation Introduction

Interpretation:

Acetylsalicyclic acid is given to be the active ingredient in aspirin. It is stated that the addition of 13.68mL of 0.5106M HCl to a flask containing aspirin and sodium hydroxide in the given amounts produced a mixture of pH=3.48 . The molar mass of acetylsalicyclic acid and its Ka value is to be calculated.

Concept introduction:

The relation between pH and pKa is given by Henderson-Hasselbach equation. According to this equation,

pH=pKa+log[Salt][Acid]

To determine: The molar mass of acetylsalicyclic acid and its Ka value.

Explanation of Solution

Explanation

To find the molar mass of acetylsalicyclic acid

Given

The volume of NaOH is 27.36mL(0.02736L) .

The molarity of NaOH is 0.5106M .

The mass of the acid is 2.51g .

The given monoprotic acid, that is acetylsalicyclic acid is assumed to be HA .

The reaction that takes place is,

HA+OHA+H2O

The number of moles is calculated by the formula,

Numberofmoles=Volume(L)×Molarity

Substitute the value of volume and the molarity of NaOH in the above expression.

Numberofmoles=0.02736L×0.5106M=1.397×102mol

The given volume of NaOH was enough to complete the reaction with the given monoprotic acid. Therefore, the number of moles of the acid and NaOH will be the same.

Therefore, the number of moles of the acid are 1.397×102mol .

The molar mass is calculated by the formula,

MolarMass=GivenMassNumberofmoles

Substitute the value of the given mass of the acid and the calculated value of the number of moles in the above expression.

MolarMass=2.51g1.397×102mol=180g/mol_

To find the Ka value of acetylsalicyclic acid

Given

The volume of HCl added is 13.68mL(0.01368L) .

The molarity of HCl added is 0.5106M .

The value of pH is 3.48 .

After the neutralization of the acid, the number of moles of A in the solution are 1.397×102mol .

The number of moles of H+ that were added to the solution is calculated by the formula,

Numberofmoles=Volume(L)×Molarity

Substitute the value of volume and molarity of the HCl added in the above expression.

Numberofmoles=0.01368L×0.5106M=6.985×103mol

The reaction that occurs is,

A+H+HA

The ICE table is formed for the given reaction

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Chapter 14 Solutions

Chemistry: An Atoms First Approach
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Ch. 14 - What are the major species in solution after...Ch. 14 - A friend asks the following: Consider a buffered...Ch. 14 - Mixing together solutions of acetic acid and...Ch. 14 - Could a buffered solution be made by mixing...Ch. 14 - Sketch two pH curves, one for the titration of a...Ch. 14 - Sketch a pH curve for the titration of a weak acid...Ch. 14 - You have a solution of the weak acid HA and add...Ch. 14 - You have a solution of the weak acid HA and add...Ch. 14 - The common ion effect for weak acids is to...Ch. 14 - Consider a buffer solution where [weak acid] ...Ch. 14 - A best buffer has about equal quantities of weak...Ch. 14 - Consider the following pH curves for 100.0 mL of...Ch. 14 - An acid is titrated with NaOH. 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