   # 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. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 95AE
Textbook Problem
1053 views

## 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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