   # A buffer solution is prepared by mixing 75.0 mL of 0.275 M fluorobenzoic acid (C 7 H 5 O 2 F) with 55.0 mL of 0.472 M sodium fluorobenzoate. The p K a of this weak acid is 2.90. What is the pH of the buffer solution? ### 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 119IP
Textbook Problem
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## A buffer solution is prepared by mixing 75.0 mL of 0.275 M fluorobenzoic acid (C7H5O2F) with 55.0 mL of 0.472 M sodium fluorobenzoate. The pKa of this weak acid is 2.90. What is the pH of the buffer solution?

Interpretation Introduction

Interpretation: The formation of a buffer solution by the mixing of Fluorobenzoic acid and Sodium fluorobenzoate is given. The value of pH of buffer solution is to be calculated.

Concept introduction: The solution of weak acid and their conjugate base which upon addition of an acid or base resists the change in pH is called buffer solution. The pH of the buffer solution remains constant.

To determine: value of pH of given buffer solution is to be calculated.

### Explanation of Solution

Explanation

Given

The concentration of Fluorobenzoic acid is 0.275M .

The concentration of Sodium fluorobenzoate is 0.472M .

The volume of Fluorobenzoic acid is 75.0mL .

The volume of Sodium fluorobenzoate is 55.0mL .

The pKa is 2.90 .

The conversion of mL into L is done as,

1mL=0.001L

The conversion of all the given species into L is done as,

75.0mL=75.0×0.001L=0.075L55.0mL=55.0×0.001L=0.055L

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres (2)

Substitute the value of concentration and volume of Fluorobenzoic acid in the equation (2).

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.275M×0.075L=0.021moles

Similarly, substitute the value of concentration and volume of Sodium fluorobenzoate in the equation (2).

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.472M×0.055L=0.026moles

The total volume =VolumeofFluorobenzoicacid+VolumeofSodiumfluorobenzoate=0

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