   # You make 1.00 L of a buffered solution (pH = 4.00) by mixing acetic acid and sodium acetate. You have 1.00 M solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered 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 84AE
Textbook Problem
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## You make 1.00 L of a buffered solution (pH = 4.00) by mixing acetic acid and sodium acetate. You have 1.00 M solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

Interpretation Introduction

Interpretation:

The volume of each of the given solutions that is acetic acid and sodium acetate, required to make a buffered solution is to be calculated.

Concept introduction:

A solution that resists a change in the pH on addition of an acid or an alkali is termed as a buffer solution.

### Explanation of Solution

Explanation

To find the value of pKa

The given value of Ka is 1.8×105 .

The value of pKa is calculated by the formula,

pKa=log(Ka)

Substitute the value of Ka in the above expression.

pKa=log(1.8×105)=4.74_

To find the ratio [Salt][Acid] at pH 4.00

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

pH=pKa+log[Salt][Acid]

The calculated value of pKa is 4.74 .

Substitute the value of pKa and the given value of pH in the Henderson-Hasselbach equation.

4.0=4.74+log[Salt][Acid]log[Salt][Acid]=0.74[Salt][Acid]=100.74[Salt][Acid]=0.18197_

To find the volume of sodium acetate and that of the acetic acid required

The concentration is calculated by the formula,

Concentration=NumberofmolesVolumeofthesolution(L)

The number of moles is calculated by the formula,

Moles=Molarity×Volume(L)

The volume of sodium acetate is assumed to be V1 and its initial molarity is assumed to be M1

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