   # Consider the acids in Table 13-2. Which acid would be the best choice for preparing a pH = 7.00 buffer? Explain how to make 1.0 L of this buffer. ### 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 47E
Textbook Problem
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## Consider the acids in Table 13-2. Which acid would be the best choice for preparing a pH = 7.00 buffer? Explain how to make 1.0 L of this buffer.

Interpretation Introduction

Interpretation: The best acid for preparing a buffer of pH=7.00 from the table 13-2 and the method for preparation of this buffer solution is to be stated.

Concept introduction: A buffer solution is an aqueous solution having the potential to maintain it’s pH even when a small amount of an acid or a base is added to it. A buffer solution is consisting of weak acid and its conjugate base or vice versa. A buffer solution can be acidic, basic and neutral depending upon its constituent compounds.

To determine: The best acid for preparing a buffer of pH=7.00 from the table 13-2 and the method for preparation of this buffer solution

### Explanation of Solution

Explanation

Given

The pH value of the buffer is 7 .

The value of pH is calculated by the formula,

pH=log[H+]

Where,

• [H+] is concentration of hydrogen ion present in the solution.

Substitute the value of pH in the above equation.

7=log[H+][H+]=Antilog[H+]=1×107

The dissociation reaction of an assumed acid HA is,

HAH++A

The value of acid dissociation constant is given by the formula,

Ka=[Concentrationofproducts][Concentrationofreactants]

For the above stated dissociation reaction the Ka is given as.

Ka=[H+][A][HA][H+]=Ka[HA][A]

Now by the values in Table 13-2 the values of Ka are calculated and listed in the table stated below.

Acid[H+]=Ka[HA][A][HA][A]HSO4107=1.2×102([HA][A])8.33×106HClO2107=1.2×102([HA][A])8.33×106HC2H2ClO2107=1.35×103([HA][A])7.40×105HF107=7.2×104([HA][A])1

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