   # Fill in the missing information in the following table. ### 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 13, Problem 52E
Textbook Problem
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## Fill in the missing information in the following table. Interpretation Introduction

Interpretation: The missing information in the given table is to be stated.

Concept introduction: The pH is used to measure the [H+] of any solution.

The pOH is used to measure the [OH] of any solution.

The [H+] represents the total hydrogen ion concentration.

The [OH] represents the total hydroxide ion concentration.

The nature of any solution is determined by its pH and pOH value.

To determine: The missing values in the table for solution a.

### Explanation of Solution

Explanation

For solution a

Given

The pH is 9.63 .

The formula to calculate pH is,

pH=log[H+]

Where

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

Substitute the  value of pH in equation.

9.63=log[H+][H+]=Antilog(pH)[H+]=Antilog(9.63)[H+]=2.34×1010M

Given

The pH is 9.63

The sum,

pH+pOH=14

Where,

• pOH is the measure of  hyroxide ion concentration.
• pH is the measure of hydrogen ion concentration.

Substitute the value of pH in the above equation.

9.63+pOH=14pOH=149.63pOH=4.37_

The [OH] is 4.27×10-4M_ .

Given

The [H+] is 2.34×1010M .

The ionic product of water is,

[H+][OH]=1.0×1014

Where,

• [H+] is the concentration of hydrogen ion.
• [OH] is the concentration of hydroxide ion.

Substitute the value of [H+] in the above equation.

2.34×1010×[OH]=1.0×1014[OH]=1.0×10142.34×1010[OH]=4.27×10-4M_

The solution a is basic.

In the given solution the [OH]>[H+] ; therefore, the given solution is basic.

For Solution b

Given

The [OH] is 3.9×106M .

The pOH is calculated by the formula,

pOH=log[OH]

Substitute the value of [OH] in the above equation.

pOH=log[3.9×106]pOH=5.41_

The [H+] is 2.57×10-9M_ .

Given

The pH is 8.59

The pH is calculated by the formula,

pH=log[H+]

Substitute the value of pH in the above equation.

8.59=log[H+][H+]=antilog[8.59][H+]=2.57×10-9_

The solution b is basic.

In the given solution, the [OH]>[H+] ; therefore, the given solution is basic.

For solution c

Given

The [H+] is 0

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