Introduction To Chemistry
Introduction To Chemistry
5th Edition
ISBN: 9781259911149
Author: BAUER, Richard C., Birk, James P., Marks, Pamela
Publisher: Mcgraw-hill Education,
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Chapter 13, Problem 78QP

(a)

Interpretation Introduction

Interpretation:

The OH concentration is to be determined and also, the acidic, basic or neutral nature of the solution is to be identified.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given value of pH is 5 .

The relation between pH and pOH is as follows:

pH+pOH=14

Substitute 5 for pH in the above expression:

pOH=14pHpOH=145pOH=9

The expression that is used to calculate the pOH of the solution is:

pOH=log10OHOH=10pOH M

Substitute 9 for pOH in the above expression:

OH=10pOH MOH=1×109 M

At neutral condition, the value of OH is 1×107 M . As the concentration of OH is less than 1×107 M , thus the solution is acidic in nature.

(b)

Interpretation Introduction

Interpretation:

The OH concentration is to be determined and also, the acidic, basic or neutral nature of the solution is to be identified.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given value of pH is 11 .

The relation between pH and pOH is as follows:

pH+pOH=14

Substitute 11 for pH in the above expression:

pOH=14pHpOH=1411pOH=3

The expression that is used to calculate the pOH of the solution is:

pOH=log10OHOH=10pOH M

Substitute 3 for pOH in the above expression:

OH=1×103 M

At neutral condition, the value of OH is 1×107 M . As the concentration of OH is greater than 1×107 M , thus the solution is basic in nature.

(c)

Interpretation Introduction

Interpretation:

The OH concentration is to be determined and also, the acidic, basic or neutral nature of the solution is to be identified.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given value of pH is 12.8 .

The relation between pH and pOH is as follows:

pH+pOH=14

Substitute 12.8 for pH in the above expression:

pOH=14pHpOH=1412.8pOH=1.2

The expression that is used to calculate the pOH of the solution is:

pOH=log10OHOH=10pOH M

Substitute 1.2 for pOH in the above expression:

OH=101.2 MOH=6.31×102 M

At neutral condition, the value of OH is 1×107 M . As the concentration of OH is greater

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Chapter 13 Solutions

Introduction To Chemistry

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