Chapter 14, Problem 49E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Calculate the [OH–] of each of the following solutions at 25°C. Identify each solution as neutral, acidic, or basic.  a. [H+] = 1.0 × 10–7 M  b. [H+] = 8.3 × 10–16 M  c. [H+] = 12 M  d. [H+] = 5.4 × 10–5 M  46. Calculate the [H+] of each of the following solutions at 25°C. Identify each solution as neutral, acidic, or basic.  a. [OH–] = 1.5 M  b. [OH–] = 3.6 × 10–15 M  c. [OH–] = 1.0 × 10–7 M  d. [OH–] = 7.3 × 10–4 M  49. Calculate the pH and pOH of the solutions in Exercises 45 and 46.

Interpretation Introduction

Interpretation:

The pH and the pOH values for thesolution in Exercises 45 and 46 are to be calculated.

Concept introduction:

The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value 7 corresponds to a neutral species, a value less than 7 corresponds to an acid and a value greater than 7 corresponds to a base.

The pOH of a solution is the measure of the [OH] .

The pH of a solution is calculated by the formula, pH=log[H+]

The pOH of a solution is calculated by the formula, pOH=log[OH]

At 25°C , pH+pOH=14

Explanation

Explanation

(45a)

Calculation of pH of the solution

The pH value is 7_ .

Given

The temperature is 25°C .

The [H+]=1.0×107M .

The pH of a solution is calculated by the formula,

pH=log[H+]

Substitute the given value of [H+] in the above expression.

pH=log[1.0×107]=7_

Calculation of pOH of the solution

The pOH value is 7_ .

At 25°C ,

pH+pOH=14

Substitute the calculated value of pH in the above expression.

pH+pOH=147+pOH=14pOH=7_

(45b)

Calculation of pH of the solution

The pH value is 15_ .

Given

The temperature is 25°C .

The [H+]=8.3×1016M .

The pH of a solution is calculated by the formula,

pH=log[H+]

Substitute the given value of [H+] in the above expression.

pH=log[8.3×1016]=15_

Calculation of pOH of the solution

The pOH value is -1_ .

At 25°C ,

pH+pOH=14

Substitute the calculated value of pH in the above expression.

pH+pOH=1515+pOH=14pOH=-1_

(45c)

The pH value is 1.1_ .

Given

The temperature is 25°C .

The [H+]=12M .

The pH of a solution is calculated by the formula,

pH=log[H+]

Substitute the given value of [H+] in the above expression.

pH=log[12]1.1_

The pOH value is 12.9_ .

At 25°C ,

pH+pOH=14

Substitute the calculated value of pH in the above expression.

pH+pOH=1.11.1+pOH=14pOH=12.9_

(45d)

The pH value is 4.3_ .

Given

The temperature is 25°C .

The [H+]=5.4×105M .

The pH of a solution is calculated by the formula,

pH=log[H+]

Substitute the given value of [H+] in the above expression.

pH=log[5.4×105]4.3_

The pOH value is 9.7_ .

At 25°C ,

pH+pOH=14

Substitute the calculated value of pH in the above expression.

pH+pOH=4.34.3+pOH=14pOH=9.7_

(46a)

The pOH value is 0

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