Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 16.4, Problem 1PPB

Practice ProblemChapter 16.4, Problem 1PPB, Practice Problem BUILD
Determine the pH of a solution at  in which the hydroxide ion concentration BUILD

Determine the pH of a solution at 25°C in which the hydroxide ion concentration is (a)  8.3 × 10 15 M , (b)  3.3 × 10 4 M ,  and (c) 1 .2 × 10 3 M .

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

The pH of the solution for each of the given concentrations of hydroxide ions is to be determined.

Concept introduction:

pH is a measure of the acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

Use the equilibrium expression Kw as follows:

Kw=[H3O+][OH-]=1.0×1014

Rearrange the above equation for hydronium ions as:

[H3O+]=1.0×1014[OH-] ……(1)

The formula to calculate the pH of a solution is:

pH=log[H3O+]=log[H+] as [H3O+]=[H+] …… (2)

Answer to Problem 1PPB

Solution:

a) 0.08

b) 10.52

c) 11.08

Explanation of Solution

8.3×1015M

Use the equilibrium expression Kw as follows:

Kw=[H3O+][OH-]=1.0×1014

Rearrange the above equation for hydronium ions as:

[H3O+]=1.0×1014[OH-]

Use equation (1) to calculate the concentration of H3O+ in the solution at 25 C and substitute the value of OH as 8.3×1015M to get:

[H3O+]=1.0×10148.3×1015=1.2

Now, substitute the value of [H3O+] as 1.2 in equation (2) as:

pH=log[H3O+]=log[H+]

pH=log[1.2]=(0.08)=0.08

3.3×104M

Use the equilibrium expression Kw as follows:

Kw=[H3O+][OH-]=1.0×1014

Rearrange the above equation for hydronium ions as:

[H3O+]=1.0×1014[OH-]

Use equation (1) to calculate the concentration of H3O+ in the solution at 25C and substitute the value of OH as 3.3×104M to get:

[H3O+]=1.0×10143.3×104=3.0×1011

Now, substitute the value of [H3O+] as 3.0×1011 in equation (2) as:

pH=log[H3O+]=log[H+]

pH=log[3.0×1011]=(10.52)=10.52

The pH of the solution is 10.52.

1.2×103M

Use the equilibrium expression Kw as follows:

Kw=[H3O+][OH-]=1.0×1014

Rearrange the above equation for hydronium ions as:

[H3O+]=1.0×1014[OH-]

Use equation (1) to calculate the concentration of H3O+ in the solution at 25C and substitute the value of OH as 1.2×103M to get:

[H3O+]=1.0×10141.2×103=8.3×1012

Now, substitute the value of [H3O+] as 8.3×1012 in equation (2) as:

pH=log[H3O+]=log[H+]

pH=log[8.3×1012]=(11.08)=11.08

The pH of the solution is 11.08.

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Chapter 16 Solutions

Chemistry

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