BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Solutions

Chapter
Section
BuyFindarrow_forward

Chemistry

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

Fill in the missing information in the following table.

Chapter 14, Problem 51E, 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 the measure of its [H+] of any solution.  The pOH is the measure of the [OH] of any solution. [H+] represents the total hydrogen ion concentration. [OH] represents the total hydroxide ion concentration.  The nature of any solution is determined by its pH and pOH value.

Explanation

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

The pOH is 7.12_ .

Given

The pH of the solution is 6.88 .

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.

6.88+pOH=14pOH=146.88pOH=7.12_

The [H+] is 1.31×10-7M_ .

Given

The pH of the solution is 6.88 .

The pH value is calculated by the formula,

pH=log[H+]

Where,

  • [H+] is the total concentration of hydrogen ions.

Rearrange the above equation to calculate the value of [H+] .

[H+]=antilog(pH)

Substitute the value of pH in the above equation.

[H+]=antilog(pH)[H+]=Antilog(6.88)[H+]=1.31×10-7M_

The [OH] is 7.6×10-8M_ .

Given

The [H+] is 1.31×107M .

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.

1.31×107×[OH]=1.0×1014[OH]=1.0×10141.31×107[OH]=7.6×10-8M_

The solution is acidic.

In the given solution, the [H+]>[OH] . Therefore, the given solution is acidic.

To determine: The missing values in the table for solution b

The [H+] is 0.119M_ .

Given

The [OH] is 8.4×1014 .

The ionic product of water is,

[H+][OH]=1.0×1014

Where,

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

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

[H+]×8.4×1014=1.0×1014[H+]=1.0×10148.4×1014[H+]=0.119M_

The pH is 0.92_

Given

The [H+] is 0.119M

The pH value is calculated by the formula,

pH=log[H+]

Where,

  • [H+] is the total concentration of hydrogen ions.

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

pH=log[0.119]pH=0.92_

The pOH is 13.08_ .

Given

The pH is 0.92 .

The sum pH+pOH=14

Substitute the value of pH in the above equation.

0.92+pOH=14pOH=140

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 14 Solutions

Show all chapter solutions add
Ch-14 P-1ALQCh-14 P-2ALQCh-14 P-3ALQCh-14 P-4ALQCh-14 P-5ALQCh-14 P-6ALQCh-14 P-7ALQCh-14 P-8ALQCh-14 P-9ALQCh-14 P-10ALQCh-14 P-11ALQCh-14 P-12ALQCh-14 P-13ALQCh-14 P-14ALQCh-14 P-15ALQCh-14 P-16ALQCh-14 P-17ALQCh-14 P-18ALQCh-14 P-19QCh-14 P-20QCh-14 P-21QCh-14 P-22QCh-14 P-23QCh-14 P-24QCh-14 P-25QCh-14 P-26QCh-14 P-27QCh-14 P-28QCh-14 P-29QCh-14 P-30QCh-14 P-31QCh-14 P-32QCh-14 P-33QCh-14 P-34QCh-14 P-35ECh-14 P-36ECh-14 P-37ECh-14 P-38ECh-14 P-39ECh-14 P-40ECh-14 P-41ECh-14 P-42ECh-14 P-43ECh-14 P-44ECh-14 P-45ECh-14 P-46ECh-14 P-47ECh-14 P-48ECh-14 P-49ECh-14 P-50ECh-14 P-51ECh-14 P-52ECh-14 P-53ECh-14 P-54ECh-14 P-55ECh-14 P-56ECh-14 P-57ECh-14 P-58ECh-14 P-59ECh-14 P-60ECh-14 P-61ECh-14 P-62ECh-14 P-63ECh-14 P-64ECh-14 P-65ECh-14 P-67ECh-14 P-68ECh-14 P-69ECh-14 P-70ECh-14 P-72ECh-14 P-73ECh-14 P-74ECh-14 P-75ECh-14 P-76ECh-14 P-77ECh-14 P-78ECh-14 P-79ECh-14 P-80ECh-14 P-81ECh-14 P-82ECh-14 P-83ECh-14 P-84ECh-14 P-85ECh-14 P-86ECh-14 P-87ECh-14 P-88ECh-14 P-89ECh-14 P-90ECh-14 P-91ECh-14 P-92ECh-14 P-93ECh-14 P-94ECh-14 P-95ECh-14 P-96ECh-14 P-97ECh-14 P-99ECh-14 P-100ECh-14 P-101ECh-14 P-102ECh-14 P-103ECh-14 P-104ECh-14 P-105ECh-14 P-106ECh-14 P-107ECh-14 P-108ECh-14 P-109ECh-14 P-110ECh-14 P-111ECh-14 P-112ECh-14 P-113ECh-14 P-114ECh-14 P-115ECh-14 P-116ECh-14 P-117ECh-14 P-118ECh-14 P-119ECh-14 P-120ECh-14 P-121ECh-14 P-122ECh-14 P-123ECh-14 P-124ECh-14 P-125ECh-14 P-126ECh-14 P-127ECh-14 P-128ECh-14 P-129ECh-14 P-130ECh-14 P-131ECh-14 P-132ECh-14 P-133ECh-14 P-134ECh-14 P-135ECh-14 P-136ECh-14 P-137ECh-14 P-138ECh-14 P-139ECh-14 P-140ECh-14 P-141ECh-14 P-142ECh-14 P-143AECh-14 P-144AECh-14 P-145AECh-14 P-146AECh-14 P-147AECh-14 P-148AECh-14 P-149AECh-14 P-150AECh-14 P-151AECh-14 P-152AECh-14 P-153AECh-14 P-154AECh-14 P-155AECh-14 P-156AECh-14 P-157AECh-14 P-158AECh-14 P-159AECh-14 P-160AECh-14 P-161AECh-14 P-162CWPCh-14 P-163CWPCh-14 P-164CWPCh-14 P-165CWPCh-14 P-166CWPCh-14 P-167CWPCh-14 P-168CWPCh-14 P-169CWPCh-14 P-170CPCh-14 P-171CPCh-14 P-172CPCh-14 P-173CPCh-14 P-174CPCh-14 P-175CPCh-14 P-176CPCh-14 P-177CPCh-14 P-178CPCh-14 P-179CPCh-14 P-180CPCh-14 P-181CPCh-14 P-182CPCh-14 P-183CPCh-14 P-184CPCh-14 P-185CPCh-14 P-186CPCh-14 P-187IPCh-14 P-188IPCh-14 P-189IPCh-14 P-190MPCh-14 P-191MP

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

List three reasons why someone might take a multivitamin supplement that does not exceed 100 percent of the RDA...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

In what ways were the models of Ptolemy and Copernicus similar?

Horizons: Exploring the Universe (MindTap Course List)

What were Matthew Maurys contributions to marine science? Benjamin Franklins?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

A testcross is a way to determine ______. a. phenotype b. genotype c. both a and b

Biology: The Unity and Diversity of Life (MindTap Course List)

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 F, and a source with Vmax =...

Physics for Scientists and Engineers, Technology Update (No access codes included)