Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 5, Problem 5.31P

(a)

Interpretation Introduction

Interpretation:

pH of 1.0×104 MH3BO3(aq) solution has to be calculated.

Concept Introduction:

The equilibrium constant of acid is given by,

  HA+H2OH3O++A Ka=[H3O+][A][HA]

(a)

Expert Solution
Check Mark

Explanation of Solution

Given,

    1.0×104 MH3BO3(aq) Acts as a monoprotic acid.

The reaction of H3BO3(aq) with water is,

  H3BO3(aq)+H2O(l)H3O(aq)++H2BO(aq)

From the above reaction the equilibrium constant is,

    Ka=[H3O+][H2BO3][H3BO3]=2.0×109

The ICE table is,

   H3BO3(aq)+H2O(l)H3O(aq)++H2BO(aq)-Initial 1.0×10-4M 0 0Change 1.0×10-4M-x +x +xEquilibrium1.0×10-4M-x xx

From the above table, the concentration of H3O(aq)+ is,

    Ka =[H3O+][H2BO3][H3BO3]=2.0×1092.0×109=x21.0×10-4M-x=x21.0×10-4M x2=2.0×109×1.0×10-4M =2.0×1013 =4.47×107

We know,

    pH=log[H3O+]=log[4.47×107]=6.3

Hence, the pH of 1.0×104 MH3BO3(aq) solution is 6.3.

(b)

Interpretation Introduction

Interpretation:

pH of 0.015 M H3PO4(aq) solution has to be calculated.

Concept Introduction:

The equilibrium constant of acid is given by,

  HA+H2OH3O++A Ka=[H3O+][A][HA]

(b)

Expert Solution
Check Mark

Explanation of Solution

Given,

    0.015 M H3PO4(aq)

The reaction of H3PO4(aq) with water is,

  H3PO4(aq)+H2O(l)H3O(aq)++H2PO4(aq)Ka1=7.6×103H2PO4(aq)+H2O(l)H3O(aq)++HPO4(aq)2Ka2=6.2×108HPO4(aq)2+H2O(l)H3O(aq)++PO4(aq)3Ka3=2.1×1013

From the above reactions the equilibrium constants are,

    Ka1=[H3O+][H2PO4][H3PO4]=7.6×103Ka2=[H3O+][HPO42][H2PO4]=6.2×108Ka3=[H3O+][PO43][HPO42]=2.1×1013

The ICE tables are,

   H3PO4(aq)+H2O(l)H3O(aq)++H2PO4(aq)Initial 0.015 M 0 0Change 0.015 M-x +x +xEquilibrium0.015 M-x xx

From the above table, the concentration of H3O(aq)+ is,

    Ka1=[H3O+][H2PO4][H3PO4]=7.6×1037.6×103=x20.015 M-x=x20.015 M x2=7.6×103×0.015 M =1.14×104 =0.0107M

   H2PO4(aq)+H2O(l)H3O(aq)++HPO4(aq)2Initial 0.0107M 0 0Change 0.0107M-x +x +xEquilibrium0.0107M-x xx

From the above table, the concentration of H3O(aq)+ is,

    Ka2=[H3O+][HPO42-][H2PO4-]=6.2×10-86.2×10-8=x20.0107M-x=x20.0107M x2=6.2×10-8×0.0107M =6.63×10-10 =2.5×10-5M

   HPO4(aq)2+H2O(l)H3O(aq)++PO4(aq)3Initial 2.5×10-5M 0 0Change 2.5×10-5M-x +x +xEquilibrium2.5×10-5M-x xx

From the above table, the concentration of H3O(aq)+ is,

    Ka3=[H3O+][PO43-][HPO42-]=2.1×10-132.1×10-13=x22.5×10-5M-x=x22.5×10-5M x2=2.1×10-13×2.5×10-5M =5.25×10-18 =2.29×10-9M

From the above calculations, the concentration of H3O(aq)+ is,

    H3O(aq)+=0.0107M+2.5×10-5M+2.29×10-9M =0.0107M

We know,

    pH=log[H3O+]=log[0.0107M]=1.97pH

Hence, the pH of 0.015 M H3PO4(aq) solution is 1.97pH.

(c)

Interpretation Introduction

Interpretation:

pH of 0.10 MH2SO3(aq) solution has to be calculated.

Concept Introduction:

The equilibrium constant of acid is given by,

  HA+H2OH3O++A Ka=[H3O+][A][HA]

(c)

Expert Solution
Check Mark

Explanation of Solution

Given,

    0.10 MH2SO3(aq)

The reaction of H2SO3(aq) with water is,

  H2SO3(aq)+H2O(l)H3O(aq)++HSO3(aq)Ka1=1.5×102HSO3(aq)+H2O(l)H3O(aq)++SO23(aq)Ka2=1.2×107

From the above reaction the equilibrium constant is,

    Ka1=[H3O+][H2SO3][H2SO3]=1.5×102Ka2=[H3O+][SO3][HSO3]=1.2×107

The ICE table is,

   H2SO3(aq)+H2O(l)H3O(aq)++HSO3(aq)Initial 0.10 M 0 0Change 0.10 M-x +x +xEquilibrium0.10 M-x xx

From the above table, the concentration of H3O(aq)+ is,

    Ka1=[H3O+][H2SO3][H2SO3]=1.5×1021.5×102=x20.10 M-x=x20.10 M x2=1.5×102×0.10 M =1.5×103 =0.038M

   HSO3(aq)+H2O(l)H3O(aq)++SO23(aq)Initial 0.10 M 0 0Change 0.10 M-x +x +xEquilibrium0.10 M-x xx

From the above table, the concentration of H3O(aq)+ is,

    Ka2=[H3O+][SO3][HSO3]=1.2×1071.2×107=x20.038M-x=x20.038M x2=1.2×107×0.038M =4.5×109 =6.7×105M

From the above calculations, the concentration of H3O(aq)+ is,

    H3O(aq)+=0.038M+6.7×105M =0.038M

We know,

    pH=log[H3O+]=log[0.038M]=1.4pH

Hence, the pH of 1.0×104 MH3BO3(aq) solution is 1.4pH.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 5 Solutions

Elements Of Physical Chemistry

Ch. 5 - Prob. 5D.1STCh. 5 - Prob. 5D.2STCh. 5 - Prob. 5D.3STCh. 5 - Prob. 5D.4STCh. 5 - Prob. 5D.5STCh. 5 - Prob. 5E.1STCh. 5 - Prob. 5E.2STCh. 5 - Prob. 5F.1STCh. 5 - Prob. 5F.2STCh. 5 - Prob. 5F.3STCh. 5 - Prob. 5F.4STCh. 5 - Prob. 5F.5STCh. 5 - Prob. 5G.1STCh. 5 - Prob. 5G.2STCh. 5 - Prob. 5G.3STCh. 5 - Prob. 5H.1STCh. 5 - Prob. 5H.2STCh. 5 - Prob. 5H.3STCh. 5 - Prob. 5I.1STCh. 5 - Prob. 5I.2STCh. 5 - Prob. 5I.3STCh. 5 - Prob. 5I.4STCh. 5 - Prob. 5I.5STCh. 5 - Prob. 5I.6STCh. 5 - Prob. 5J.1STCh. 5 - Prob. 5J.2STCh. 5 - Prob. 5J.3STCh. 5 - Prob. 5J.4STCh. 5 - Prob. 5J.5STCh. 5 - Prob. 5A.1ECh. 5 - Prob. 5A.2ECh. 5 - Prob. 5A.3ECh. 5 - Prob. 5A.4ECh. 5 - Prob. 5A.5ECh. 5 - Prob. 5A.6ECh. 5 - Prob. 5A.7ECh. 5 - Prob. 5A.8ECh. 5 - Prob. 5A.9ECh. 5 - Prob. 5A.10ECh. 5 - Prob. 5A.11ECh. 5 - Prob. 5A.12ECh. 5 - Prob. 5A.13ECh. 5 - Prob. 5B.1ECh. 5 - Prob. 5B.2ECh. 5 - Prob. 5B.3ECh. 5 - Prob. 5B.4ECh. 5 - Prob. 5B.5ECh. 5 - Prob. 5C.1ECh. 5 - Prob. 5C.2ECh. 5 - Prob. 5C.3ECh. 5 - Prob. 5C.4ECh. 5 - Prob. 5D.1ECh. 5 - Prob. 5D.2ECh. 5 - Prob. 5D.3ECh. 5 - Prob. 5D.4ECh. 5 - Prob. 5E.1ECh. 5 - Prob. 5E.2ECh. 5 - Prob. 5F.4ECh. 5 - Prob. 5G.1ECh. 5 - Prob. 5G.2ECh. 5 - Prob. 5G.3ECh. 5 - Prob. 5H.1ECh. 5 - Prob. 5H.2ECh. 5 - Prob. 5H.3ECh. 5 - Prob. 5H.4ECh. 5 - Prob. 5I.1ECh. 5 - Prob. 5I.2ECh. 5 - Prob. 5I.3ECh. 5 - Prob. 5I.4ECh. 5 - Prob. 5J.1ECh. 5 - Prob. 5J.2ECh. 5 - Prob. 5J.3ECh. 5 - Prob. 5J.4ECh. 5 - Prob. 5.1DQCh. 5 - Prob. 5.2DQCh. 5 - Prob. 5.3DQCh. 5 - Prob. 5.4DQCh. 5 - Prob. 5.5DQCh. 5 - Prob. 5.6DQCh. 5 - Prob. 5.7DQCh. 5 - Prob. 5.8DQCh. 5 - Prob. 5.9DQCh. 5 - Prob. 5.10DQCh. 5 - Prob. 5.11DQCh. 5 - Prob. 5.12DQCh. 5 - Prob. 5.13DQCh. 5 - Prob. 5.14DQCh. 5 - Prob. 5.15DQCh. 5 - Prob. 5.16DQCh. 5 - Prob. 5.17DQCh. 5 - Prob. 5.1PCh. 5 - Prob. 5.2PCh. 5 - Prob. 5.3PCh. 5 - Prob. 5.4PCh. 5 - Prob. 5.5PCh. 5 - Prob. 5.6PCh. 5 - Prob. 5.7PCh. 5 - Prob. 5.8PCh. 5 - Prob. 5.9PCh. 5 - Prob. 5.10PCh. 5 - Prob. 5.11PCh. 5 - Prob. 5.12PCh. 5 - Prob. 5.13PCh. 5 - Prob. 5.14PCh. 5 - Prob. 5.15PCh. 5 - Prob. 5.16PCh. 5 - Prob. 5.17PCh. 5 - Prob. 5.18PCh. 5 - Prob. 5.19PCh. 5 - Prob. 5.20PCh. 5 - Prob. 5.21PCh. 5 - Prob. 5.22PCh. 5 - Prob. 5.23PCh. 5 - Prob. 5.24PCh. 5 - Prob. 5.25PCh. 5 - Prob. 5.27PCh. 5 - Prob. 5.28PCh. 5 - Prob. 5.29PCh. 5 - Prob. 5.30PCh. 5 - Prob. 5.31PCh. 5 - Prob. 5.32PCh. 5 - Prob. 5.33PCh. 5 - Prob. 5.34PCh. 5 - Prob. 5.35PCh. 5 - Prob. 5.36PCh. 5 - Prob. 5.37PCh. 5 - Prob. 5.38PCh. 5 - Prob. 5.39PCh. 5 - Prob. 5.40PCh. 5 - Prob. 5.41PCh. 5 - Prob. 5.42PCh. 5 - Prob. 5.43PCh. 5 - Prob. 5.44PCh. 5 - Prob. 5.45PCh. 5 - Prob. 5.46PCh. 5 - Prob. 5.47PCh. 5 - Prob. 5.48PCh. 5 - Prob. 5.49PCh. 5 - Prob. 5.50PCh. 5 - Prob. 5.51PCh. 5 - Prob. 5.52PCh. 5 - Prob. 5.53PCh. 5 - Prob. 5.54PCh. 5 - Prob. 5.55PCh. 5 - Prob. 5.56PCh. 5 - Prob. 5.58PCh. 5 - Prob. 5.59PCh. 5 - Prob. 5.60PCh. 5 - Prob. 5.61PCh. 5 - Prob. 5.62PCh. 5 - Prob. 5.63PCh. 5 - Prob. 5.64PCh. 5 - Prob. 5.66PCh. 5 - Prob. 5.67PCh. 5 - Prob. 5.68PCh. 5 - Prob. 5.69PCh. 5 - Prob. 5.70PCh. 5 - Prob. 5.71PCh. 5 - Prob. 5.72PCh. 5 - Prob. 5.73PCh. 5 - Prob. 5.74PCh. 5 - Prob. 5.76PCh. 5 - Prob. 5.77PCh. 5 - Prob. 5.78PCh. 5 - Prob. 5.79PCh. 5 - Prob. 5.1PRCh. 5 - Prob. 5.2PRCh. 5 - Prob. 5.3PRCh. 5 - Prob. 5.4PR
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY