OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781285460420
Author: John W. Moore; Conrad L. Stanitski
Publisher: Cengage Learning US
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium reaction is given below.

  HNO2(aq)+H2O(l)NO2(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

  Ka=[NO2][H3O+][HNO2]

A table can be set up as shown below.

  HNO2(aq)H3O+(aq)NO2(aq)Initialconc.(M)0.101×1070Changeinconc.(M)x+x+xEquilibriumconc.(M)0.10xxx

The concentration of H3O+ comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10x)=7.4×104

Then,

  Ka=(x)(x)(0.15x)=7.4×104x2=(7.4×104)×(0.10x)x2+7.4×104x7.4×105=0x=8.2×103=[H3O+]

The pH of the solution can be calculated as shown below.

  pH=log[H3O+]=log(8.2×103)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium reaction is given below.

  NH4+(aq)+H2O(l)NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

  Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

  NH4+(aq)H3O+(aq)NH3(aq)Initialconc.(M)0.101×1070Changeinconc.(M)x+x+xEquilibriumconc.(M)0.10xxx

The concentration of H3O+ comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10x)=5.6×1010

Assuming x is very small, it can be written 0.10x0.10.

Then,

  Ka=(x)(x)(0.15x)=5.6×1010x2=(5.6×1010)×(0.10)x=5.6×1011=7.5×106M=[H3O+]

The pH of the solution can be calculated as shown below.

  pH=log[H3O+]=log(7.5×106)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium reaction is given below.

  F(aq)+H2O(l)HF(aq)+OH(aq)

The base ionization constant can be written as given below.

  Kb=[HF][OH][F]

A table can be set up as shown below.

  F(aq)HF(aq)OH(aq)Initialconc.(M)0.1001×107Changeinconc.(M)x+x+xEquilibriumconc.(M)0.10xxx

The concentration of OH comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10x)=1.5×1011

Assuming x is very small, it can be written 0.10x0.10.

Then,

  Kb=(x)(x)(0.15x)=1.5×1011x2=(1.5×1011)×(0.10)x=1.5×1012=1.2×106M=[OH]

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(1.2×106)=5.91.pH=14.00pOH=14.005.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution
Check Mark

Explanation of Solution

The concentration of CH3COO can be calculated as shown below.

  [CH3COO]=0.10MMg(CH3COO)2×2molCH3COO1molMg(CH3COO)2=0.20M.

The equilibrium reaction is given below.

  CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)

The base ionization constant can be written as given below.

  Kb=[CH3COOH][OH][CH3COO]

A table can be set up as shown below.

  CH3COO(aq)CH3COOH(aq)OH(aq)Initialconc.(M)0.2001×107Changeinconc.(M)x+x+xEquilibriumconc.(M)0.20xxx

The concentration of OH comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20x)=5.6×1010

Assuming x is very small, it can be written 0.20x0.20.

Then,

  Kb=(x)(x)(0.20x)=5.6×1010x2=(5.6×1010)×(0.20)x=(5.6×1010)×(0.20)=1.1×105M=[OH]

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(1.1×105)=4.98.pH=14.00pOH=14.004.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution
Check Mark

Explanation of Solution

The reaction of BaO with water is given below.

  O2(aq)+H2O(l)2OH(aq)

The concentration of OH can be calculated as shown below.

  [OH]=0.10MO2×2molOH1molO2=0.20M.

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(0.20)=0.70.pH=14.00pOH=14.000.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution
Check Mark

Explanation of Solution

HSO4 acts as both acid and base.

  HSO4(aq)+H2O(aq)H3O+(aq)+SO42(aq)Ka=[H3O+][SO42][HSO4]=1.1×102HSO4(aq)+H2O(aq)H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH][HSO4]=verysmall

HSO4 is more acidic than basic.

A table can be set up as shown below.

  HSO4(aq)H3O+(aq)SO42(aq)Initialconc.(M)0.101×1070Changeinconc.(M)x+x+xEquilibriumconc.(M)0.10xxx

The concentration of H3O+ comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10x)=1.1×102

Then,

  x2=(1.1×102)×(0.10x)x2+1.1×102x1.1×103=0x=2.8×102M=[H3O+]

The pH of the solution can be calculated as shown below.

  pH=log[H3O+]=log(2.8×102)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution
Check Mark

Explanation of Solution

HCO3 acts as both acid and base.

  HCO3(aq)+H2O(aq)H3O+(aq)+CO32(aq)Ka=[H3O+][CO32][HCO3]=4.7×1011HCO3(aq)+H2O(aq)OH(aq)+H2CO3(aq)Kb=[H2CO3][OH][HCO3]=2.3×108

HCO3 is more basic than acidic.

A table can be set up as shown below.

  HCO3(aq)H2CO3(aq)OH(aq)Initialconc.(M)0.1001×107Changeinconc.(M)x+x+xEquilibriumconc.(M)0.10xxx

The concentration of OH comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10x)=2.3×108

Then,

Assuming x is very small, it can be written 0.20x0.20.

  Kb=(x)(x)(0.10x)=2.3×108x2=(2.3×108)×(0.10)x=(2.3×108)×(0.10)=4.8×105M=[OH]

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(4.8×105)=4.32.pH=14.00pOH=14.004.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution
Check Mark

Explanation of Solution

BaCl2Ba2++2Cl

Neither of these ions affect the  pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl is a weak base.  So  pH is 7.00 and no calculation is needed.

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 14 Solutions

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

Ch. 14.4 - Calculate the pH of a 0.040-M NaOH solution. Ch. 14.4 - In a hospital laboratory the pH of a bile sample...Ch. 14.4 - Prob. 14.8CECh. 14.4 - Prob. 14.9ECh. 14.4 - Prob. 14.10ECh. 14.5 - Write the ionization equation and ionization...Ch. 14.5 - Write the ionization equation and the Kb...Ch. 14.5 - Prob. 14.11CECh. 14.5 - Prob. 14.12CECh. 14.5 - Prob. 14.13ECh. 14.6 - Prob. 14.14CECh. 14.6 - Prob. 14.15CECh. 14.6 - Prob. 14.16CECh. 14.6 - Prob. 14.17CECh. 14.6 - Prob. 14.18CECh. 14.7 - Lactic acid is a monoprotic acid that occurs...Ch. 14.7 - Prob. 14.9PSPCh. 14.7 - Prob. 14.19ECh. 14.7 - Prob. 14.10PSPCh. 14.7 - Prob. 14.20ECh. 14.8 - Prob. 14.11PSPCh. 14.8 - Prob. 14.21CECh. 14.8 - Prob. 14.12PSPCh. 14.8 - Prob. 14.22ECh. 14.8 - Prob. 14.23CECh. 14.8 - Prob. 14.24CECh. 14.9 - Predict whether each of these is a Lewis acid or a...Ch. 14.9 - Prob. 14.26ECh. 14.9 - Prob. 14.27ECh. 14.10 - Prob. 14.28ECh. 14.10 - Prob. 14.13PSPCh. 14.10 - Prob. 14.29ECh. 14.10 - Calculate the pH of 5.2-M aqueous sodium...Ch. 14 - Lactic acid, CH3CH(OH)COOH, is a weak monoprotic...Ch. 14 - Define a Brnsted-Lowry acid and a Brnsted-Lowry...Ch. 14 - Prob. 2QRTCh. 14 - Prob. 3QRTCh. 14 - Prob. 4QRTCh. 14 - Prob. 5QRTCh. 14 - Prob. 6QRTCh. 14 - Prob. 7QRTCh. 14 - Prob. 8QRTCh. 14 - Write a chemical equation to describe the proton...Ch. 14 - Write a chemical equation to describe the proton...Ch. 14 - Prob. 11QRTCh. 14 - Prob. 12QRTCh. 14 - Prob. 13QRTCh. 14 - Prob. 14QRTCh. 14 - Prob. 15QRTCh. 14 - Prob. 16QRTCh. 14 - Prob. 17QRTCh. 14 - Prob. 18QRTCh. 14 - Prob. 19QRTCh. 14 - Prob. 20QRTCh. 14 - Prob. 21QRTCh. 14 - Prob. 22QRTCh. 14 - Prob. 23QRTCh. 14 - Formic acid, HCOOH, is found in ants. Write a...Ch. 14 - Milk of magnesia, Mg(OH)2, has a pH of 10.5....Ch. 14 - A sample of coffee has a pH of 4.3. Calculate the...Ch. 14 - Calculate the pH of a solution that is 0.025-M in...Ch. 14 - Calculate the pH of a 0.0013-M solution of HNO3....Ch. 14 - Prob. 29QRTCh. 14 - Prob. 30QRTCh. 14 - A 1000.-mL solution of hydrochloric acid has a pH...Ch. 14 - Prob. 32QRTCh. 14 - Prob. 33QRTCh. 14 - Prob. 34QRTCh. 14 - Figure 14.3 shows the pH of some common solutions....Ch. 14 - Figure 14.3 shows the pH of some common solutions....Ch. 14 - The measured pH of a sample of seawater is 8.30....Ch. 14 - Prob. 38QRTCh. 14 - Valine is an amino acid with this Lewis structure:...Ch. 14 - Leucine is an amino acid with this Lewis...Ch. 14 - Prob. 41QRTCh. 14 - Prob. 42QRTCh. 14 - Prob. 43QRTCh. 14 - Prob. 44QRTCh. 14 - Prob. 45QRTCh. 14 - Prob. 46QRTCh. 14 - Prob. 47QRTCh. 14 - Prob. 48QRTCh. 14 - Prob. 49QRTCh. 14 - Prob. 50QRTCh. 14 - Prob. 51QRTCh. 14 - Prob. 52QRTCh. 14 - Prob. 53QRTCh. 14 - Prob. 54QRTCh. 14 - A 0.015-M solution of cyanic acid has a pH of...Ch. 14 - Prob. 56QRTCh. 14 - The pH of a 0.10-M solution of propanoic acid,...Ch. 14 - Prob. 58QRTCh. 14 - Prob. 59QRTCh. 14 - Prob. 60QRTCh. 14 - Prob. 61QRTCh. 14 - Amantadine, C10H15NH2, is a weak base used in the...Ch. 14 - Prob. 63QRTCh. 14 - Lactic acid, C3H6O3, occurs in sour milk as a...Ch. 14 - Prob. 65QRTCh. 14 - Complete each of these reactions by filling in the...Ch. 14 - Complete each of these reactions by filling in the...Ch. 14 - Predict which of these acid-base reactions are...Ch. 14 - Predict which of these acid-base reactions are...Ch. 14 - Prob. 70QRTCh. 14 - Prob. 71QRTCh. 14 - Prob. 72QRTCh. 14 - Prob. 73QRTCh. 14 - Prob. 74QRTCh. 14 - Prob. 75QRTCh. 14 - Prob. 76QRTCh. 14 - Prob. 77QRTCh. 14 - Prob. 78QRTCh. 14 - Prob. 79QRTCh. 14 - Prob. 80QRTCh. 14 - Prob. 81QRTCh. 14 - Trimethylamine, (CH3)3N, reacts readily with...Ch. 14 - Prob. 83QRTCh. 14 - Prob. 84QRTCh. 14 - Prob. 85QRTCh. 14 - Prob. 86QRTCh. 14 - Common soap is made by reacting sodium carbonate...Ch. 14 - Prob. 88QRTCh. 14 - Prob. 89QRTCh. 14 - Prob. 90QRTCh. 14 - Prob. 91QRTCh. 14 - Prob. 92QRTCh. 14 - Prob. 93QRTCh. 14 - Several acids and their respective equilibrium...Ch. 14 - Prob. 95QRTCh. 14 - Prob. 96QRTCh. 14 - Does the pH of the solution increase, decrease, or...Ch. 14 - Does the pH of the solution increase, decrease, or...Ch. 14 - Prob. 99QRTCh. 14 - Prob. 100QRTCh. 14 - Prob. 101QRTCh. 14 - Prob. 102QRTCh. 14 - Prob. 103QRTCh. 14 - Prob. 104QRTCh. 14 - Prob. 105QRTCh. 14 - Prob. 106QRTCh. 14 - When all the water is evaporated from a sodium...Ch. 14 - Prob. 108QRTCh. 14 - Prob. 109QRTCh. 14 - Prob. 110QRTCh. 14 - Prob. 111QRTCh. 14 - Prob. 112QRTCh. 14 - Prob. 113QRTCh. 14 - Prob. 114QRTCh. 14 - Prob. 115QRTCh. 14 - Prob. 116QRTCh. 14 - Home gardeners spread aluminum sulfate powder...Ch. 14 - Prob. 118QRTCh. 14 - Prob. 119QRTCh. 14 - Prob. 120QRTCh. 14 - Prob. 121QRTCh. 14 - Prob. 122QRTCh. 14 - Prob. 123QRTCh. 14 - Prob. 124QRTCh. 14 - Prob. 125QRTCh. 14 - A chilled carbonated beverage is opened and warmed...Ch. 14 - Prob. 127QRTCh. 14 - Explain why BrNH2 is a weaker base than ammonia,...Ch. 14 - Prob. 129QRTCh. 14 - Prob. 130QRTCh. 14 - At 25 C, a 0.10% aqueous solution of adipic acid,...Ch. 14 - Prob. 132QRTCh. 14 - Prob. 133QRTCh. 14 - Prob. 134QRTCh. 14 - Prob. 135QRTCh. 14 - Prob. 14.ACPCh. 14 - Develop a set of rules by which you could predict...
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781285199030
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY