MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 19, Problem 19.56P

(a)

Interpretation Introduction

Interpretation:

pH of 2.65mLof0.0750Mpyridine  at the equivalence point and the volume of 0.447MHNO3 needed to the reach the point’s in titrations has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

Relationship between Kaand Kb

  Ka×Kb=Kw

(a)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a solution made by mixing 2.65mLof0.0750Mpyridine with 0.447MHNO3 is 3.21.

Explanation of Solution

Given,

The balance equation is given below,

  HNO3(aq) +C5H5N(aq)C5H5NH+(aq) +NO3+(aq)

The nitrate ions on the product side are written as separate spices because they have no effect on pH of the solution.

Calculate the required volume of HNO3,

Volume mL of HNO3,

HNO3=(0.0750molC5H5NL)(2.65mL)(1molHNO31molC5H5N)(L0.447molHNO3)(1mL103L)=444.63087=445mLofHNO3.

Therefore, the required volume of HNO3=445mL

Determination the moles of C5H5NH+produced,

MolesofC5H5NH+=(0.0750molC5H5N1mL)(2.65mL)(1molC5H5NH+1molC5H5N)=0.19875molofC5H5NH+

Determination the liters of solution present at the equivalent point,

Volume

   =[2.65L+(444.63087mL)](103L/1mL)=3.09463L

Concentration of C5H5NH+ at equivalent point,

Molarity,

  =(0.19875molC5H5NH+)/(3.09463L)=0.064224M

CalculateKafor C5H5NH+

The equilibrium value for Kb(C5H5N)=1.7×109

  Kb=KW/Ka=(1.0×1014)(1.7×109)=5.88253×106 

Determination the hydrogen ion concentration from the Ka and then determine the pH.

  Ka=5.88253×106=[H3O+][C5H5N][C5H5NH+]=[x][x][0.064224x]=x2[0.064224][H3O+]=x=6.1464×104M

Calculation for pH

  pH=log[H3O+]=log(6.1464×104)pH=3.211379=3.21

Therefore, the volume of 0.447MHNO3 pH reachable point is 3.21_.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.188Lof0.250M ethylenediamine at the equivalence point and the volume of 0.447MHNO3 needed to the reach the point’s in titrations has to be calculated.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  pH=-log[H3O+]

Acid ionization constantKa:

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

Relationship between Kaand Kb

  Ka×Kb=Kw

(b)

Expert Solution
Check Mark

Answer to Problem 19.56P

The pH of a solution made by mixing 0.188Lof0.250M ethylenediamine with 0.447MHNO3 is 5.36.

The second equivalent pH of a given buffer solution after the addition of 0.188Lof0.250M ethylenediamine and 0.447MHNO3 is 3.89.

Explanation of Solution

The balance equation is given below,

  HNO3(aq) +H2NCH2CH2NH2(aq)H2NCH2CH2NH3+(aq) +NO3(aq)HNO3(aq) +H2NCH2CH2NH3+(aq)H3NCH2CH2NH32+(aq) +NO3(aq)

The nitrate ions on the product side are written as separate spices because they have no effect on the pH of the solution.

Calculate the volume of HNO3 needed,

Volume of HNO3 in mL,

=(0.250molH2NCH2CH3NH2L)(0.188L)(1molHNO31molH2NCH2CH3NH2)(L0.447molHNO3)(1mL103L)=105.1454=105mLofHCl.

The solution requires an equal volume to reach second equivalence point 105mLofHCl.

Determination the moles of H2NCH2CH2NH3+present,

MolesofH2NCH2CH2NH3+=(0.250molH2NCH2CH2NH2L)(0.188mL)(1molH2NCH2CH2NH3+1molH2NCH2CH2NH2)=0.0470molofH2NCH2CH2NH3+

An equal number of moles of H2NCH2CH2NH3+ will be present at the second equivalent point.

Determination the liters of solution at the first equilibrium point:

Volume

   =[0.188L+(105.1454mL)](103L/1mL)=0.293145L

Determination the liters of solution at the second equilibrium point:

Volume

   =[0.188L+2(105.1454mL)](103L/1mL)=0.39829L

Concentration of H2NCH2CH2NH3+ at equivalent point,

Molarity,

  =(0.0470molH2NCH2CH2NH3+)(0.293145L)=0.16033M

Concentration of H3NCH2CH2NH32+ at equivalent point,

Molarity,

  =(0.0470molH3NCH2CH2NH32+)(0.39829L)=0.11800M

Calculate Kafor H2NCH2CH2NH3+

The equilibrium value for Kb(H2NCH2CH2NH2)=8.5×105

  Kb=KW/Ka=(1.0×1014)(8.5×105)=1.17674×1010 

Next calculate Kafor H2NCH2CH2NH32+

The equilibrium value for Kb(H2NCH2CH2NH3+)=7.1×108

  Kb=KW/Ka=(1.0×1014)(7.1×108)=1.40845×107 

Determination the hydrogen ion concentration from the Ka and then determine the pH for the first equilibrium point.

Calculation for first equilibrium point:

  Ka=1.17647×1010=[H3O+][H2NCH2CH2NH2][H3NCH2CH2NH3+]=[x][x][0.16033x]=x2[0.16033]x=[H3O+]=4.3430780×106M

Calculation for pH

  pH=log[H3O+]=log(4.3430780×106)=5.36220pH=5.36

Determination the hydrogen ion concentration from Ka' and then determine the pH for the second equilibrium point.

Calculation for second equilibrium point:

  Ka=1.40845×107=[H3O+][H2NCH2CH2NH3+][H3NCH2CH2NH32+]=[x][x][0.11800x]=x2[0.11800]x=[H3O+]=1.2891745×104M

Calculation for pH

  pH=log[H3O+]=log(1.2891745×104M)=3.889688pH=3.89

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 19 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 19.3 - Prob. 19.6AFPCh. 19.3 - Prob. 19.6BFPCh. 19.3 - Prob. 19.7AFPCh. 19.3 - Prob. 19.7BFPCh. 19.3 - Prob. 19.8AFPCh. 19.3 - Prob. 19.8BFPCh. 19.3 - Prob. 19.9AFPCh. 19.3 - Prob. 19.9BFPCh. 19.3 - Prob. 19.10AFPCh. 19.3 - Prob. 19.10BFPCh. 19.3 - Prob. 19.11AFPCh. 19.3 - Prob. 19.11BFPCh. 19.3 - Prob. 19.12AFPCh. 19.3 - Prob. 19.12BFPCh. 19.3 - An environmental technician collects a sample of...Ch. 19.3 - A lake that has a surface area of 10.0 acres (1...Ch. 19.4 - Cyanide ion is toxic because it forms stable...Ch. 19.4 - Prob. 19.13BFPCh. 19.4 - Prob. 19.14AFPCh. 19.4 - Calculate the solubility of PbCl2 in 0.75 M NaOH....Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - Prob. 19.4PCh. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Does the pH increase or decrease with each of the...Ch. 19 - The scenes below depict solutions of the same...Ch. 19 - The scenes below show three samples of a buffer...Ch. 19 - What are the [H3O+] and the pH of a propanoic...Ch. 19 - What are the [H3O+] and the pH of a benzoic...Ch. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Find the pH of a buffer that consists of 0.50 M...Ch. 19 - A buffer consists of 0.22 M KHCO3 and 0.37 M...Ch. 19 - A buffer consists of 0.50 M NaH2PO4 and 0.40 M...Ch. 19 - What is the component concentration ratio,...Ch. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - A buffer that contains 0.40 M of a base, B, and...Ch. 19 - A buffer that contains 0.110 M HY and 0.220 M Y−...Ch. 19 - A buffer that contains 1.05 M B and 0.750 M BH+...Ch. 19 - A buffer is prepared by mixing 204 mL of 0.452 M...Ch. 19 - A buffer is prepared by mixing 50.0 mL of 0.050 M...Ch. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Choose specific acid-base conjugate pairs to make...Ch. 19 - An industrial chemist studying bleaching and...Ch. 19 - Oxoanions of phosphorus are buffer components in...Ch. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.41PCh. 19 - What species are in the buffer region of a weak...Ch. 19 - Prob. 19.43PCh. 19 - Prob. 19.44PCh. 19 - Prob. 19.45PCh. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Prob. 19.53PCh. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Use figure 19.9 to find an indicator for these...Ch. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Prob. 19.66PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Write the ion-product expressions for (a)...Ch. 19 - Write the ion-product expressions for (a) calcium...Ch. 19 - Prob. 19.70PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.72PCh. 19 - Prob. 19.73PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Prob. 19.77PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Prob. 19.83PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Prob. 19.87PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - Prob. 19.91PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Prob. 19.95PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - Prob. 19.99PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 19 - Prob. 19.103PCh. 19 - When 0.84 g of ZnCl2 is dissolved in 245 mL of...Ch. 19 - When 2.4 g of Co(NO3)2 is dissolved in 0.350 L of...Ch. 19 - Prob. 19.106PCh. 19 - A microbiologist is preparing a medium on which to...Ch. 19 - As an FDA physiologist, you need 0.700 L of formic...Ch. 19 - Tris(hydroxymethyl)aminomethane [(HOCH2)3CNH2],...Ch. 19 - Water flowing through pipes of carbon steel must...Ch. 19 - Gout is caused by an error in metabolism that...Ch. 19 - In the process of cave formation (Section 19.3),...Ch. 19 - Phosphate systems form essential buffers in...Ch. 19 - The solubility of KCl is 3.7 M at 20°C. Two...Ch. 19 - It is possible to detect NH3 gas over 10−2 M NH3....Ch. 19 - Manganese(II) sulfide is one of the compounds...Ch. 19 - The normal pH of blood is 7.40 ± 0.05 and is...Ch. 19 - A bioengineer preparing cells for cloning bathes a...Ch. 19 - Sketch a qualitative curve for the titration of...Ch. 19 - Prob. 19.120PCh. 19 - The scene at right depicts a saturated solution of...Ch. 19 - Prob. 19.122PCh. 19 - The acid-base indicator ethyl orange turns from...Ch. 19 - Prob. 19.124PCh. 19 - Prob. 19.125PCh. 19 - Prob. 19.126PCh. 19 - Prob. 19.127PCh. 19 - Prob. 19.128PCh. 19 - Prob. 19.129PCh. 19 - Calcium ion present in water supplies is easily...Ch. 19 - Calculate the molar solubility of Hg2C2O4 (Ksp =...Ch. 19 - Environmental engineers use alkalinity as a...Ch. 19 - Human blood contains one buffer system based on...Ch. 19 - Quantitative analysis of Cl− ion is often...Ch. 19 - An ecobotanist separates the components of a...Ch. 19 - Some kidney stones form by the precipitation of...Ch. 19 - Prob. 19.137PCh. 19 - Prob. 19.138PCh. 19 - Because of the toxicity of mercury compounds,...Ch. 19 - A 35.0-mL solution of 0.075 M CaCl2 is mixed with...Ch. 19 - Rainwater is slightly acidic due to dissolved CO2....Ch. 19 - Prob. 19.142PCh. 19 - Ethylenediaminetetraacetic acid (abbreviated...Ch. 19 - Buffers that are based on...Ch. 19 - NaCl is purified by adding HCl to a saturated...Ch. 19 - Scenes A to D represent tiny portions of 0.10 M...Ch. 19 - Prob. 19.147PCh. 19 - Prob. 19.148PCh. 19 - Prob. 19.149P
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
Acid-Base Titration | Acids, Bases & Alkalis | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=yFqx6_Y6c2M;License: Standard YouTube License, CC-BY