Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change
7th Edition
ISBN: 9780078129865
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 19, Problem 19.60P

(a)

Interpretation Introduction

Interpretation:

pH65.5mL of 0.234MNH3 at the equivalence point and the volume of 0.125MHCl 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.

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

(a)

Expert Solution
Check Mark

Answer to Problem 19.60P

The pH of a solution made by mixing 65.5mL and 0.234MNH3 with 0.125MHCl is 5.71.

Explanation of Solution

Given,

The balance equation is given below,

  HCl(aq) +NH3(aq)NH4+(aq) +Cl(aq)

The chlorine 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 HCl,

Volume mL of HCl,

HCl=(0.234molNH3L)(103L1mL)(65.5mL)(1molHCl1molNH3)(L0.125molHCl)(1mL103L)=122.616=123mLofHCl

Therefore, the required volume of HCl=123mL.

Determination the moles of NH3present,

MolesofNH3=(0.234molNH31mL)(103L1mL)(65.5mL)=0.015327molNH3

At the equivalent point, 0.015327molNH3 will be added so, the moles acid = moles base.

The HCl will react with an equal amount of the acid 0molNH3 will remain and 0.015327molofNH4+ will be formed.

ICE Table of equilibrium,

 HCl(aq) +NH3(aq)NH4+(l) +Cl(aq)Initial (M)0.015327mol0.015327mol   0Change(M)0.015327mol0.015327mol+0.015327molFinal  00+0.015327mol

Determination the liters of solution present at the equivalent point,

Volume

   =[(65.5+122.616)mL](103L/1mL)=0.188116L

Concentration of NH4+ at equivalent point,

Molarity,

  =(0.015327molNH4+)/(0.188116L)=0.081476M

Next calculate Kafor NH4+

The equilibrium value for KbNH3=1.76×105

So,

  Kb=KW/Ka=(1.0×1014)(1.76×105)=5.6818×1010 

Using a reaction table for the equilibrium reaction of NH4+

   NH4++H2ONH3 +H3O+Initial (M):0.081476M  00Change(M):x+x+xEquilibrium: 0.081476x+x+x

Determination the hydroxide ion concentration from the Ka and then determine the pH from the pOH.

  Ka=5.6818×1010=[H3O+][NH3][NH4+]=[x][x][0.081476x]=x2[0.081476][H3O+]=x=6.803898×106M

Calculation for pH and pOH

  pH=log[H3O+]=log(6.803898×106)pH=5.1672=5.71

(b)

Interpretation Introduction

Interpretation:

21.8mL of 1.11MCH3NH2 at the equivalence point and the volume of 0.125MHCl 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

(b)

Expert Solution
Check Mark

Answer to Problem 19.60P

The pH of a solution made by mixing 21.8mL and 1.11MCH3NH2 with 0.125MHCl is 5.80.

Explanation of Solution

The balance equation is given below,

  HCl(aq)+CH3NH2(aq)CH3NH3+(aq)+Cl(aq)

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

Next calculate the volume of HCl,

Volume of HCl in mL,

HCl=(1.11molCH3NH2L)(103L1mL)(21.8mL)(1molKOH1molH2CO3)(L0.125molKOH)(1mL103L)=193.584=194mLofHCl

The solution requires an equal volume to reach second equivalence point 194mLofHCl.

Determination the moles of CH3NH2present,

  MolesofCH3NH2=(1.11molCH3NH2L)(103L1mL)(21.8mL)=0.024198molofCH3NH2

At the equivalence point, 0.024198molHCl will be added the moles acid=moles of base.

The HCl will react with an equal amount of the base 0molCH3NH2 will remain and 0.024198molofCH3NH3+ will be formed.

ICE Table of equilibrium,

 HCl(aq) +CH3NH2(aq)CH3NH3+(aq) +Cl(aq)Initial (M)0.024198mol0.024198mol   0Change(M)0.024198mol0.024198mol+0.024198molFinal  00+0.024198mol-

Determination the liters of solution present at the equivalent point,

Volume

   =[(21.8+193.584)mL](103L/1mL)=0.0215384L.

Concentration of HCO3 at equivalent point,

Molarity,

  =(0.022249molHCO3)(0.055548L)=0.0404875M.

Concentration of CH3NH3+ at equivalent point,

Molarity,

  =(0.024198molCH3NH3+)(0.215384L)=0.1123482M.

Next calculate Kafor CH3NH3+

The equilibrium value for Kb(CH3NH3+)=4.4×104

  Kb=KW/Ka=(1.0×1014)(4.5×107)=2.222×108 

The equilibrium value for,

  Kb=KW/Ka=(1.0×1014)(4.4×104)=2.2727×1011

Following the reaction table for the equilibrium reaction of CH3NH3+

 CH3NH3+(aq) +H2O(aq)CH3NH2(aq) +H3O+(aq)Initial (M)0.1123482mol  00Change(M)x00Final  0.1123482-xxxx

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

Calculation for first equilibrium point:

  Ka=2.2727×1011=[H2O+][CH3NH2][CH3NH3+]=[x][x][0.1123482x]=x2[0.1123482]x=[H3O+]=1.5979×106M

Calculation for pH

  pH=log[H3O+]=log(1.5979×106)=5.7964pH=5.80.

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

Connect 2-Year Access Card for Chemistry: The Molecular Nature of Matter and Change

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 - The scenes below depict solutions of the same...Ch. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Does the pH increase or decrease with each of the...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. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 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. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 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. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 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 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.44PCh. 19 - What species are in the buffer region of a weak...Ch. 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 - Use figure 19.9 to find an indicator for these...Ch. 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 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.84PCh. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.88PCh. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.96PCh. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.100PCh. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 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. 19.105PCh. 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 - A solution contains 0.10 M ZnCl2 and 0.020 M...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 - A 0.050 M H2S solution contains 0.15 M NiCl2 and...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. 19.138PCh. 19 - Prob. 19.139PCh. 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. 19.143PCh. 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. 19.148PCh. 19 - Prob. 19.149PCh. 19 - Prob. 19.150P
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