Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 6, Problem 6H.10E
Interpretation Introduction

Interpretation:

The pH at the stoichiometric point of the titration of 30.0 mL of 0.0172MC17H19O3N(aq) with 0.0160MHCl(aq) has to be calculated.

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+]

On rearranging, the concentration of hydrogen ion [H+] can be calculated using pH as follows,

  [H+]=10-pH

Expert Solution & Answer
Check Mark

Answer to Problem 6H.10E

The pH at the stoichiometric point of the titration of 30.0 mL of 0.0172MC17H19O3N(aq) with 0.0160MHCl(aq) is 5.54.

Explanation of Solution

The initial amount of C17H19O3N present in the analyte solution is

  n(C17H19O3N)=(0.030×10-3L)×(0.0172molL-1)=5.16×10-4mol=0.516×10-3mol=0.516mmol

The amount of H3O+ required to react with C17H19O3N is calculated and their reaction stoichiometry is

1molH3O+=1molC17H19O3N.

  n(H3O+)=(initial amount ofC17H19O3Nin the analyte)×1molH3O+1molNH3=(0.516×10-3mol)×1molH3O+1molC17H19O3N=0.516×10-3mol=0.516mmol

Therefore, the H3O+ ion required to react with C17H19O3N is 0.516 mmol.

From the stoichiometric point the amount of C17H19O3NH+ is equal to the amount of OH added.

  n(NH4+)=0.5mmol

The amount of H3O+ ions present in the volume of titrant is calculated as.

  Vadded=H3O+ion required to react withNH3concentrationofH3O+ion=0.5×10-3mol0.015mol×L-1=3.33×10-2L=33.3mL

Hence, the amount of H3O+ ions present in the volume of titrant is 33.3 mL.

The total volume of the solution at the stoichiometric point is calculated as,

  Vfinal=Vinitial+Vadded=25.0+33.3mL=58.3mL

The concentration of ClO ions at the stoichiometric point is

  [NH4+]=amount of NH4+presentintheanalytesolutiontotal volume of the solution=0.5×10-3mol3.33×10-2L=0.015molL-1

Therefore, the concentration of NH4+ is 0.015molL-1.

The chemical equilibrium reaction is given below.

  C17H19O3N(aq)+H2O(l)C17H19O3NH+(aq)+OH(aq)

  Ka=[NH3][H3O+][NH4+]

 NH4+NH3H3O+
Initial concentration0.01500
Change in concentration-x+x+x
Equilibrium concentration0.015-xxx

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  Ka=x×x0.015-x

The above equation, assume that the x present in 0.015-x is very small than 0.015 then it can be negligible and it follows,

  Ka=x20.015

Ammonia Ka value is 5.6×10-10 substitute this in above equation.

  5.6×10-10=x20.015x2=(5.6×10-10)×0.015x=(5.6×10-10)×0.015=2.90×10-6

Hence, the concentration of H3O+ is 2.90×10-6.

Now, we can calculate the pH of the solution,

  pH=-log(H3O+)=-log(2.90×10-6)=5.54

Hence, the pH at the stoichiometric point of the titration of 25.0 mL of 0.020MNH3(aq) with 0.015MHCl(aq) is 5.54.

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Chapter 6 Solutions

Chemical Principles: The Quest for Insight

Ch. 6 - Prob. 6A.5ECh. 6 - Prob. 6A.6ECh. 6 - Prob. 6A.7ECh. 6 - Prob. 6A.8ECh. 6 - Prob. 6A.9ECh. 6 - Prob. 6A.10ECh. 6 - Prob. 6A.11ECh. 6 - Prob. 6A.12ECh. 6 - Prob. 6A.13ECh. 6 - Prob. 6A.14ECh. 6 - Prob. 6A.15ECh. 6 - Prob. 6A.16ECh. 6 - Prob. 6A.17ECh. 6 - Prob. 6A.18ECh. 6 - Prob. 6A.19ECh. 6 - Prob. 6A.20ECh. 6 - Prob. 6A.21ECh. 6 - Prob. 6A.22ECh. 6 - Prob. 6A.23ECh. 6 - Prob. 6A.24ECh. 6 - Prob. 6B.1ASTCh. 6 - Prob. 6B.1BSTCh. 6 - Prob. 6B.2ASTCh. 6 - Prob. 6B.2BSTCh. 6 - Prob. 6B.3ASTCh. 6 - Prob. 6B.3BSTCh. 6 - Prob. 6B.1ECh. 6 - Prob. 6B.2ECh. 6 - Prob. 6B.3ECh. 6 - Prob. 6B.4ECh. 6 - Prob. 6B.5ECh. 6 - Prob. 6B.6ECh. 6 - Prob. 6B.7ECh. 6 - Prob. 6B.8ECh. 6 - Prob. 6B.9ECh. 6 - Prob. 6B.10ECh. 6 - Prob. 6B.11ECh. 6 - Prob. 6B.12ECh. 6 - Prob. 6C.1ASTCh. 6 - Prob. 6C.1BSTCh. 6 - Prob. 6C.2ASTCh. 6 - Prob. 6C.2BSTCh. 6 - Prob. 6C.3ASTCh. 6 - Prob. 6C.3BSTCh. 6 - Prob. 6C.1ECh. 6 - Prob. 6C.2ECh. 6 - Prob. 6C.3ECh. 6 - Prob. 6C.4ECh. 6 - Prob. 6C.5ECh. 6 - Prob. 6C.6ECh. 6 - 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