Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 16, Problem 16.44QP
Interpretation Introduction

Interpretation: The pH of the solution of trimethylamine and hydrochloric acid at different volume of acid is to be calculated.

Concept introduction: The pH of the solution is calculated by using the formula,

pH=log[H+]

To determine: The pH of the solution of trimethylamine and hydrochloric acid at different volume of base.

Expert Solution & Answer
Check Mark

Answer to Problem 16.44QP

Solution

The pH of the solution at 10.0mL volume of acid is 9.79_ .

The pH of the solution at 20.0mL volume of acid is 5.52_ .

The pH of the solution at 30.0mL volume of acid is 1.64_ .

Explanation of Solution

Explanation

Given

The volume of trimethylamine is 25.0mL .

The concentration of trimethylamine is 0.100M .

The concentration of hydrochloric acid is 0.125M .

The volume of hydrochloric acid added is 10.0mL,20.0mL and 30.0mL .

At 10.0mL volume of hydrochloric acid:

The concentration of trimethylammonium chloride formed after addition of hydrochloric acid is calculated by multiplying the ratio of hydrochloric acid present in the solution with the concentration of hydrochloric acid solution.

Concentrationof[(CH3)3NHCl]=10.010.0+25.0×0.125M=0.036M

The concentration of trimethylamine left in the solution after formation of trimethylammonium chloride is calculated by subtracting the concentration of trimethylammonium chloride from the volume of trimethylamine in the solution.

Concentrationof[(CH3)3N](insolution)=25.025.0+10.0×0.100M0.036M=0.035M

The pH of the given buffer solution is calculated by using the formula,

14pH=pKb+log[(CH3)3NHCl][(CH3)3N]

Where,

  • pH is the hydrogen content of solution.
  • pKb is the dissociation constant of base.
  • [(CH3)3NHCl] is the concentration of trimethylammonium chloride solution.
  • [(CH3)3N] is the concentration of trimethylamine.

Substitute the values of [(CH3)3NHCl] , [(CH3)3N] and pKb in the above equation.

14pH=4.19+log0.0360.035pH=9.79_

At 20.0mL volume of hydrochloric acid:

At 20.0mL volume of hydrochloric acid the whole acid present in the solution gets neutralized and the solution will reach at the equivalence point. The equivalent concentration of trimethylammonium chloride salt in the solution is calculated as,

Concentrationof[(CH3)3NHCl]=20.020.0+25.0×0.125M=0.056M

The concentration of hydroxyl ions in the given solution is calculated by using the dissociation constant of trimethylammonium chloride by using the following equation.

(CH3)3NH+(aq)+H2O(l)(CH3)3N(aq)+H3O+(aq)

The concentration of trimethylamine and hydrogen ions formed in the given equilibrium reaction is considered to be x . The dissociation constant of the trimethylammonium chloride is calculated by using the formula,

Ka=[H3O+][(CH3)3N][(CH3)3NH+]

Where,

  • Ka is the dissociation constant of the acid.
  • [H3O+] is the hydrogen ion concentration.
  • [(CH3)3N] is the concentration of trimethylamine.
  • [(CH3)3NH+] is the concentration of trimethylammonium ion.

Substitute the values of Ka , [H3O+] , [(CH3)3N] and [(CH3)3NH+] in the above equation.

1.56×1010=(x)(x)(0.056x)

Since, the value of dissociation constant is very less than the concentration of the solution. Therefore, the value of x<<<0.056 due to which 0.056x0.056 .

1.56×1010=(x)(x)(0.056)x=2.96×106

Therefore, the concentration of hydrogen ions in the solution is 2.96×106 .

The pH of the solution is calculated by using the formula,

pH=log[H+]

Where,

  • [H+] is the concentration of hydrogen ions in the solution.

Substitute the values of [H+] in the above equation.

pH=log(2.96×106)pH=5.52_

At 30.0mL volume of hydrochloric acid:

At 30.0mL volume of hydrochloric acid the volume of hydrochloric acid is present in excess in the solution. Therefore, the pH of the solution corresponds to the excess acid present in the solution. The concentration of excess acid present in the solution is calculated as,

ConcentrationofHCl(excess)=10.030.0+25.0×0.125M=0.023M

The pH of the solution is calculated by using the formula,

pH=log[H+]

Where,

  • [H+] is the concentration of hydrogen ions in the solution.

Substitute the values of [H+] in the above equation.

pH=log(0.023)pH=1.64_

Conclusion

The pH of the solution at 10.0mL volume of acid is 9.79_ .

The pH of the solution at 20.0mL volume of acid is 5.52_ .

The pH of the solution at 30.0mL volume of acid is 1.64_

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Chapter 16 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 16.4 - Prob. 11PECh. 16.4 - Prob. 12PECh. 16.5 - Prob. 13PECh. 16.6 - Prob. 14PECh. 16.8 - Prob. 15PECh. 16.8 - Prob. 16PECh. 16.8 - Prob. 17PECh. 16.8 - Prob. 18PECh. 16.8 - Prob. 19PECh. 16 - Prob. 16.1VPCh. 16 - Prob. 16.2VPCh. 16 - Prob. 16.3VPCh. 16 - Prob. 16.4VPCh. 16 - Prob. 16.5VPCh. 16 - Prob. 16.6VPCh. 16 - Prob. 16.7VPCh. 16 - Prob. 16.8VPCh. 16 - Prob. 16.9VPCh. 16 - Prob. 16.10VPCh. 16 - Prob. 16.11QPCh. 16 - Prob. 16.12QPCh. 16 - Prob. 16.13QPCh. 16 - Prob. 16.14QPCh. 16 - Prob. 16.15QPCh. 16 - Prob. 16.16QPCh. 16 - Prob. 16.17QPCh. 16 - Prob. 16.18QPCh. 16 - Prob. 16.19QPCh. 16 - Prob. 16.20QPCh. 16 - Prob. 16.21QPCh. 16 - Prob. 16.22QPCh. 16 - Prob. 16.23QPCh. 16 - Prob. 16.24QPCh. 16 - Prob. 16.25QPCh. 16 - Prob. 16.26QPCh. 16 - Prob. 16.27QPCh. 16 - Prob. 16.28QPCh. 16 - Prob. 16.29QPCh. 16 - Prob. 16.30QPCh. 16 - Prob. 16.31QPCh. 16 - Prob. 16.32QPCh. 16 - Prob. 16.33QPCh. 16 - Prob. 16.34QPCh. 16 - Prob. 16.35QPCh. 16 - Prob. 16.36QPCh. 16 - Prob. 16.37QPCh. 16 - Prob. 16.38QPCh. 16 - Prob. 16.39QPCh. 16 - Prob. 16.40QPCh. 16 - Prob. 16.41QPCh. 16 - Prob. 16.42QPCh. 16 - Prob. 16.43QPCh. 16 - Prob. 16.44QPCh. 16 - Prob. 16.45QPCh. 16 - Prob. 16.46QPCh. 16 - Prob. 16.47QPCh. 16 - Prob. 16.48QPCh. 16 - Prob. 16.49QPCh. 16 - Prob. 16.50QPCh. 16 - Prob. 16.51QPCh. 16 - Prob. 16.52QPCh. 16 - Prob. 16.53QPCh. 16 - Prob. 16.54QPCh. 16 - Prob. 16.55QPCh. 16 - Prob. 16.56QPCh. 16 - Prob. 16.57QPCh. 16 - Prob. 16.58QPCh. 16 - Prob. 16.59QPCh. 16 - Prob. 16.60QPCh. 16 - Prob. 16.61QPCh. 16 - Prob. 16.62QPCh. 16 - Prob. 16.63QPCh. 16 - Prob. 16.64QPCh. 16 - Prob. 16.65QPCh. 16 - Prob. 16.66QPCh. 16 - Prob. 16.67QPCh. 16 - Prob. 16.68QPCh. 16 - Prob. 16.69QPCh. 16 - Prob. 16.70QPCh. 16 - Prob. 16.71QPCh. 16 - Prob. 16.72QPCh. 16 - Prob. 16.73QPCh. 16 - Prob. 16.74QPCh. 16 - Prob. 16.75QPCh. 16 - Prob. 16.76QPCh. 16 - Prob. 16.77QPCh. 16 - Prob. 16.78QPCh. 16 - Prob. 16.79QPCh. 16 - Prob. 16.80QPCh. 16 - Prob. 16.81QPCh. 16 - Prob. 16.82QPCh. 16 - Prob. 16.83QPCh. 16 - Prob. 16.84QPCh. 16 - Prob. 16.85QPCh. 16 - Prob. 16.86QPCh. 16 - Prob. 16.87QPCh. 16 - Prob. 16.88QPCh. 16 - Prob. 16.89QPCh. 16 - Prob. 16.90QPCh. 16 - Prob. 16.91QPCh. 16 - Prob. 16.92QPCh. 16 - Prob. 16.93QPCh. 16 - Prob. 16.94QPCh. 16 - Prob. 16.95QPCh. 16 - Prob. 16.96QPCh. 16 - Prob. 16.97QPCh. 16 - Prob. 16.98QPCh. 16 - Prob. 16.99QPCh. 16 - Prob. 16.100QPCh. 16 - Prob. 16.101QPCh. 16 - Prob. 16.102QPCh. 16 - Prob. 16.103QPCh. 16 - Prob. 16.104QPCh. 16 - Prob. 16.105QPCh. 16 - Prob. 16.106QPCh. 16 - Prob. 16.107QPCh. 16 - Prob. 16.108QPCh. 16 - Prob. 16.109QPCh. 16 - Prob. 16.110QPCh. 16 - Prob. 16.111APCh. 16 - Prob. 16.112APCh. 16 - Prob. 16.113APCh. 16 - Prob. 16.114APCh. 16 - Prob. 16.115APCh. 16 - Prob. 16.116APCh. 16 - Prob. 16.117APCh. 16 - Prob. 16.118AP
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