Solution Manual for Quantitative Chemical Analysis
Solution Manual for Quantitative Chemical Analysis
9th Edition
ISBN: 9781464175633
Author: Daniel Harris
Publisher: Palgrave Macmillan Higher Ed
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Chapter 11, Problem 11.25P
Interpretation Introduction

Interpretation:

The pH at 10mL intervals( from 0mL to 100 mL ) in titration of 40.0mL of piperazine with 0.100M HCl has to be calculated and a graph of pH vs Va has to be drawn where Va is volume of acid added.

Concept introduction:

Acid-base titration is titration between acid and base. It is also known as neutralization reaction.

There are primarily four types of acid-base titrations –

  • Strong base vs strong acid
  • Strong base vs weak acid
  • Weak base vs strong acid
  • Weak base vs weak acid

Equivalence point ensures the titration reaction becomes completed and at this point the number of moles of titrant and the number of moles of analyte becomes equal.

Expert Solution & Answer
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Explanation of Solution

Piperazine is a weak base and HCl is a strong acid.  So the titration of piperazine with HCl is titration of weak base with strong acid.  The reaction of piperazine with HCl is written as,

Solution Manual for Quantitative Chemical Analysis, Chapter 11, Problem 11.25P , additional homework tip  1

Piperazine has two N-atoms and each N-atom has a pair of non-bonding electrons available for bonding which makes the molecule basic. The protons from HCl protonates piperazine.  The pH of the titration varies at each different stage of titration depending upon the volume of the base and acid present.  Let ‘B’ be the term that denotes base here.

Given data:

volume of piperazine = 40.0 mLMolarity of piperazine = 0.100MMolarity of HCl = 0.100M

When volume of acid added Va is 0mL, only base is present in aqueous medium.  The base gets hydrolyzed by water as,

B+H2OBH+OH

Then, [BH+]=[OH]=x, [B]=100x 

The base has two Ka values as there are two basic N-atoms.  One of the Ka value is 1.86×1010 and another one is 4.65×106 .  The equation of Kb for the above reaction is,

Kb = x20.100xKwKa2 = 1.0×10141.86×1010=x20.100x

Solving for ‘x’,

x=2.29×103M=[OH]

Therefore pH is calculated as,

pH = -log(Kw[OH]) =  -log(1×10142.29×103M)=11.36

Va=10.0mL

10mL of acid consumes 10mL of base and so volume of base remaining unreacted is 30mL.

Therefore pH is calculated as,

pH = pK2+log([B][BH+]) =  9.73+log(3010)=10.21

Va=20.0mL

20mL of acid consumes 20mL of base and so volume of base remaining unreacted is 20mL.

Therefore pH is calculated as,

pH = pK2+log([B][BH+]) =  9.73+log(2020) =   9.73+0 = 9.73

Va=30.0mL

30mL of acid consumes 30mL of base and so volume of base remaining unreacted is 10mL.

Therefore pH is calculated as,

pH = pK2+log([B][BH+]) =  9.73+log(3010) =   9.25

Va=40.0mL

40mL of acid consumes 40mL of base and so no volume of base is remaining unreacted.  Thus B is converted to BH+ at a formal concentration of,

F = (40.0mL40.0mL+40.0mL)(0.100M) =  0.0500M

[H+] is calculated as,

[H+] = K1K2F+K1KWK1+F =  (4.65×106×1.86×1010×0.0500M)+(4.65×106×1.0×1014)4.65×106+0.0500M

Solving the above equation for [H+] and calculating pH using the formula pH=log[H+],

pH=7.53

Solution Manual for Quantitative Chemical Analysis, Chapter 11, Problem 11.25P , additional homework tip  2

Va=50.0mL

50mL of acid consumes 20mL of BH+ and volume of BH+ remaining unreacted is 50mL20mL=30mL and so volume of BH22+ formed is 10mL

Therefore pH is calculated as,

pH = pK1+log([BH+][BH22+]) =  5.33+log(3010) =   5.81

Va=60.0mL

60mL of acid consumes 30mL of BH+ and so volume of BH22+ formed is 6030=30mL and volume of BH+ remaining unreacted is 30mL.

Therefore pH is calculated as,

pH = pK1+log([BH+][BH22+]) =  5.33+log(3030) =   5.33

Va=70.0mL

70mL of acid consumes 40mL of BH+ and so volume of BH22+ formed is 7040=30mL and volume of BH+ remaining unreacted is 10mL.

Therefore pH is calculated as,

pH = pK1+log([BH+][BH22+]) =  5.33+log(1030) =  4.86

Va=80.0mL

All of B is converted to BH22+ at formal concentration,

F=(40mL40mL+80mL)(0.100M)=0.0333M

Therefore pH is determined by dissociation of BH22+

BH22+BH+H+

Then, [BH+]=[H+]=x, [BH22+]=0.0333-x

Dissociation constant Ka is,

Ka =4.65×106Ka = x20.0333xx20.0333x = 4.65×106

Solving for ‘x’,

x=3.91×104M=[H+]

Therefore pH is calculated as,

pH = -log[H+] = -log(3.91×104M) =  3.4 

Va=90.0mL

[H+] =(100mL90mL90mL+40mL)×0.100M = 0.0077M

Therefore pH is calculated as,

pH = -log[H+] = -log(0.0077) =  2.11 

Va=100.0mL

[H+] =(100mL80mL100mL+40mL)×0.100M = 0.0143M

Therefore pH is calculated as,

pH = -log[H+] = -log(0.0143) =  1.84 

For each volume of acid added corresponding pH is calculated.  The graph of pH vs Va is plotted as,

Solution Manual for Quantitative Chemical Analysis, Chapter 11, Problem 11.25P , additional homework tip  3

Figure 1

Conclusion

The pH at 10mL intervals( from 0mL to 100 mL ) in titration of 40.0mL of piperazine with 0.100M HCl was calculated and a graph of pH vs Va was drawn.

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Chapter 11 Solutions

Solution Manual for Quantitative Chemical Analysis

Ch. 11 - Prob. 11.HECh. 11 - Prob. 11.IECh. 11 - Prob. 11.JECh. 11 - Prob. 11.KECh. 11 - Prob. 11.1PCh. 11 - Prob. 11.2PCh. 11 - Prob. 11.3PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Prob. 11.6PCh. 11 - Prob. 11.7PCh. 11 - Prob. 11.8PCh. 11 - Prob. 11.9PCh. 11 - Prob. 11.10PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Prob. 11.13PCh. 11 - Prob. 11.14PCh. 11 - Prob. 11.15PCh. 11 - Prob. 11.16PCh. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Prob. 11.19PCh. 11 - Prob. 11.20PCh. 11 - Prob. 11.21PCh. 11 - Prob. 11.22PCh. 11 - Prob. 11.23PCh. 11 - Prob. 11.24PCh. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Prob. 11.27PCh. 11 - Prob. 11.28PCh. 11 - Prob. 11.29PCh. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - Prob. 11.32PCh. 11 - Prob. 11.33PCh. 11 - Prob. 11.34PCh. 11 - Prob. 11.35PCh. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Prob. 11.40PCh. 11 - Prob. 11.41PCh. 11 - Prob. 11.42PCh. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Prob. 11.47PCh. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Prob. 11.53PCh. 11 - Prob. 11.54PCh. 11 - Prob. 11.55PCh. 11 - Prob. 11.56PCh. 11 - Prob. 11.57PCh. 11 - Prob. 11.58PCh. 11 - Prob. 11.60PCh. 11 - Prob. 11.61PCh. 11 - Prob. 11.62PCh. 11 - Prob. 11.63PCh. 11 - Prob. 11.64PCh. 11 - Prob. 11.65PCh. 11 - Prob. 11.66PCh. 11 - Prob. 11.67PCh. 11 - Prob. 11.68PCh. 11 - Prob. 11.69PCh. 11 - Prob. 11.70PCh. 11 - Prob. 11.71PCh. 11 - Prob. 11.72PCh. 11 - Prob. 11.73PCh. 11 - Prob. 11.74PCh. 11 - Prob. 11.75P
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