Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 11, Problem 11.DE
Interpretation Introduction

Interpretation:

The pH values of the reaction medium during titration of 0.0500 M Malonic acid- 0.100 M NaOH system has to be calculated for the given volumes of NaOH .  A graph of pH versus volume of NaOH added has to be drawn.

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

The term pH refers to concentration of H+ ion in a solution.

Expert Solution & Answer
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Answer to Problem 11.DE

The pH values of the reaction medium during titration of 0.0500 M Malonic acid- 0.100 M NaOH system are calculated for the given volumes of NaOH as,

S.NoVolume of NaOH , Vb (mL) pH
1. 0.0 2.11
2. 8.0 2.52
3. 12.5 2.85
4. 19.3 3.38
5. 25.0 4.28
6. 37.5 5.70
7. 50.0 9.05
8. 56.3 11.77

The graph of pH versus volume of NaOH is plotted as,

Quantitative Chemical Analysis, Chapter 11, Problem 11.DE , additional homework tip  1

Explanation of Solution

Given that volume of base, NaOH and it is denoted by Vb. Given the strength ( M1 ) and volume of Malonic acid HO2CCH2CO2H solution ( V1 ) and strength of KOH ( M2 ), Since malonic acid is a diprotic acid two equivalence points do occur.

The reaction occurring in titration of malonic acid with NaOH is represented as,

HO2CHCH2CO2H+OHO2CHCH2CO2H+H2OO2CHCH2CO2H + OH  O2CHCH2CO2

The volume of NaOH added to reach first equivalence point ( V1 )is calculated as,

V1M1 = V2M250.0 mL× 0.0500 M =  V2×0.100MV2 =  50 mL× 0.0500 M0.100M =  25.0 mL

Thus at first equivalence point 50mL of acid is converted to 50mL of monobasic anion.  Upon continual addition of base another proton left in monobasic anion is abstracted and dibasic anion is formed.  Thus at second equivalence point, 50mL of monobasic anion is converted to 50mL of dibasic anion.

The volume of NaOH added to reach second equivalence point ( V2 )is calculated as,

V1M1 = V2M250.0 mL× 0.0500 M =  V2×0.100MV2 =  50 mL× 0.0500 M0.100M =  25.0 mL

Thus after the addition of 25.0mL of base we further need to add 25.0mL of base to reach the second equivalence point.  Hence totally 25.0mL+25.0mL=50.0mL of base is required to reach second equivalence point from the beginning of the titration.

Let H2M be malonic acid.  When volume of base added is 0.00mL , that is in absence of base, the pH depends upon the dissociation of malonic acid.  Since it is a weak acid it doesn’t dissociate completely into individual ions.  The acid has two Ka values – K1 and K2 since it is diprotic acid.

H2MH++HM

[H+]=[HM]=x, [H2M]=0.0500x and K1 of H2M is 1.42×103

Ka for the above reaction is,

Ka = x20.0500x1.42×103 = x20.0500x

Solving for ‘x’,

x=7.75×103M=[H+]

pH is determined as,

pH  =  - log[H+] = -log(7.75×103) =   2.11

When volume of NaOH added is 8.0mL , the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed increases with addition of base at each step. Using Henderson-Hasselbalch equation pH can be determined.

The titration reaction at this instant is,

H2M+OHHM+H2O

Their respective concentration is,

  H2M           +  OH   HM   +         H2O_______________________________________________________Initial:  25 mL              8 mL                -                        -Final:    25-8 mL          -                    8 mL                   -

Substitute the values known to determine pH ,

pH  =  pK1log[HM][H2M] = 2.847 +log(817) =   2.52

When volume of NaOH added is 12.5mL , the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed increases with addition of base at each step. Using Henderson-Hasselbalch equation pH can be determined.

The titration reaction at this instant is,

H2M+OHHM+H2O

Their respective concentration is,

  H2M           +  OH   HM   +         H2O_______________________________________________________Initial:  25 mL              12.5 mL                -                  -Final:    25-12.5 mL          -                    12.5 mL           -

Substitute the values known to determine pH ,

pH  =  pK1log[HM][H2M] = 2.847 +log(12.512.5) =   2.85

When volume of NaOH added is 19.3mL , the base begins to react with acid.  Upon reaction with base the acid forms conjugate base.  The titrating mixture then becomes buffer.  The concentration of conjugate base formed increases with addition of base at each step. Using Henderson-Hasselbalch equation pH can be determined.

The titration reaction at this instant is,

H2M+OHHM+H2O

Their respective concentration is,

  H2M           +  OH   HM   +         H2O_______________________________________________________Initial:  25 mL              19.3 mL                -                  -Final:    25-19.3 mL          -                    19.3 mL           -

Substitute the values known to determine pH ,

pH  =  pK1log[HM][H2M] = 2.847 +log(19.35.7) =  3.38

When volume of NaOH added is 25.0mL , 25.0mL of the acid has been converted to its conjugate base, monobasic anion as the first equivalence point is reached.

formal concentration of malonic acid at this instant is,

F = (50mL50mL+25mL)×0.0500M=0.0333M

Hence [H+] is determined as,

[H+] = K1+K2F+K1KwK1+F =  (1.42×103)+(2.01×106×0.0333)+(1.42×103×1×1014)(1.42×103)+0.0333

Solving the above equation,

[H+]=5.23×105M

Substitute the values known to determine pH ,

pH  =  -log[H+] = -log(5.23×105) =  4.28

After reaching the first equivalence point, continual addition of base forms dibasic anion.

When volume of base added is 37.5mL ,

Volume of base is equivalent to, Vb=32Ve

Therefore,

pH  =  pK2log[HM][H2M] = 5.70 +log(37.537.5) =  5.70

When volume of base added is 50mL , second equivalence point is reached as calculated previously.  All of the monobasic anion, HM- is completely converted to dibasic anion, M2-.  Hydrolysis of dibasic anion determines the pH. The reaction is

M2+H2OHM+OH

Formal concentration of malonic acid is,

(50mL50mL+50mL)×0.0500M=0.025M

Then [HM]=[OH]=x, [M2]=0.025x

Kb of malonic acid is,

Kb = KwKa2 = 10142.01×106

Also,

Kb =  x20.0250x10142.01×106 =  x20.0250x

Solving the above equation for ‘x’,

x=[OH]=1.12×105M

Substitute the values known to determine pH ,

pH+pOH = 14pH = 14 - pOH =  14-(-log [OH]) =  14-(-log(1.12×105))  =   14 - 4.95 = 9.05

When volume of acid added is 56.3mL , concentration of [OH] by excess addition of sodium hydroxide determines the pH.

[OH] =(56.350 mL106.3 mL)×0.100M = 5.93×103M

Substitute the values known to determine pH ,

pH+pOH = 14pH = 14 - pOH =  14-(-log [OH]) =  14-(-log(5.93×103))  =   14 - 2.23 = 11.77

The graph of pH versus volume of NaOH is plotted as,

Quantitative Chemical Analysis, Chapter 11, Problem 11.DE , additional homework tip  2

Conclusion

The pH values of the reaction medium during titration of 0.0500 M Malonic acid- 0.100 M NaOH system was calculated using relevant formula and the graph of pH versus volume of NaOH was drawn.

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Chapter 11 Solutions

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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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