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 11.17P
Interpretation Introduction

Interpretation:

The pH values of the reaction medium during titration of 0.0319 M benzylamine- 0.0500 M HCl system has to be calculated for the given volumes of HCl .

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 11.17P

The pH values of the reaction medium during titration of 0.0100 M NaOH- 0.100 M HCl system are calculated for the given volumes of HCl as,

S.NoVolume of HCl , Va (mL) pH
1. 0.00 10.92
2. 12.0 9.57
3. 1/2 Ve (half the volume at equivalence point - 15.95 ) 9.35
4. 30.0 8.15
5. Ve (31.9) 5.53
6. 35.0 2.74

Explanation of Solution

Given that volume of acid, HCl and it is denoted by Va. Given the strength ( M1 ) and volume of benzylamine solution ( V1 ) and strength of HCl ( M2 ), the volume of HCl at equivalence point ( Ve )is calculated as,

V1M1 = VeM250 mL× 0.0319 M =  Ve×0.0500MVe =  50 mL× 0.0319 M0.0500M =  31.9 mL

Proton from HCl protonates the base benzylamine (B) and forms conjugate acid (BH+).

But when no acid is added (Va is 0.00mL ,), the base initially reacts with water in the reaction medium.

The equation for this reaction is written as,

B+H2OBH++OH

Thus, [BH+]=[OH]=x,  [B]=0.0319x

Dissociation constant Kb for the above reaction is known to be 2.2×105 and hence,

Kb = [[BH+][OH][B]]=2.2×1052.2×105 =  x20.0319x

Solving for ‘x’

x=[OH]=8.31×104M

Determining pH ,

pH+pOH =14pOH =log[OH] =log(8.31×104)=3.08

Then as we know,

pH+pOH =14pH + 3.08 =14pH =14-3.08 = 10.92

When volume of acid added is 12.0mL ,

Reaction:         B    +        H+                 BH+_________________________________________Initial conc.    31.9            12.0                           0Final conc.     31.9-12.0      0                            12.0

pKa for benzylamine is known to be 9.35. Substitute the values known to determine pH.

pH  =  pKalog[B][BH+] = 9.35+log[19.912.0] =   9.57

When volume of acid added is half the volume of equivalence point, 1/2 Ve=31.9mL2=15.95mL,

Reaction:         B    +        H+                 BH+_________________________________________Initial conc.    31.9            15.95                           0Final conc.     31.9-15.95      0                            15.95

pKa for benzylamine is known to be 9.35. Substitute the values known to determine pH.

pH  =  pKalog[B][BH+] = 9.35+log[15.9515.95] =   9.35+0 = 9.35

When volume of acid added is 30.0mL ,

Reaction:         B    +        H+                 BH+_________________________________________Initial conc.    31.9            30.0                           0Final conc.     31.9-30.0      0                            30.0

pKa for benzylamine is known to be 9.35. Substitute the values known to determine pH ,

pH  =  pKalog[B][BH+] = 9.35+log[1.930] =   9.35-1.20 = 8.15

When volume of acid added is 31.9mL which is the volume of acid at equivalence point of the titration, at this stage the base benzylamine is completely converted to its conjugate acid BH+.  Concentration of BH+ at equivalence point is,

50.0mL50.0+31.9mL×0.0319=0.0195M

The equation at this stage is written as,

BH+B+H+

Thus, [H+]=[B]=x,  [BH+]=0.0195x

Dissociation constant Ka for the above reaction is known to be 4.7×1010 and hence,

Ka = [[H+][B][BH+]]=4.7×10104.7×1010 =  x20.0195x

Solving for ‘x’

x=[H+]=2.96×106M

Determining pH ,

pH = -log[H+] = -log(2.96×106M) =   5.53

35.0 mL of HCl is beyond the equivalence point which is 31.9 mL. thus volume of HCl at this stage is 35.0 mL-31.9 mL = 3.1 mL. pH of the solution has to be determined not exactly at equivalence point but after reaching the equivalence point – that is post-equivalence region.  At this stage the concentration of base is nil.  pH is determined by measuring the concentration of excess acid added beyond the equivalence point. Therefore,

[H]+ = (3.1mL50.0mL+35.0mL)×0.0500M = 0.00182 M

Determining pH ,

pH =log[H+] =log(0.00182) = 2.74

Conclusion

The pH values of the reaction medium during titration of 0.0319 M benzylamine- 0.0500 M HCl system was calculated using relevant formula.

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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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