Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 16, Problem 16.AE
Interpretation Introduction

Interpretation:

The potential at the given volumes of Ce4+ has to be calculated and the titration curve has to be drawn.

Expert Solution & Answer
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Answer to Problem 16.AE

The potential for given volume of Ce4+ are,

At 0.100mL : - 0.161V

At 1.00mL : -0.130 V

At 5.00mL : -0.102 V

At 9.50mL : -0.064 V

At 10.0mL : 0.342 V

At 10.10mL : 1.11 V

At 12.0mL : 1.19 V

The titration curve is,

Quantitative Chemical Analysis, Chapter 16, Problem 16.AE , additional homework tip  1

Figure 1

Explanation of Solution

Given

Amount of solution of Sn2+ = 20.0 mL

Strength of Sn2+ solution = 0.00500 M

Strength of Ce4+ solution = 0.0200 M

To explain: The potential for given volume of Ce4+

Titration:Sn2++2Ce4+Sn4++2Ce3+Ve=10.0mL

At 0.100mL :

Ve=0.100mL/10.00mL

E+= 0.139-0.59162log[Sn2+][Sn4+]= 0.139-0.59162log9.900.100= 0.080VE = E+-E-= 0.080-0.241= - 0.161V

At 10.0mL :

2E+=2(0.139)- 0.05916 log[Sn2+][Sn4+]E+=1.47- 0.05916 log[Ce3+][Ce4+]_3E+=1.748 - 0.05916 log [Sn2+][Ce3+][Sn4+][Ce4+]

At the equivalence point,

[Sn4+] =12[Ce3+] and [Sn2+] = 12[Ce4+] , which forms the log term zero, So 3E+=1.748  and E+= 0.583

E=E+-E-= 0.583 - 0.241= 0.342V

At 10.10 mL : The first 10.10 mL used to making Ce3+ . Next 0.10 mL remains unreacted Ce4+ .

E+= 1.47- 0.5916 log[Ce3+][Ce4+]= 1.47- 0.5916log10.00.10= 1.352VE = E+- E-= 1.352 - 0.241= - 0.11V

To write: titration curve

mL                        E (V)
0.100                 -0.161
1.00                   -0.130
5.00                   -0.102
9.50                   -0.064
10.00                0.342
10.10                 1.11
12.00                 1.19

Quantitative Chemical Analysis, Chapter 16, Problem 16.AE , additional homework tip  2

Figure 1

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