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Derive a titration curve for 50 cm³ of 0.0250 M U4+ with 0.100 M Ce+. The solution is 1.0 M H2SO4 throughout the titration (assume H' concentration for this solution is 1.0 M). U+ + 2Ce+ + 2H2O UO22+ + 2Ce3++ 4H. After addition of the following volumes: (i) 20% of equivalence point; (ii) at equivalence point; (iii) at 100.4% of equivalence point. UO₂+ + 4H* + 2e - U4+ E=0.334 V Ce4+ + e Ce3+ E = 1.44 V (15)

Derive a titration curve for 50 cm³ of 0.0250 M U4+ with 0.100 M Ce+. The solution is 1.0 M H2SO4 throughout the titration (assume H' concentration for this solution is 1.0 M). U+ + 2Ce+ + 2H2O UO22+ + 2Ce3++ 4H. After addition of the following volumes: (i) 20% of equivalence point; (ii) at equivalence point; (iii) at 100.4% of equivalence point. UO₂+ + 4H* + 2e - U4+ E=0.334 V Ce4+ + e Ce3+ E = 1.44 V (15)

Fundamentals Of Analytical Chemistry
Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter19: Applications Of Standard Electrode Potentials
Section: Chapter Questions
Problem 19.13QAP
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Derive a titration curve for 50 cm³ of 0.0250 M U4+ with 0.100 M Ce+. The solution is 1.0 M H2SO4 throughout the titration (assume H' concentration for this solution is 1.0 M). U+ + 2Ce+ + 2H2O UO22+ + 2Ce3++ 4H. After addition of the following volumes: (i) 20% of equivalence point; (ii) at equivalence point; (iii) at 100.4% of equivalence point. UO₂+ + 4H* + 2e - U4+ E=0.334 V Ce4+ + e Ce3+ E = 1.44 V (15)
Transcribed Image Text:Derive a titration curve for 50 cm³ of 0.0250 M U4+ with 0.100 M Ce+. The solution is 1.0 M H2SO4 throughout the titration (assume H' concentration for this solution is 1.0 M). U+ + 2Ce+ + 2H2O UO22+ + 2Ce3++ 4H. After addition of the following volumes: (i) 20% of equivalence point; (ii) at equivalence point; (iii) at 100.4% of equivalence point. UO₂+ + 4H* + 2e - U4+ E=0.334 V Ce4+ + e Ce3+ E = 1.44 V (15)
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