3) A redox titration: 160.00 mL of 0.0500 M Sn2+ solution was titrated with 0.200 M Ce according to thereaction below.Buret containing CeCalomelreferenceelectrode.0.00500 M SnMagneticstirring barTitration reaction:2 Ce++ Sn2+-------> 2 Ce3+ + Sn4+EºCee-------> Ce3+Sn+ + 2e------> Sn²+1.470 V0.139 Va) What is V. (the equivalence volume) (of Ce") needed to reach the endpoint?b) Calculate the potential (voltage), when 10.00 mL Ce" is added.c) What is the cell potential at the equivalence point?d) What is the cell potential when 60 mL Ce4+ is added?SCE = +0.241 V

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Answered: 3) A redox titration: 160.00 mL of…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.40QAP

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A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139Va. What is the VE (the equivalence volume ) (of Ce4+) needed to reach the endpoint?
A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139VCalculate the potential (voltage), when 10.00mL of Ce4+ is added. (SCE= +0.241 V)
A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139VWhat is the cell potential at the equivalence point?
A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139V\What is the cell potential when 60mL Ce4+ is added?
+ H (a) CuCN (b) CaCO3 (c) Ag₂CO3 (d) PbBr₂ (e) Y(OH)3 Periodic Table B W Chapter... Supporting Materials Additional Materiala [Ca²+] = 5.2 x 104 M, [CO32- ] = 6.5x 10-6 M * G Please use the values in the resources listed below instead of the textbook values. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated. [Cu] = 1.9x10-10 M, [CN] = 1.9x10-¹0 M [Ag+] = 1.8x10-5 M, [CO32] = 2.7x10-² M [Pb²+] = 1.2x 10-2 M, [Br] = 2.4x10-2 M [Y³+] = 3.5x 10-16 M, [OH-] = 6.6x 10-³ M Constants and Factors №. webassign.net ● F 893 Supplemental Data L C C a Ⓒ + D r
A 0.5131-g sample containing KBr is dissolved in 50 mL distilled water. Titrating with 0.04614 M AgNO3 requires 25.13 mL to reach the Mohr endpoint. A blank titration requires 0.65 mL to reach the same endpoint. Report the % (w/w) KBr in the sample. [Ans. 26.19 % (w /w)]
A 1.000-g sample containing KHC204, H2C204, and impurities required 38.22 mL of 0.1000 M NaOH for acid-base titration. Another 1.000-g sample required 27.02 mLof 0.04000 M KMNO, for redox titration in acid solution, according to the following reaction: Mno, + 5C,0,2- + 16H* 2 2MN2* + 10co, + 8H,0 How many mmols of KHC,0, are there in the sample? 1.351 mmols KHC204 1.582 mmols KHC204 O 1.120 mmols KHC204 1.911 mmols KHC204
A 20-tablet sample of soluble saccharin was treated with 20.00 mL of 0.08181 M AgNO3 . The reaction is : • R-N-Na + Ag+  R-N-Ag(s) + Na+ • After removal of the solid, titration of the filtrate and washings required 2.81 mL of 0.04124 M SCN- . Ag+ + SCN−  AgSCN(s) (Ag+ titrated with SCNrepresents the unreacted Ag+ in the above solution.) • Calculate the average number of milligrams of saccharin (205.17) in each tablet.
A 0.64 g sample containing KCl ( mw = 74.6 ) is dissolved in 50mL of water and titrated to the Ag2CrO4 end point, requiring 26.2 mL of 0.15 M AgNO3 .The %w/w KCl in the sample is ?
e) via:. Lisbeth Myla CHEM 3321 A Back titration - Homework A solution containing Cl- was analyzed as follows in a back titration. A solution containing (excess) Ag + is added, and the AgCl precipitate is filtered and removed. The excess Ag is then titrated with SCN- Ag+ (aq) + Cl(aq) - Ag+ (aq) + SCN (aq) -------> AgCl (s) --------> AgSCN (aq) In one analysis, 25.00 mL of 0.01840 M AgNO3 (aq) was added to 50.00 mL of a solution containing dissolved chloride. The resulting AgCl (s) was filtered and removed. The excess Ag+ (aq) remaining required 14.25 mL of 0.02008 M SCN (aq) to reach the endpoint. What is the original concentration of Cl (aq) in the original 50.00 mL soln?
A 4.912-g sample of a petroleum product was burnedin a tube furnace, and the SO2produced was collectedin 3% H2O2.Reaction:SO2(g)+H2O2→H2SO4A 25.00-mL portion of 0.00873 M NaOH was introducedinto the solution of H2SO4, following whichthe excess base was back-titrated with 15.17 mL of0.01102 M HCl. Calculate the sulfur concentrationin the sample in parts per million.
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