6 . A 100.0-mL sample of spring water was treated to convert any iron present to Fe*.Titration with 25.00-mL of 0.002497 M K2Cr2O7 resulted in the reaction6FE2+ + Cr0, 2- + 14H* → 6FE3* 1 2Cr** + 7H2OThe excess K2Cr207 was back-titrated with 7.45 mL of 0.00953 M Fe2* solution.Calculate the concentration of iron in the sample in parts per million.

Answered: Image /qna-images/answer/dd00c912-3cb0-482d-aa53-4fcdc796c1f4.jpg

A 100.0-mL sample of spring water was treated to convert any iron present to Fe. Titration with 25.00-mL of 0.002497 M K2C12O7 resulted in the reaction 6Fe+ + Cr20, 2- + 14H* → 6Fe* 1 2Cr** + 7H2O The excess K2Cr207 was back-titrated with 7.45 mL of 0.00953 M Fe2* solution. Calculate the concentration of iron in the sample in parts per million.

A 100.0-mL sample of spring water was treated to convert any iron present to Fe. Titration with 25.00-mL of 0.002497 M K2C12O7 resulted in the reaction 6Fe+ + Cr20, 2- + 14H* → 6Fe* 1 2Cr** + 7H2O The excess K2Cr207 was back-titrated with 7.45 mL of 0.00953 M Fe2* solution. Calculate the concentration of iron in the sample in parts per million.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter13: Titrations In Analytical Chemistry

Section: Chapter Questions

Problem 13.27QAP

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