A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- Nickel ion required 11.70 mL of 0.002146 M EDTA What is the percentage of NaBr (102.894) in the 1.000 g sample?

A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- Nickel ion required 11.70 mL of 0.002146 M EDTA What is the percentage of NaBr (102.894) in the 1.000 g sample?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter17: Complexation And Precipitation Reactions And Titrations

Section: Chapter Questions

Problem 17.34QAP

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A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA. Calculate mmol Ni in the 50.00 mL aliquot. Calculate mmol Br- in the 50.00 mL aliquot. Calculate the percentage of NaBr (102.894) in the 1.000 g sample.
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