A sample of Al2(SO4); weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-mL of buffer solution, 50-mL of alcohol and 2.0-ml dithizone TS were added. The resulting solution required B.7-mL of 0.031M ZnSO, to reach the endpoint. (Note: Each ml of 0.050M EDTA is equivalent to 16.66-mg Al2(SO.)h•18H;0). Compute for the percent purity of the sample.
A sample of Al2(SO4); weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL. 100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-mL of buffer solution, 50-mL of alcohol and 2.0-ml dithizone TS were added. The resulting solution required B.7-mL of 0.031M ZnSO, to reach the endpoint. (Note: Each ml of 0.050M EDTA is equivalent to 16.66-mg Al2(SO.)h•18H;0). Compute for the percent purity of the sample.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:Answer provided 2.5% Al2(SO4)3
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Transcribed Image Text:A sample of Al;(SO.); weighing 7.52-g was dissolved in enough water and diluted to 250.0-mL.
100.0-mL of the dilution was transferred to another flask, 10.0-mL of 0.050M EDTA, 20-ml of buffer
solution, 50-mL of alcohol and 2.0-ml dithizone TS were added. The resulting solution required
8.7-mL of 0.031M ZnSO, to reach the endpoint. (Note: Each ml of 0.050M EDTA is equivalent to
16.66-mg Al:(SO.);•18H;0). Compute for the percent purity of the sample.
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