CA 0.4569 G s C Chegg s University of Cincinnati CX Guided S /ibiscms/mod/ibis/v

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CA 0.4569 G s C Chegg s University of Cincinnati CX Guided S /ibiscms/mod/ibis/view.php?id-2366010 WWW sapling learning.com Apps For q book ks b rt book 3/16/2016 11:55 PM A 49.3/100 3/15/2016 03:02 PM Gradebook Attempts Score Periodic Table Print Calculator 20 uestion 7 of 7 Map 00 A 0.4383 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as Sno2 4H20 and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 250.0 mL. A 15.000 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.40 m of 0.001475 M EDTA. Thiosulfate was used to mask the copper in a second 20.00 mL aliquot. Titration of the lead and zinc in this aliquot required 34.25 mL of the EDTA solution 00 Finally, cyanide was used to mask the copper and the zinc in a third 25.00 mL aliquot. Titration of the lead in this aliquot required 25.25 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample. 00 25 4.9 Cu 2.9 Zn 14.65 77.55 Sn Pb Incorrect. Previous 3 Give Up & View solution Try Again Next SExit Explanation 7:05 PM 3/16/2016 Assignment Information Available From 3/9/2016 07:00 PM 3/16/2016 11:55 PM Due Date: Points Possible 100 Grade Category: Default Description: Policies: Homework heck y You can view solutions when vou complete or give up on any question You can keep trying to answer each question until you get it right or give up You lose 5% of the points available to each answer in your question for each incorrect attempt at that eTextbook O Help With This Topic O Web Help & Videos Technical Support and Bug Reports

Explanation / Answer

Original mass of mixture (Sn, Pb, Cu, Zn) = 0.4383 g

15 ml of aliquot needed = 0.001475 M x 35.40 ml = 0.0522 mmol EDTA

moles of Pb+Cu+Zn present = 0.0522 mmol in 15 ml

So in 250 ml = 0.87 mmol of Pb+Cu+Zn present

20 ml of aliquot (Pb+Zn) required = 0.001475 M x 34.25 ml = 0.050 mmol EDTA

moles of Pb+Zn present = 0.050 mmol in 20 ml

So in 250 ml = 0.625 mmol of Pb+Zn present

Amount of Cu present in original mixture = (0.87 - 0.625)mmol x 63.546 g/mol/1000 = 0.0156 g

% Cu in original mixture = 0.0156 x 100/0.4383 = 3.56%

25 ml of aliquot (Pb) needed = 0.001475 M x 25.25 ml = 0.037 mmol EDTA

moles of Pb present = 0.037 mmol in 25 ml

So in 250 ml = 0.37 mmol Pb present

Amount of Pb present in original mixture = 0.37 mmol x 207.2 g/mol/1000 = 0.077 g

%Pb in original mixture = 0.077 x 100/0.4383 = 17.57%

moles of Zn present = 0.625 - 0.37 = 0.255 mmol in 250 ml

Amount of Zn present in original mixture = 0.255 mmol x 65.38 g/mol/1000 = 0.017 g

%Zn in original mixture = 0.017 x 100/0.4383 = 3.88%

Amount of Sn in original mixture = 0.4343 - (0.0156+0.077+0.017) = 0.3247 g

%Sn in original mixture = 0.3247 x 100/0.4383 = 74.76%

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