0.8153 g of a sample containing Pb(NO3)2 was taken, dissolved in water, and 40.20 mL of 0.06 M EDTA was added. If the excess EDTA was back-titrated with 23.10 mL of 0.02 M EDTA, which of the following is the percentage of Pb(NO3)2? (Pb(NO3)2: 331 g/mol, Pb: 207 g/mol) A. 85,23 B. 38.25 C. 35.63 D. 79.17

0.8153 g of a sample containing Pb(NO3)2 was taken, dissolved in water, and 40.20 mL of 0.06 M EDTA was added. If the excess EDTA was back-titrated with 23.10 mL of 0.02 M EDTA, which of the following is the percentage of Pb(NO3)2? (Pb(NO3)2: 331 g/mol, Pb: 207 g/mol) A. 85,23 B. 38.25 C. 35.63 D. 79.17

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter19: Complex Ions

Section: Chapter Questions

Problem 51QAP: A child eats 10.0 g of paint containing 5.0% Pb. How many grams of the sodium salt of EDTA,...

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0.8153 g of a sample containing Pb(NO3)2 was taken, dissolved in water, and 40.20 mL of 0.06 M EDTA was added. What is the percentage of Pb(NO3)2 since the excess EDTA is back-titrated with 23.10 mL of 0.02 M EDTA? (Pb(NO3)2: 331 g/mol, Pb: 207 g/mol)
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