4. The cadmium and lead ions in a 50.00 mL sample required 40.09 mL of a 0.005000 M EDTA for titration. A 75.00 mL portion of the same sample was made basic and treated with excess KCN masking the cadmium as Cd(CN),2. This solution required 21.42 mL of the EDTA for titration. Calculate the concentration of Cd²* and Pb²* in the sample ppm and in M.

4. The cadmium and lead ions in a 50.00 mL sample required 40.09 mL of a 0.005000 M EDTA for titration. A 75.00 mL portion of the same sample was made basic and treated with excess KCN masking the cadmium as Cd(CN),2. This solution required 21.42 mL of the EDTA for titration. Calculate the concentration of Cd²* and Pb²* in the sample ppm and in M.

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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Similar questions

To analyze the amount of iron (Fe; Mw = 55.85 g/mol) contained in an ore sample, the sample was digested with acid and diluted to 50 mL with water. This solution was then treated with 25.00 mL of 0.2922 M EDTA. The excess EDTA was back titrated with 6.47 mL of 0.0843 M Zn2+ to reach the equivalence point. How many grams of Fe contained in the ore sample?
A 0.7352g sample of ore containing Fe3+, Al3+ and Sr2+ was dissolved and made up to 500.00 mL. The analysis of metals was performed by a chemistry using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard solution of EDTA 0.02145 mol/L, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Determine the percentage of each of the metals in the sample Given the molar masses: Fe=55.845 g/mol; Al=26.982 g/mol and Sr=87.620 g/mol.
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