A 25.00 mL sample containing an unknown amount of Al³+; + and Pb2+required 15.54 mL of 0.05734 M EDTA to reach the end point. A 50.00 mL sample of the unknown was then treated with F¯ to mask the Al³+. To the 50.00 mL sample, 25.00 mL of 0.05734 M EDTA was added. The excess EDTA was then titrated with 0.02256 M Mn² +. A total of 25.7 mL was required to reach the methylthymol blue end point. Determine pAl³+ and pPb2+ in the unknown sample. PA1³ +: pPb²+ =
A 25.00 mL sample containing an unknown amount of Al³+; + and Pb2+required 15.54 mL of 0.05734 M EDTA to reach the end point. A 50.00 mL sample of the unknown was then treated with F¯ to mask the Al³+. To the 50.00 mL sample, 25.00 mL of 0.05734 M EDTA was added. The excess EDTA was then titrated with 0.02256 M Mn² +. A total of 25.7 mL was required to reach the methylthymol blue end point. Determine pAl³+ and pPb2+ in the unknown sample. PA1³ +: pPb²+ =
Chapter17: Complexation And Precipitation Reactions And Titrations
Section: Chapter Questions
Problem 17.22QAP
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