A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 19.13 mL of 0.04842 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 0.04842 M EDTA was added. The excess EDTA was then titrated with 0.02147 M Mn2+. A total of 21.1 mL was required to reach the methylthymol blue end point. Determine pAl3+ and pPb2+ in the unknown sample. PA3+ pPb2+
A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 19.13 mL of 0.04842 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 0.04842 M EDTA was added. The excess EDTA was then titrated with 0.02147 M Mn2+. A total of 21.1 mL was required to reach the methylthymol blue end point. Determine pAl3+ and pPb2+ in the unknown sample. PA3+ pPb2+
Chapter17: Complexation And Precipitation Reactions And Titrations
Section: Chapter Questions
Problem 17.3QAP
Related questions
Question
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 6 steps with 6 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you