化学题解答
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发布时间：2022-04-23 14:02
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热心网友
时间：2023-09-27 23:06
Sorry for I can't type in Chinese on my Mac computer.
Analysis:
This is an example of back titration. First, 50.00 mL of 0.05??? M EDTA was added to complex all Zn2+ ions and Al3+ ions, and the excess EDTA was then titrated with 5.08 mL of 0.05000 M standard Zn2+ solution.
Addition of excess F- is to complex Al3+ and to release the amount of EDTA which was complexed with Al3+ previously. This amount of EDTA was then titrated by the standard Zn2+ solution.
Solution:
mmoles of (Zn2+ + Al3+) = (50.00 mL x 0.05??? M)- (5.08 mL x 0.05000 M) = 2.246 mmoles
mmoles of Al3+ = 20.70 mL x 0.05000 M = 1.037 mmol
so mmoles of Zn2+ = 2.246 - 1.037 = 1.209 mmol
So W(Zn) = (1.209 x 65.38 )/1000 = 0.07904 g
W(Al) = (1.037 x 26.98)/1000 = 0.02798 g
