To 200.0 ml of a solution of Al(NO3)3 with concentration (2.000x10^-3) mol L-1, (3.0x10^2) ml of (1.00x10^-1) mol L-1 NaF is added. Determine the concentration of free aluminum ions after equilibrium has been established.(K+(AIF63-) = 7.0 x 1019) Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer units

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter15: Complex Ion And Precipitation Equilibria
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To 200.0 ml of a solution of Al(NO3)3 with concentration (2.000x10^-3) mol L-1,
(3.0x10^2) ml of (1.00x10^-1) mol L-1 NaF is added. Determine the concentration of
free aluminum ions after equilibrium has been established.(K+(AIF63-) = 7.0 x 1019)
Note: Your answer is assumed to be reduced to the highest power possible.
Your Answer:
x10
Answer
units
Transcribed Image Text:To 200.0 ml of a solution of Al(NO3)3 with concentration (2.000x10^-3) mol L-1, (3.0x10^2) ml of (1.00x10^-1) mol L-1 NaF is added. Determine the concentration of free aluminum ions after equilibrium has been established.(K+(AIF63-) = 7.0 x 1019) Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: x10 Answer units
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