Calculate the concentration of free Zn2+ ion at equilibrium when 1.58x10-2 mol zinc sulfate is added to 1.00 L of solution that is buffered at pH 12.30. For Zn(OH)4²", Ky = 4.6x1017, [Zn2+] = M

Chemistry: Principles and Reactions
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Author:William L. Masterton, Cecile N. Hurley
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Chapter14: Equilibria In Acid-base Solutions
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Calculate the concentration of free Zn2+ ion at equilibrium when 1.58x10-2 mol zinc sulfate is added to 1.00 L of
solution that is buffered at pH 12.30. For Zn(OH),2, K, 4.6x1017.
[Zn2+] =
M.
Transcribed Image Text:Calculate the concentration of free Zn2+ ion at equilibrium when 1.58x10-2 mol zinc sulfate is added to 1.00 L of solution that is buffered at pH 12.30. For Zn(OH),2, K, 4.6x1017. [Zn2+] = M.
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