Calculate the number of milliliters of 0.779 M KOH required to precipitate all of the Zn2+ ions in 150 mL of 0.571 M ZnCl2 solution as Zn(OH)2. The equation for the reaction is: ZnCl2(aq) + 2KOH(aq) -> Zn(OH)2(s) + 2KCl(aq)

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.98QE
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Calculate the number of milliliters of 0.779 M KOH required to precipitate all of the Zn2+ ions in 150 mL of 0.571 M ZnCl2 solution as Zn(OH)2. The equation for the reaction is:

ZnCl2(aq) + 2KOH(aq) -> Zn(OH)2(s) + 2KCl(aq)

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