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Give the rxns for: Zn^2+ + OH^- ----> and Zn^2+ + OH^- (excess) ----> What...
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Zn^2+ + OH^- ----> Zn(OH)2 (s)
Zn^2+ + OH^- (excess) ----> Zn(OH)4 2˄-
The second reaction may be depicted in a stepwise manner as a result of two successive syeps: Zn^2+ + OH^- ----> Zn(OH)2 (s) --- (i)
Zn(OH)2 + 2OH^- ----> Zn(OH)4˄2- (aq.) --- (ii)
When hydroxide ions are added to a solution of Zn˄2+ ions, white precipitate of Zn(OH)2 is produced. This hydroxide is soluble in acids. Upon addition of excess of alkali (at hydroxide concentration of 0.1 to 1.0M), the precipitate goes into solution, through the formation of a soluble zincate complex, Zn(OH)4 2˄-. The colorless complex ion is tetrahedral in geometry and structurally similar to other ions of zinc e.g. Zn[NH3]4 or Zn[CN]4˄2-, with hydroxide ions in place of NH3 or CN- ions.
Posted by llltkl on May 1, 2013 at 6:14 AM (Answer #1)
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