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Because NH3 reacts with water to form NH4+ and OH-, so in weak ammonia solutions, you will have the same precipitation of M(OH)z as you would in weakly basic solutions. As the concentration of NH3 goes up, the amount of unreacted NH3 increases as well, and NH3 is a great ligand for many transition metals, so you will actually see the solids formed redissolve, as they form complex ions, usually of form M(NH3)6 (x+).

The same effect can happen in highly basic solutions, as OH- can also act as a ligand to form M(OH)6 (y-).

2006-10-03 17:44:25 · answer #1 · answered by TheOnlyBeldin 7 · 0 0

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