Solving the Mystery of AlCl3 & NaOH/NH4OH Precipitation

[ChemDunce]

Hey Chem geniuses!

I am a struggling high school chem student who couldn't combine water with water without blowing something up so some help would definitley be appreciated.

My chem final has been issued already and it is as follows:

Put 5 mL of 3 M aluminum chloride solution in each of two test tubes. To one test tube add a few drops of 3 M sodium hydroxide solution. Then add more sodium hydroxide solution, with stirring, until the precipitate dissolves. Treat the other test tube similarly, but use 3 M ammonum hydroxide solution to obtain precipitation and then add more ammonium hydroxide solution. Why did the precipitate dissolve in one case and not the other?


I calculated the amounted needed to be 2.00 grams of aluminum chloride. Since I need to put that amount in TWO test tubes, I requested 4.00 grams for my experiment. The 3M of Sodium Hydroxide and 3 M of Ammonium Hydroxide were requested in 2.5 mL each. If anyone sees error in this, please let me know! Thanks


The question remains... why did one case precipitate and not in the other?

Thanks a bunch.

 

[ferrios25]

complexing ions...

Hi ChemDunce,

The reason that the precipitate dissolves in one of the solutions is because the equilibrium moves to the right until Al(OH)3(s) is precipitated; this easily reacts with excess NaOH to give the aluminate ion, Al(OH)4-(aq). Basically, the Al(OH)3 is complexed to create an ion that dissolves in solution.

The ammonium hydroxide solution is unable to complex the Al(OH)3 precipitate and, therefore, remains as a precipitate. Hope this information helps. Good luck on your final!

Fernando

 

[ChemDunce]

Thanks a bunch! Now, for the chemical equations, the AlCl3+ NH4OH is not a double replacement, then what would it be? While the AlCl3+NaOH ---> Al(OH)4-?
Are there any practical uses for this reaction in everyday life?