Re: Reaction of the magnesium ion with ammonium hydroxide

Dear Tom
This behavior, the failure of Mg[OH2 to ppt. in the presence of excess 
ammonium salts, is commonly used in some qual. analysis schemes to separate 
Mg from Ca, Sr, and Ba. It is used here to separate Mg from Al and Zn. 

[1]Al and Zn are amphoteric, while Mg is not. Also, Zn readily forms good 
stable complexes with NH3, while Al and Mg do not[Mg forms a weak complex].

[2]When NH4OH is first added, all metals ppt. as their hydroxide. The pH at 
which they precipitate is about 4 for Al, 6.8 for Zn, and about 10 for Mg 
[Look up the different Ksp for these 3 hydroxides]. 

[3]When more NH4OH is added, the colorless Zn tetra-ammine complex is 
formed which dissolves, but Al and Mg do nt form strong ammine complexes so 
they remain as the insoluble hydroxide.

[4]When acidified with HNO3, the Al and Mg hydroxides are dissolved as the 
H of the acid reacts with the OH of the precipitates and the metal ions are 
in solution as nitrates. The Zn ammine complex is also destroyed and the Zn 
is also in solution. 

[5]Now in the solution there are the Zn, Mg, and Al cations. and also 
much ammonium nitrate and some nitric acid. 

[6] Now NH4OH is added in excess to neutralize the nitric acid - now there 
is a lot of ammonium nitrate in the solution.  The Al and Zn precipitate  
again as they should, but the Mg does not.

[7] The answer is in the large anmount of ammonium nitrate in the solution. 
 This is a salt of a weak base and a strong acid [pure ammonium nitrate in 
solution has a pH of about 5.4] So the solution is now strongly buffered, 
look at the pHs of the hydroxides in [2] above. Maybe if you made the 
solution VERY basic with [say] NaOH, the Mg[OH] would ppt. again.

[8] When  Mg[OH]2 is contacted by an ammonium salt solution,[say, NH4Cl]the 
NH4 ions react with the OH to form the poorly ionized NH4OH [there are 
excess NH4 ions present which further supress the ionization of the NH4OH  
by the common ion effect]. [P.S.,  I show NH4CL here, since it is easier to 
type than the nitrate,  but any NH4 salt would do.] 

[9]The reaction is then:  Mg[OH]2  +  2 NH4Cl  =  MgCl2 + 2NH4OH

Here is a calculation to show this:
  A solution contains 0.21 M MgCL2,  2.0 M NH4CL, AND 0.30 M NH4OH
   The ionization constant for NH4OH is Kb = 1.8 x 10 [exp-5]
 The Ksp for Mg{OH]2 is 8.9 x 10[exp-12]

    From this, calculate the IP for Mg[OH]2. Will the Mg precipitate? 


Remember "A strong acid will displace a weak acid from it's salt", this is 
a similar situation, with bases. The VERY concentrated base [Mg[OH]2] 
displaces the dilute weak base from it's salt [NH4Cl]. An interesting 
problem in equilibria.

                               Regards, 
                               Charlie Cruitchfield