Mg(OH)2 ==> Mg^2+ + 2OH^-
Ksp = (Mg^2+)(OH^-)^2
When NH3 is added to the solution of MgCl2 (which is quite soluble) the NH3 reacts with the H2O to form NH4^+ + OH^-.
That increases the OH^- enough to form a Mg(OH)2 ppt because Ksp for Mg(OH)2 is exceeded. (SO YOU SHOULD HAVE OBSERVED A WHITE PPT AT THAT POINT. THE FACT YOU DIDN'T MEANS YOU DIDN'T ADD ENOUGH NH3.)
NH3 + HOH ==> NH4^ + OH^-
Then when you add NH4Cl, that increases the NH4^+ from the NH3 solution, the NH3 equilibrium is shifted to the left which decreases the OH^- to the point that the Ksp for Mg(OH)2 is no longer exceeded. As a result the Mg(OH)2 ppt that was there dissolves. This is the chemistry behind the qualitative separation of group III cations (Al, Fe, Cr, etc) from group V cations (Mg especially). If not for the addition of NH4Cl, Mg from group V ppts in group III. So you get a ppt in group III, it doesn't pan out to be anything so you use the eenie, meenie, miney, mo method and over report some metal in group III. Then when you get to group V, you under report Mg since all of it pptd earlier.
1.a) What did you observe upon the addition of NH3 (aq) to the magnesium chloride solution? Explain using relevant chemical reactions.
I observed a colorless liquid. (Not sure how to explain)
b) What happens when NH4Cl was added? Explain observation with respect to the equilibrium reactions that exist in the test tube using any relevant chemical reactions.
I observed a clear liquid (no change)
Not sure how to explain.
I appreciate the help!
1 answer
1 answer