Question

Confirmatory test for magnesium ion with magneson reagent equation blue lake test

Answer 1

A common problem with magnesium is, that all tests only work well and specific if you have a pure solution. Most heavier elements react more readily with the chemicals, than magnesium itself does. Another problem is, that salts from magnesium have no flame colouration at all.

I am again loosely translating from chemgapedia.de.*

You can precipitate magnesium with 8-hydroxyquinoline:


It forms a yellow-green complex. This is only a specific test, if you know you have no heavier elements in your test tube. For example, aluminium forms the complex Tris(8-hydroxyquinolinato)aluminium:
Al3++3C9H6NOH⟶Al(C9H6NO)3+3H+
AlX3++3CX9HX6NOH⟶Al(CX9HX6NO)X3+3HX+
Another possibility is precipitation as magnesium ammonium phosphate (struvite). And yes, you guessed it, this happens with all other alkali and earth alkali elements (except lithium). But if you are certain you only have got magnesium, you can grow nice crystals. The pH for this reaction should be buffered at around 8-9 with ammonia/ ammonium chloride.
Mg2++NH4++−OH+HPO42−−→−−−−−−NH3/NH4ClH2OMg(NH4)PO4⋅6H2O↓
MgX2++NHX4X++X−X22−OH+HPOX4X2−→NHX3/NHX4ClHX2OMg(NHX4)POX4⋅6HX2O↓
There are two more tests, that I would consider best suitable, because of the pretty colour.

The reaction of magnesium hydroxide with titan yellow in presence of sodium hydroxide will produce a red precipitate. It is important to do a control/ blind experiment, as older solutions of titan yellow will form a precipitate with sodium hydroxide alone.

The reaction of magnesium hydroxide with azo violet (Magneson) or Magneson II (Magnezon II) in presence of sodium hydroxide will produce a blue precipitate. Again test with sodium hydroxide to observe the violet solution. This reaction also occurs with Ca2+,Al3+CaX2+,AlX3+.

Testing for magnesium is as far as I remember is some kind of a lottery when in combination with other elements (qualitative analysis). Good luck.

* Please note that the first reaction equation has depicted hydroxyquinoline with two circles for the aromatic system, which is wrong. In benzene one circle stands for 6 ππ electrons. So that would lead to hydroxyquinoline having 12 ππ electrons, two too many.