I'm working on designing a lab on how to determine the Mg 2+ content in a sample of rock. Does anyone have any ideas as to how this can be done?

well I would exploit the fact that it`s Sulphate is quite soluble compared to the Other group 2 metals.

that`ll be a step in the right direction

well I would exploit the fact that it`s Sulphate is quite soluble compared to the Other group 2 metals.

that`ll be a step in the right direction

 

Thanks. That was helpful. Here is what I have figured out so far.

 

Here is the general scheme:

1.) rock (containing Mg 2+) + HF (able to dissolve rocks)---->MgF2 (and if Ca, Sr, and Ba are in the rock they will precipitate as well)

2.) To be sure that the above mentioned ions are not part of the MgF2 precipitate

(MgF2+ precipitate) + Sulfate ---->MgSulfate(aq) + possible Ca, Sr, Ba,sulfate precipitate

3.) I would then discard the precipitate and use the aq MgSulfate sln to determine the initial Mg content in the rock.

 

Does this seem plausible to you? Also, do you have any suggestions as to how I can determine the Mg content from the MgSulfate aqueous solution?

For this to work you would need to use hot, concentrated (98%) H2SO4 to form MgSO4 (liberation of HF) so you couldnt get the MgSO4 to dissolve without the rest of the metal sulpahtes also doing so. Please note that HF is an extremly corrosive acid and I wouldnt consider trying this unless you are suitably skilled.

 

So maybe you could co-percipitate the Ca and Mg ions using strontium sulphate collect the filtrate and determine magnesium content using an EDTA titration to determine the Ca content first (EDTA will chelate with Ca and Mg) using the fact that the Ca2+ is more tightly bound to EDTA then Mg2+. Consequently, the colour change using Calmagite as an indicator (has a red colour when complexed with Mg2+ and a blue color when Mg2+ is chelated by the EDTA) for Mg2+ occurs after all Ca2+ is chelated. Then you can subtract this from the total amout of Ca and Mg to get the Mg concentration

For this to work you would need to use hot, concentrated (98%) H2SO4 to form MgSO4 (liberation of HF) so you couldnt get the MgSO4 to dissolve without the rest of the metal sulpahtes also doing so. Please note that HF is an extremly corrosive acid and I wouldnt consider trying this unless you are suitably skilled.

 

So maybe you could co-percipitate the Ca and Mg ions using strontium sulphate collect the filtrate and determine magnesium content using an EDTA titration to determine the Ca content first (EDTA will chelate with Ca and Mg) using the fact that the Ca2+ is more tightly bound to EDTA then Mg2+. Consequently, the colour change using Calmagite as an indicator (has a red colour when complexed with Mg2+ and a blue color when Mg2+ is chelated by the EDTA) for Mg2+ occurs after all Ca2+ is chelated. Then you can subtract this from the total amout of Ca and Mg to get the Mg concentration

 

I talked to my teacher and he said no HF so i was actually in the process of creating a lab based on what you just said. I do need to clarify something though; based on the solubility rules, if you add sulfuric acid to Mg2+

wouldn't you get and aqueous sln of MgSulfate and the other sulfates would form a precipate (this is based on the Mg2+ being selectively soluble with Sulfate). I guess I'm just confused about how the addition of SrSulfate will help. Could you clarify?

Okay my bad, i should have said you can percipitate out the calcium sulphate (along with strontium sulphate), filter it and the use an EDTA titration on the filtrate which contains the Mg2+ ions to determine magnesium content.

you can use HCl instead of HF if your teacher considers it too dangerous, both will work quite nicely

 

then you can add your dilute sulphuric to make the sulphates and leave HCl again, which is driven off with heat.

 

what remains should be soluble sulphates then, and all being well your Mg ions if they were present.

HF is used to dissolve rocks for analysis because it disolves silica.

You don't need to do that so, as has been said, treating the (crushed) rock with HCl will leach out most of the metals.

Then it's a matter of cleaning up the Mg from the other metals and measuring it (and that's relatively straightforward analysis).