hello, I'm working on the calibration for a potentiometer following the ASTM 3230-13 (salts in crude oil) but I have a question about the equivalence between a anhydrous salt and a hydrated salt.
We have a hydrated MgCl2*6H2O and we must find the equivalent weight for the anhydrous MgCl2 in the laboratory based on this:
Magnesium Chloride (MgCl2) Solution (10g/L): Transfer 1g of MgCl2, OR THE EQUIVALENT WEIGHT OF A HYDRATED SALT, into a 100mL volumetric flask and dissolve in 25mL of water.. Dilute to the mark (100mL) with mixed alcohol solvent.
I want to find that equivalent weight of MgCl2*6H2O
Thanks in advance.
MW of MgCl2= 95.2, MW og MgCl2.6H20 = 203.4 . 203.4/95.2 = 2.14. So 2.14 g of MgCl2.6H20 contains 1 g of Mg Cl2 and 1.14 G of H20.
Add 2.14 g of the Hydrate in stead of 1 g MgCl2 and consider reducing the H2O by about 1.14 ml.
MW of MgCl2= 95.2, MW og MgCl2.6H20 = 203.4 . 203.4/95.2 = 2.14. So 2.14 g of MgCl2.6H20 contains 1 g of Mg Cl2 and 1.14 G of H20.
Add 2.14 g of the Hydrate in stead of 1 g MgCl2 and consider reducing the H2O by about 1.14 ml.
thank you very much for your response! I was thinking about those calculations. However, I had not thouhgt about the reduction in 1.14ml of water. Is that absolutely necessary for precision? I thought by doing the weight equivalence between anhydrous and hydrated the amount of water would not be necessary to alter.
I ask this because I need to do something similar for CaCl2.2H2O and I'd do the appropiate calculations.
Once again thank for your answer.
If the instructions are to add water until the total volume in th flask is 25 ml than that is clear if the instruction is add about 24 ml to get a total of 25 ml it should be changed to add about 23 ml of water to get 25 ml total because the hydrated mgcl2 already contains about 1.14 ml of water. The amount of water you add is less but the total volume you end up with is the same
reasonable and helpful.
I suppose you equal 1.14g of H2O to 1.14 mL of water based on the density 1g/mL, that makes sense.
Thank you Mr Manner.
There are several magnesium chloride hydrates with the general formula MgCl2(H2O)x, with x = 2, 4, 6, 8, 12. If we initially have pure hexahydrate, MgCl2(H2O)6, in closed atmosphere, its relative humidity will tend to a stable value, determined by the water vapour pressure of MgCl2(H2O)6. The salt is hygroscopic and deliquescent. It should be kept in a well-stoppered flask and stored in a relatively dry environment. If left in open atmosphere of higher relative humidity, the next higher hydrate, MgCl2(H2O)8, can be expected to form. Chemical equilibrium then tends to be established between both hydrates. The effect of the atmosphere in the equilibrium thus approached depends on temperature and relative humidity. Eventually, conversion to the octahydrate may be completed, and being the relative humidity sufficiently high, the highest hydrate can be expected to form. Eventually, the highest hydrate may become wet and start to dissolve. If the original hexahydrate somehow uptaked moisture, its original stoichiometry can be recovered by drying in hot dry air at 115 ºC, being hereafter cooled in desiccator and weighted in a relatively dry environment.
- Badges
- Science topic
- Similar topics
- Engineering
- Petroleum Engineering