and MgHPO₄ + H₂SO₄ = H₃PO₄ + MgSO₄.
The magnesia, when in the form of oxid, is capable of producing a
reversion of the monocalcium phosphate, as is shown below:
CaH₄(PO₄)₂ + MgO = CaMgH₂(PO₄)₂ + H₂O.
One part by weight of magnesia can render three and one-half parts of
soluble monocalcium phosphate insoluble.
=141. Determination of Quantity of Sulfuric Acid Necessary for Solution
of a Mineral Phosphate.=—The theoretical quantity of sulfuric acid
required for the proper treatment of any phosphate may be calculated
from its chemical analysis and by the formulas and reactions already
given. For the experimental determination the method of Rümpler may be
followed.[123]
Twenty grams of the fine phosphate are placed in a liter flask with a
greater quantity of accurately measured sulfuric acid than is necessary
for complete solution. The acid should have a specific gravity of 1.455
or 45° B. The mixture is allowed to stand for two hours at 50°. It is
then cooled, the flask filled with water to the mark, well shaken, and
the contents filtered. Fifty cubic centimeters of the filtrate are
treated with tenth normal soda-lye until basic phosphate begins to
separate. The excess of acid used is then calculated. Example: Twenty
grams of phosphate containing 28.3 per cent of phosphoric acid, 10.0
per cent of calcium carbonate, 5.5 per cent of calcium fluorid, and
2.4 per cent of calcium chlorid were treated as above with sixteen
cubic centimeters of sulfuric acid containing 10.24 grams of sulfur
trioxid. In titrating fifty cubic centimeters of the filtrate obtained
as described above, 10.4 cubic centimeters of tenth normal soda-lye
were used, equivalent to 0.0416 gram of sulfur trioxid. Then 10.24 × 50
÷ 1000 = 0.5120 = total sulfur trioxid in fifty cubic centimeters of
the filtrate, and 0.5120 - 0.0416 = 0.4704 gram, the amount of sulfur
trioxid consumed in the decomposition.
Therefore the sulfur trioxid required for decomposition is 47.04 per
cent of the weight of the phosphate employed. One hundred parts of the
phosphate would therefore require 47.04 parts of sulfur trioxid = to
73.6 parts of sulfuric acid of 1.710 specific gravity or 92.1 parts of
1.530 specific gravity.
A more convenient method than the one mentioned above consists in
treating a small quantity of the phosphate, from one-half to one
kilogram, in the laboratory, or fifty kilograms in a lead box, just
as would be practiced on a large scale. A few tests with these small
quantities, followed by drying and grinding will reveal to the skilled
operator the approximate quantity and strength of sulfuric acid to
be used in each case. The quantities of sulfuric acid as determined
by calculation from analyses and by actual laboratory tests agree
fairly well in most instances. There is, however, sometimes a marked
disagreement. The general rule of practice is to use always an amount
of sulfuric acid sufficient to produce and maintain water-soluble
phosphoric acid in the fertilizer, but the sulfuric acid must not be
used in such quantity as to interfere with the subsequent drying,
grinding, and marketing of the acid phosphate.
For convenience the following table may be used for calculating the
quantity of oil of vitriol needed for each unit of weight of material
noted: