Calcium Sulfate
CaSO4 136.14
CaSO4·2H2O 172.17
Sulfuric acid, calcium salt (1:1);
Calcium sulfate (1:1)
[7778-18-9].
Dihydrate[10101-41-4].
(kal' see um sul' fate).
CaSO4 136.14
CaSO4·2H2O 172.17
Sulfuric acid, calcium salt (1:1);
Calcium sulfate (1:1)
[7778-18-9].Dihydrate
[10101-41-4].DEFINITION
Calcium Sulfate is anhydrous or contains two molecules of water of hydration. It contains NLT 98.0% and NMT 101.0% of CaSO4, calculated on the dried basis.
IDENTIFICATION
• A. Identification Tests—General, Calcium 
191
191
Sample solution:
Dissolve 200 mg by warming in a mixture of 4 mL of 3 N hydrochloric acid and 16 mL of water.
Acceptance criteria:
Meets the requirements
• B. Identification Tests—General, Sulfate 191
191
Sample solution:
Dissolve 200 mg by warming in a mixture of 4 mL of 3 N hydrochloric acid and 16 mL of water.
Acceptance criteria:
Meets the requirements
ASSAY
• Procedure
Sample:
Dissolve 300 mg of Calcium Sulfate in 100 mL of water and 4 mL of 3 N hydrochloric acid. Boil, if necessary, to dissolve, and cool before titrating.
Titrimetric system
(See Titrimetry 541.)
541.)
Mode:
Direct titration
Titrant:
0.05 M edetate disodium VS
Blank:
100 mL of water and 4 mL of 3 N hydrochloric acid
Endpoint detection:
Visual
Analysis:
While stirring, preferably with a magnetic stirrer, add to the Sample solution, in the order named, 0.5 mL of triethanolamine, 300 mg of hydroxy naphthol blue, and, from a 50-mL buret, 30 mL of 0.05 M edetate disodium VS. Add sodium hydroxide solution (45 in 100) until the initial red color changes to clear blue, continue to add dropwise until the color changes to violet, and add an additional 0.5 mL. The pH is 12.3–12.5. Continue the titration dropwise with 0.05 M edetate disodium VS to the appearance of a clear-blue endpoint that persists for NLT 60 s.
Perform a blank determination. Calculate the percentage of Calcium Sulfate (CaSO4) in the Sample taken:
Result = [(V 
B) × N × F × 100]/W
B) × N × F × 100]/W
| V | = | = volume of titrant consumed by the Sample (mL) |
| B | = | = volume of titrant consumed by the Blank (mL) |
| N | = | = actual normality of the titrant (mEq/mL) |
| F | = | = equivalency factor, 136.14 mg/mEq |
| W | = | = weight of the Sample (mg) |
Acceptance criteria:
98.0%–101.0% on the dried basis
IMPURITIES
• Iron 241
241
Sample solution:
Dissolve 100 mg in 8 mL of 3 N hydrochloric acid, and dilute with water to 47 mL.
Acceptance criteria:
NMT 100 ppm
• Heavy Metals, Method I 231:
NMT 10 ppm
231:
NMT 10 ppm
Test preparation:
Mix 2.0 g with 20 mL of water, add 25 mL of 3 N hydrochloric acid, and heat to boiling to dissolve the test specimen. Cool, and add ammonium hydroxide to a pH of 7. Filter, evaporate to a volume of about 25 mL, and refilter, if necessary, to obtain a clear solution.
SPECIFIC TESTS
• Loss on Drying 731
731
Samples:
Neat
Analysis:
Dry a sample at a temperature NLT 250
to constant weight.
to constant weight.
Acceptance criteria:
NMT 1.5% for the anhydrous form and NMT 19.0%–23.0% for the dihydrate
ADDITIONAL REQUIREMENTS
• Packaging and Storage:
Preserve in well-closed containers.
• Labeling:
Label it to indicate whether it is anhydrous or the dihydrate.
Auxiliary Information—
Please check for your question in the FAQs before contacting USP.
| Topic/Question | Contact | Expert Committee |
|---|---|---|
| Monograph | Robert H. Lafaver, M.S.
Scientific Liaison 1-301-816-8335 |
(EXC2010) Monographs - Excipients |
USP35–NF30 Page 1724
Pharmacopeial Forum: Volume No. 27(6) Page 3337