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Potassium Sodium Tartrate
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C4H4KNaO6·4H2O 282.22

Butanedioic acid, 2,3-dihydroxy-, [R-(R*,R*)]-, monopotassium monosodium salt, tetrahydrate.
Monopotassium monosodium tartrate tetrahydrate [6100-16-9; 6381-59-5].

Anhydrous 210.16 [304-59-6].
» Potassium Sodium Tartrate contains not less than 99.0 percent and not more than 102.0 percent of C4H4KNaO6, calculated on the anhydrous basis.
Packaging and storage— Preserve in tight containers.
Identification—
A: Ignite it: it emits the odor of burning sugar and leaves a residue that is alkaline to litmus and that effervesces with acids.
B: To 10 mL of a solution (1 in 20) add 10 mL of 6 N acetic acid: a white, crystalline precipitate is formed within 15 minutes.
C: A solution (1 in 10) responds to the tests for Tartrate 191.
Alkalinity— A solution of 1.0 g in 20 mL of water is alkaline to litmus, but after the addition of 0.20 mL of 0.10 N sulfuric acid no pink color is produced by the addition of 1 drop of phenolphthalein TS.
Water, Method I 921: between 21.0% and 27.0%.
Limit of ammonia— Heat a 5-mL portion of a solution (1 in 10) with 5 mL of 1 N sodium hydroxide: the odor of ammonia is not noticeable.
Heavy metals, Method II 231: 0.001%.
Residual solvents 467: meets the requirements.
(Official January 1, 2007)
Assay— Weigh accurately about 2 g of Potassium Sodium Tartrate in a tared porcelain crucible, and ignite, gently at first, until the salt is thoroughly carbonized, protecting the carbonized salt from the flame at all times. Cool the crucible, place it in a glass beaker, and break up the carbonized mass with a glass rod. Without removing the glass rod or the crucible, add 50 mL of water and 50.0 mL of 0.5 N sulfuric acid VS, cover the beaker, and boil the solution for 30 minutes. Filter, and wash with hot water until the last washing is neutral to litmus. Cool the combined filtrate and washings, add methyl red-methylene blue TS, and titrate the excess acid with 0.5 N sodium hydroxide VS. Perform a blank determination (see Residual Titrations under Titrimetry 541). Each mL of 0.5 N sulfuric acid is equivalent to 52.54 mg of C4H4KNaO6.
Auxiliary Information— Staff Liaison : Elena Gonikberg, Ph.D., Scientist
Expert Committee : (MDGRE05) Monograph Development-Gastrointestinal Renal and Endocrine
USP29–NF24 Page 1777
Pharmacopeial Forum : Volume No. 31(3) Page 787
Phone Number : 1-301-816-8251