2nd Order Terms in Purity Calculations of Reference Standards

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2nd order terms cannot be excluded in purity calculations of pharmaceutical reference standards.

Text of 2nd Order Terms in Purity Calculations of Reference Standards

  • 1. Second Order Terms in PurityCalculations of Pharmaceutical ReferenceStandards!jbladon! 1!

2. CASE I! Protocol requires using HPLC purity and!the water content to determine analyte purity! %Purity = 100% %Water % HPLC Impurities?!jbladon! 2! 3. CASE I!Protocol requires using HPLC purity and!the water content to determine analyte purity!%Purity = F1 F2 100%F1 is the fraction of entire sample that is not water.F2 is the fraction of sample excluding water that is analyte.jbladon!3! 4. Let,A = Analyte*Im = Impurities* W = Water Then, A + Im A F1 = , F2 = A + Im+ W A + Im A + Im A A F1F2 = = A + Im+ W A + Im A + Im+ W ____________________Where each response factor is known or it is assumedthey are identicaljbladon! 4! 5. %HPLC Impurities Im =100% A + Im Rearrange F2 A A + Im Im Im F2 = = = 1 A + Im A + Im A + Im %HPLC Impurities F2 = 1 100% jbladon!5! 6. %WaterW= 100% A + Im+ WRearrange F1 A + Im A + Im+ W W F1 = = A + Im+ W A + Im+ W A + Im+ W W = A + Im+ W A + Im+ W %Water F1 = 1 100% jbladon!6! 7. jbladon! 7! 8. The Second Order Term %Water %HPLC Impurities %Purity = 1 1 100% 100% 100% %Water %HPLC Impurities %Water %HPLC Impurities %Purity = 1 + 100% 100%100% 100% 100% jbladon! 8! 9. Example Let%Water = 2.5% %HPLC Impurities = 5.0% Then%Purity = {1- 0.025 - 0.050 + (0.025)(0.050)}100% %Purity = 92.6% If the second order term is omitted, %Purity = 92.5% jbladon! 9! 10. CASE IIProtocol requires using the HPLC purity,!the water and solvent content, and! the inorganic ion content !to determine analyte purity!jbladon!10! 11. 1. Water Content by TGA!2. Solvent Content by GasChromatography!3. Inorganic Ions by Ion Chromatography!4. Fraction of what is left by HPLC Purity! jbladon!11! 12. %Purity = 100% %Water % Solvent % Inorganic % HPLC Impurities ?! jbladon! 12! 13. %Purity = F1 F2 100%F1 is the fraction of sample that is not water/solvent/inorganicF2 is the fraction of sample excluding water/solvent/inorganic that is analyte.jbladon! 13! 14. Let,A = Analyte*IMP = Impurities*W = WaterS = SolventIN = InorganicThen, A + IMPA F1 = , F2 = A + IMP + IN + W + S A + IMP ____________________Where each response factor is known orit is assumed they are identical jbladon!14! 15. A + IMP A F1F2 = A + IMP + W + S + IN A + IMP A F1F2 = A + IMP + W + S + IN jbladon!15! 16. %HPLC Impurities IMP =100% A + IMPRearrange F2 A A + IMP IMP F2 = = A + IMP A + IMP A + IMP IMP F2 = A + IMP A + IMP %HPLC ImpuritiesF2 = 1 100%jbladon!16! 17. %Water / SolventW +S = 100%A + IMP + W + S + IN%ININ=100% A + IMP + W + S + INRearrange F1 A + IMPF1 = = A + IMP + W + S + IN A + IMP + W + S + IN W S IN F1 = A + IMP + W + S + IN A + IMP + W + S + INW +S A + IMP + W + S + IN A + IMP + W + S + IN F1 = IN A + IMP + W + S + IN %Water / Solvent %INF1 = 1 100%100% jbladon!17! 18. %Water / Solvent % IN %Purity = 1 100%100% %HPLC Impurities x 1 100%100% jbladon!18! 19. The Second Order Terms %Water/Solvents %IN %HPLC Impurities 1100%100%100% %Water/Solvents %HPLC Impurities %Purity = + 100% 100%100% %IN %HPLC Impurities + 100% 100% jbladon!19! 20. Example Let %Water/Solvent = 2.5%%Inorganic = 5.0%%HPLC Impurities = 5.0%Then%Purity = {1- 0.025 - 0.050 - 0.050 + (0.025)(0.050) + (0.050)(0.050)}100%%Purity = 87.9%If both second order terms are omitted,%Purity = 87.5%jbladon! 20! 21. Meet the Boys of the FDADr. Wiley and the Division of Chemistry Staff in 1883. Wiley, third from right, was 37.Courtesy Wallace F. JanssenFDA Historian jbladon! 21! 22. If you have a question about anequation to calculate the purity of areference standard in a protocol or method, see your supervisor andbring my handouts for this presentation jbladon! 22!