Assay Of Lactic Acid 622723

Prepared by: Mas, Mylyn L. Ong, Dyan S. PhCh 135-PHC

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ASSAY OF LACTIC ACID

PRINCIPLES INVOLVED Type of Titration: Alkalimetric Residual with Blank Type of Reaction: Neutralization Indicator: Phenolphthalein TS Endpoint: colorless  faint pink Faint Pink  colorless Titration Condition: Room Temperature

(lactic acid) (sodium lactate) Factor: 1 since 1 mole of NaOH is needed to neutralize 1 mole of lactic acid. Neutralization reaction: 2NaOH (aq) + H2SO4 (aq)  Na2SO4 (aq) + H2O V.

SCHEMATIC DIAGRAM (USP 35 p. 3630)

2.5 mL Lactic Acid

 Add 50mL 1N NaOH  Boil for 20 minutes  Add Phenolphthalein TS

 Titrate excess alkali with 1N Sulfuric Acid TS Volume Reading Trial 1 Initial Reading

0.00 mL

COMPUTATIONS 1. Pre-Lab Computation: Calculate for weight of Lactic Acid needed: Wt (g)= NT X VT X MWreagent fx1000 = 1.0 N X (50.0 mL-20.0 mL) X 90.08g/mol 1 X 1000 = 2.7024 g of Lactic Acid Since volume is needed: 2.7024 g of Lactic Acid X ((%A.I /100 ) Xspec.gr.) 2.7024 g X (90/100 X 1.2) X = 2.919 mL of Lactic Acid to be used 2. Post-Lab Computations Volume of Excess Titrant Used (1.0 N NaOH); Nf= 1.0

Trial 2 0.00 mL 50.00 mL 50.00 mL

Final Reading 50.00 mL *Perform a Volume of Back Titrant Total Volume Used 50.00 mL Used (1.0 N H2SO4); Nf= 1.0 blank Volume Trial 1 Trial 2 Blank determination Reading Reasons for Important Steps: % A.I Initial 0.00 mL 0.00 mL 0.00 mL  The solution is boiled for 20 = [NT Reading X Nf Final Reading 21.50 mL 21.65 mL 48.00 mL minutes to speed up the X Total Volume 21.50 mL 21.65 mL 48.00 mL (V blan reaction Used k The titration is done in room Vactual )]backtitrant] X MWanalyte temperature because many f x 1000 sample wt (g) indicators give different Trial 1: indications at different [1.0 N X 1.0 N x (48 mL-21.50mL) x 90.08/1000)] temperatures. 2.7024

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OFFICIAL REQUIREMENT Lactic Acid is a mixture of lactic acid (C3H6O3) and lactic acid lactate (C6H10O5) equivalent to a total of not less than 88.0 percent and not more than 92.0 percent, by weight, of C3H6O3. CHEMICAL EQUATION

+ NaOH  +H2O

C3H5O3H (aq) + NaOH(aq) C3H5O3Na(aq)+ H2O

=88.33 % Trial 2: [1.0 N X 1.0 N x (48 mL-21.65mL) x 90.08/1000)] 2.7024 =87.83% Average: 88.08 % of lactic acid contained in sample Acceptance criteria: 88.0 % - 92.0 % ED % deviation b78etween trials: |88.33% -87.83%| [(88.33%+87.83%)/2] =0.568% References: Knevel, A. and DiGangi, F. (1977). Jenkins’ quantitative pharmaceutical chemistry (7th ed.). USA: McGraw-Hill Inc.

Prepared by: Mas, Mylyn L. Ong, Dyan S. PhCh 135-PHC

ASSAY OF LACTIC ACID

Skoog, D., West, D., Holler, F., Crouch, S. (2014). Fundamentals of analytical chemistry (9th ed.).

California: Cengage Learning. United States Pharmacopeial Convention, Inc. (2012). United States Pharmacopeia 35 th Revision and National Formulary 30th Edition (p.3630). Maryland: USPCI.