Question

How many milliliters of phenytoin sodium suspension containing 30 mg per 5 mL and 100 mg per 5 mL should be used in preparing 500 mL of a suspension containing 10 mg of phenytoin sodium per mL?

Answer 1

This can be solved by using Allegation method, which is used to calculate the required proportion of two solutions to be mixed to get required final volume and concentration.

1) THE GIVEN DATA HIGHER CONCENTRATION = 100mg|sm) LOWER CONCENTRATION = RequiRER CONCENTRATION = 10mg im! FINAL VOLUME = 500ml sml RED 10 CALCULATE ALLIGATION RATIO AND ALLIGATION VOLUMES THE INITIAL CONCENTRATIONS NEED TO BE CONVERTED TO SIMILAR FORMS AND UNITS. IE. HOW MANY mg/ml THUS, HIGHER CONCENTRATION IS loomg/5ml =20mglim LOWER CONCENTRATION 1S BY APPLYING ALLIGATION, 20 > 4 [10-6] SO THE ALLIGATION RADIO is=4:10 10 [20-10] = 2:5 THEREFORE 2 PARTS OF Toomg 15m] HIGHER CONCENTR- ARON AND 3 PARTS OF F30mg/5mi] LOWER CONCENTRADON OF PHENYTOIN SUSPENSION SHOULD BE MIXED TO GET DESIRED CONCENTRATION OF Omglimla THE VOLUME REQUIRED HIGHER CONCENTRATION = -xsoomL 2+5 VOLUME = .142.857 ML - 143mL

LOWER CONCENTRATION - 5 X 500ml VOLUME - 2+5 = 357.14 =357 mL WHERE D AND 5 ARE ALLIGATION RAD 10 OF HIGHER AND LOWER CONCENTRTION RESPECTIVELY AND 500mL IS THE TOTAL DESIRED VOLUME. THEREFORE, TO'GET 500ml Of 10mglimb PHENYTOIN SUSPENSION, MIX 143 mL OF 100mg/5mL HIGHER CONCENTRATION AND 357 mL of 30mg 15ml [LOWER CONCENTRATION.