How To Make Dilutions For Standard Curve
Channel: PharmEase
Date: 03/08/2020

This video explain how dilution factor applied to the drug analysis i.e. making standard curve or calibration curves for HPLC/ LCMS or spectroscopic or photometeric methods ( scale may vary from milligram/ml or (microgram/ml=ug/mL) or nanogram/ml or even extremely low levels). This video elaborates how to make dilution to prepare the standard curve of a model drug (paracetamol) where known concentrations (conc) of drug are prepared ( in this example viz. 5.0,10.0,15.0, 20.0 to 25.0ug/ml and if Stock solution be supplied with concentration of 1000ug/ml =1mg/ml. 

In step-I: Stock solution of 1000ug/ml is provided or prepared by dissolving 100mg of powdered drug dissolved in 0.2-0.4ml of ethanol and make up the volume to 100.0ml in 100ml capacity volumetric container.



In step-II: Stock solution is diluted 10 times and resulted to give concentration of 100ug/ml: ( by taking 2ml of stock and 18ml of diluting fluid).  Dilution Factor (DF) in this step=10

In step-III :  2.0mL Aliquot (100ug/ml) from step-II solution is taken and diluted to 4 times and resulted to produce concentration of 25ug/ml ( by taking 3ml solution from above step and mixed with 9ml of diluting fluid).
DF in this step=4; (12 ml/ 3ml)

Overall Dilution Factor = 10 * 4= 40


In step-IV: Again pipette out the aliqouts from solution prepared at step-III (of 25ug/ml strength) taken 1.0ml, 2.0mL, 3.0mL, 4.0mL and 5.0mL in five separate testubes (1 ,2, 3..5) and mixed with 4.0, 3.0, 2.0 and 1.0ml diluting fluid respectively so as to make up the volume 5.0ml in each test tube.  Please note that no need to add dilution in test tube which already filled with aliquot from step-III.
wow in this way, u prepare 5,10, 15, 20 and 25 microgram/ml of stock solution.


Dilution factors calculation of step-IV can be explained below


a) Aliquot 1.0mL (25ug/mL) is taken from step-III solution and mixed with 4.0ml of water.
DF= Total volume/aliquot taken= 5/1;  hence DF= 5 

so the concentration in test tube (1) at step IV 

[Concentration after diltuon= Initial conc/DF] so that  it is 25/5= 5ug/mL


b)  Aliquot 2.0mL (25ug/mL) is taken from step-III and mixed with 3.0ml of water.
DF =Total volume/aliquot taken= 5/2;  hence DF= 2.5
 so the concentration in the test tube (2) is 25/2.5= 10ug/mL


c) Aliquot 3.0mL (25ug/mL) is taken from step-III and mixed with 2.0ml of water.
Total volume/aliquot taken= 5/3;  hence DF= 1.66
 so the concentration in test tube (3) is 25/1.667= 15ug/mL


d) Aliquot 4.0mL (25ug/mL) is taken from step-III solution and mixed with 1.0ml of water.
DF=Total volume/aliquot taken= 5/4;  hence DF= 0.8 

so the concentration in test tube (4) is 25/0.8= 20ug/mL


e) No dilution took place in 5th test tube, it has same conc  as that of step=3 ie. 25ug/mL


happy learning
i) you can try making standards of concentrations 2,4, 6, 8, 10 ug/ml
ii) you can also trying making standards of 10, 100, 1000, 10000ug/mL 



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