Important Role of Dilutions in Quantitative Estimations

Sample Dilution Laboratory Ware

Dilutions play a crucial role in quantitative estimations. The concentration of the analyte whenever high requires a single or multi stage dilution before estimation. Similarly dilutions are essential for preparing standard solutions for generation of calibration plots.

In this article critical concepts on dilutions are clarified with illustrative calculations.

Volume or Weight Requirement

Before you start any quantitative estimation you have to do elementary calculations to arrive at weight or volume requirements of standards.

Example – weight requirement

What weight of sodium chloride is required to make 100 ppm of sodium ion solution from NaCl standard?


Formula weight of NaCl = 23+35.5 = 58.5

23 g of Na is present in 58.5 g of NaCl

Therefore, 1 g of Na is present in 58.5/23 = 2.5 g of NaCl

1 ppm solution will contain 1 mg or 0.001 g Na/liter of solution

Therefore 100 ppm solution will contain 100 mg or 0.1 g Na / per litre of solution

1 g of Na is present in 2.5 g NaCl

Therefore 0.1 g of Na is present in 2.5/1.0 X 0.1 = 0.25 g of NaCl

Thus 0.25 g of NaCl dissolved in 1 L volume will give rise to 100 ppm Na solution

Example – volume requirements

You have 1000 ppm Ca standard solution. How would you proceed to make 1, 2, 5 and 10 ppm solutions


The basic relationship for dilutions is

C1 x V1 = C2 x V2


C1 is initial concentration of standard solution (1000 ppm)

V1 is the volume of stock solution(1ml) to be diluted to get 100 ml of 10 ppm solution

C2  will be 1, 2, 5 or 10 ppm for respective dilution standards

V2 will be final volume of standards obtained after completion of dilutions

The individual dilutions calculations are illustrated below:

Stage – 1 1000 ppm standard to 100 ppm (10 fold dilution)

1.0 X 1000 ppm = V2  X 100 ppm

V2 = 1 X 1000 /100 = 10 ml

Stage – 2 dilutions of 100 ppm to 1, 2, 5 and 10 ppm concentrations using 10 ml volumetric flasks

100X V (10 ppm) = 10 X 10

V (10 ppm) = 10 X 10 / 100 = 1ml

100 X V (5ppm ) = 5 X 10

V (5ppm) = 5 X10 / 100 = 0.5 ml

100X V(2 ppm) = 2 X10

V (2 ppm) = 2X10 / 100 = 0.2 ml

100X V(1 ppm) = 1 X 10

V (1 ppm) = 1 X10 /100 = 0.1 ml

The above dilutions stages are illustrated in the video

Suggested precautions for dilutions

  • Before proceeding with dilutions ensure that the standard solution is within its validity period and has been stored under prescribed conditions
  • Make use of micropipette for dilutions
  • Wear disposable gloves to prevent contamination of the sample
  • Use clean and dry volumetric  glassware for dilutions
  • It is advisable to carry diluted sample flasks in trays to avoid breakages
  • Never return unused standard back into the standard bottle

Remember that the quality of your estimations in terms of accuracy and precision is in direct proportion to the care taken at each stage of serial dilution. Error at any stage gets multiplied in subsequent stages so utmost care should be exercised during dilutions operations.

Related Articles


Your email address will not be published. Required fields are marked *

  1. how to calculate impurity concentration with using sample and std concentration


Dont Get left Out!

over 20,000 scientists read our weekly Newsletter!