Important Role of Dilutions in Quantitative Estimations

Sample-Dilution-Laboratory-Ware

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?

Solution

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

Solution

The basic relationship for dilutions is

C_i_n_i_t_i_a_l X  V_i_n_i_t_i_a_l = C_f_i_n_a_l X V_f_i_n_a_l

where

C_i_n_i_t_i_a_l is initial concentration of standard solution (1000 ppm)

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

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

V_f_i_n_a_l  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 = V_f_i_n_a_l  X 100 ppm

V_f_i_n_a_l = 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.

About Dr. Deepak Bhanot

Dr Deepak Bhanot is a seasoned professional having nearly 30 years expertise beginning from sales and product support of analytical instruments. After completing his graduation and post graduation from Delhi University and IIT Delhi he went on to Loughborough University of Technology, UK for doctorate research in analytical chemistry. His mission is to develop training programs on analytical techniques and share his experiences with broad spectrum of users ranging from professionals engaged in analytical development and research as well as young enthusiasts fresh from academics who wish to embark upon a career in analytical industry.

Comments

  1. This page definitely has all the information I wanted about this subject and didn’t know who to ask.

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