I want to prepare irrigation solutions of EC (Electrical conductivity) 05 and 08 ds/m of NaCl. Can anyone guide me with a a detailed procedure of making these saline solutions.
What consideration should I give to prepare these solution?
I can't give a detailed, accurate method. So if you don't get one I suggest that you prepare NaCl solutions of the same molarity as the KCl ones that would satisfy the target EC values. That will get you close. Then you can simply adjust the EC by adding small amounts of water or NaCl solution. Remember EC is temp sensitive.
you could dissolve 640 mg of sodium chloride in water.this will give you 1ds-m.from this you could dilute to get the required EC levels. you could also make a solution of 10mM sodium chloride which is equivalent to 1ds-m.from this you could dilute with distilled water to get the required EC levels.
I need your suggestion.Dr. M A Mojid has given me following details. When I go on molar solution basis and compare with this equation (Given below), it gives a different answer.
Develop a calibration equation between NaCl concentration and EC. You prepare 5 to 6 NaCl solutions of different concentrations (say, 0.5, 1.0, 3.0, 5.0, 8.0 ... g/litre). Then you do regression analysis and get the calibration equation. But note that, EC is temperature dependent; EC increases by approximately 2% for each degree increase in temperature.
The calibration equation at 25 degrees Celcius will be as below:
EC = 1.3443C + 0.1682 (EC = electrical conductivity, dS/m; C = NaCl salt concentration, g/l)
You can use this equation. Just do back calculation. Put 2 or 6 for EC and you will get the value of C (NaCl concentration). Then make the solution of that concentration.
Use the already established relation ship: 640mg per liter of either Na Cl = 1ds/m. OR 10mM of NaCL=1ds/m. Try to homogenise the water after dissolution of salts and test EC.