EC and total soluble salts

DETERMINATION OF EC AND TOTAL SOLUBLE SALTS

Principle

         We know that solutions offer some resistance to the passage of electric current through them depending upon the concentration of the type of ions present. Higher the salt content, lesser the resistance to the flow of current (more is the conductance). The resistance is defined by the Ohm’s Law as the ratio of electrical potential in volts and strength of the current in amperes.

Resistance on Ohms ‘R’

=Volts (V) / Current (I)

Electrical conductivity or conductance is reverse of resistance

= 1 / R = 1 / Ohm

         It is expressed in reciprocal ohms or mhos per cm. As the values of electrical conductance obtained for soil solution and most of the irrigation waters are very small, it is therefore convenient to express them in milli mhos (m.mohs) per cm or micro mhos (mmhos) per cm.

Apparatus required

1.      Conductivity meter

2.      Conductivity cell

Procedure

         Dip the cell in the water sample under test. Then take the cell out and dip it in the same water sample contained in beaker. Switch on the instrument and take the conductance of this water with Conductivity Bridge.

iii. TOTAL SOLUBLE SALTS

         The gravimetric method of determination may be adopted where the number of samples is few while the conductivity method comes in handy when large numbers of samples are to be analyzed in a day.

Procedure

         Pipette out 100 ml of water sample (after filtration if necessary) to a weighed porcelain dish/potash dish. Evaporate to dryness on a water bath. Finish drying in an air oven at 100-1050C cool the dish in a desiccator and weigh. The weight of the residue represents the total salts in 100 ml of water sample. Express the result as parts per million (ppm).

Observation & calculation

Weight of the dish                   =                 W1 g

Weight of the dish + residue  =                   W2 g

Weight of the residue      =             (W2–W1) g

This represents total salts in 100 ml of water

Therefore total salts in 1 litre        =       ( W2–W1 ) x 1000/100

In terms of ppm                  =         ( W2–W1 ) x 1000 x 1000/100