Quantitative Estimation of Cell Sap Chloride Ion

In this article we will discuss about the tests for quantitative estimation of cell sap chloride ion in plants.

Among essential elements chloride is one of the important elements, which is required in trace quantity. But in saline habitat, where the soil have very high value of available chloride, it may pose detrimental effects on plants. However, there are certain specially adapted plants whose cell sap contain a very high value of chloride and they are very much chloride dependent.

Requirements:

1. Plant materials:

Fresh healthy leaves of Acanthus illicifolius.

2. Chemicals and reagents:

(a) 0.02 (N) AgNO₃ solution — 0.34 gm. AgNO₃ dissolved in 1,000 ml distilled water.

(b) 0.5 N NaCl standard solution.

(c) Potassium chromate indicator — 5 gms in 100 ml distilled water.

3. Glassware:

Beaker, burette, pipette, funnel, conical flask.

4. Miscellaneous item:

Mortarpestel, filter paper, charcoal, neutral sand, tan balances, weight box etc.

Test Procedure:

1. At first 0.02 (N) AgNO₃ solution was titrated against 10 ml of standard N/10 NaCl solution using potassium chromate indicator.

2. Now 10 gms of washed leaves of Acanthus illicifolius was taken in a mortar and then crushed with a small amount of neutral sand and distilled water. After crushing, the extract was diluted with distilled water and filtered through absorbent cotton.

3. The chlorophyll pigment of the extract was then removed by treating with activated charcoal for overnight. Finally, cleared extract was obtained after filtration of treated extract. The final filtrate volume was made 100 ml.

4. Finally, 10 ml of extract was titrated against 0.02 (N) AgNO₃ solution using dichromate indicators.

5. The end-point of titration was detected by the presence of reddish-brown colour of silver chromate (AgCrO₄).

Results:

1. The standardization result of 0.02 (N) AgNO₃ against 0.5 (N) NaCl solution is given:

Calculations:

0.5 (N) NaCl solution is prepared by dissolving 29.25 mgm of NaCl in 1,000 ml

So, 29.25 NaCl contains — 35.5/58.5 × 29.25 = 2.37 mg of chlorine

Thus, 1,000 ml standard NaCl contains 2.37 mg of chlorine

Thus 1 ml solution contains = 2.37/1,000 mg chlorine

Again suppose x ml AgNO₃ is needed to titrate 10 ml standard NaCl solution, so 1 ml AgNO₃ solution can be equivalent to 10/x ml of NaCl soln.

1 ml of NaCl solution contains = 2.37 × 10^-3 mgm chlorine

So 10/x ml NaCl soln. s 2.37 × 10^-2 mgm chlorine

Thus, 1 ml AgNO₃ slon. = 2.37 × 10^-2 mgm chlorine

Finally, if 10 ml of plant extract is required, suppose y ml of AgNO₃ solution

Then y ml AgNO₃ contains = y × 2.37 × 10^-2/10mg chlorine

10 ml extract contains = y × 2.37 × 10^-2/10mg chlorine

100 ml extract contains = y × 2.37 × 10^-2 × 100/10mg chlorine = y × 2.37/10 mg chlorine

So 5 gm. of leaf tissue contains = y × 2.37/10 mg chlorine

The chlorine content can be expressed as y × 2.37 /10 mg chlorine/gm. of leaf tissue