DETERMINATION OF AVAILABLE CHLORINE IN CHLORINE STERILIZER

Principle:

The chemical sterilization of some of the dairy equipment is carried out by hypochlorite solution. Different strengths are suggested for different types of operations. The stock solution of sodium hypochlorite with about 5 percent available chlorine is available. As the strength of chlorite sterilizers decreases on storing it is necessary to check the strength. This method is based on the reaction between available chlorine from hypochlorite solution and acidified potassium Iodine solution in which Iodine is liberated.

            NaOCL + 2 KI + 2CH₂COOH                      NaCl + 2CH₂COOK + I₂ + H₂O

The liberated Iodine is titrated against N/10 sodium thiosulphate using starch as indicator. From the reading of sodium thiosulphate the quantity of available chlorine can be found out.

Procedure:

1.      Pipette 10 ml of the sodium hypochlorite sample into 250 ml volumetric flask.

2.      Make up the volume upto the mark with distilled water and mix well.

3.      Pipette 25 ml of diluted solution in a conical flask.

4.      Add 2 g. potassium Iodine crystals to the solution and dissolve them followed by 10 ml glacial acetic acid.

5.      Available chlorine which is liberated from hypochlorite by the action of acid liberates an equivalent amount of Iodine from potassium Iodine which produces a yellowish brown colour.

6.      Immediately titrate the mixture against N/10 sodium thiosulphate until the brown colour changes to light straw yellow.

7.      Immediately add 1 percent of freshly prepared starch solution and titrate till the colour disappears.

8.      Note the reading of N/10 sodium thiosulphate as V ml.

9.      Make the blank determination using the same reagents and deduct from     V ml.

10. Record the values of both the titration in appropriate table.

Observation:

Burette reading	Titration I	Titration II	Titration III
Final reading
Initial reading
Volume delivered in ml.

Calculation:

1 ml of N/10 sodium thiosulphate = 0.003546 g. available chlorine

V ml of N/10 sodium thiosulphate = (V) (0.003546) g. available chlorine

This quantity of chlorine is from 25 ml diluted solution

    V X 0.003546 X 250

Therefore, available chlorine in 250 ml diluted solution =                                  = x

                                                                                                                 25

Which is equal to 10 ml of stock solution  = x

                                                                         x  X 100

Therefore, percent of available chlorine =                        = 10 x

                                                                             10

Note: If necessary the dilution of stock solution can be changed and the above formula will have to be modified accordingly.

Result: The available chlorine in chlorine sterilizer is ________ %