Determine amount of free Chlorine

                               EXPERIMENT NO. 03

 

OBOBJECT: To determine the amount of free chlorine present in the given water sample using N/200 hypo solution and starch as an indicator (Iodometric titration).

Observation Table:

Titration between given water sample and sodium thiosulphate solution

 

S. No.	Volume of Water Sample (mL)	Burette Reading (mL)		Concordant Reading (Y) (mL)
		Initial	Final
1	20	0.0	25.6	25.6

Calculation:

1 mole of Cl₂ = 1 mole of I₂ = 2 moles of Na₂S₂O₃

Hence, 1 mole of Na₂S₂O₃ = ½ mole of Cl₂ = 35.5 g of Cl₂

1 mL of Na₂S₂O₃ = 35.5 mg of Cl₂

According to the law of equivalence:

 

N1V1 of Sample	N2V2 of Hypo solution
N1 × 20	= N/200 × 25.6
N1	= 25.6/4000

 

 

Amount of free chlorine = Normality × Equivalent weight of chlorine

= 25.6/4000 × 35.5 g/L

= 25.6/4000 × 35.5 x 1000 mg/L or ppm.

= 227.2 mg/L or ppm.

RESULT:

The amount of free chlorine present in a given water sample is 227.2 ppm.