Discussion
When 50 ml of water sample was treated with 10 ml of 10% KI solution and then titrated with \(\frac{N}{50}\) Hypo solution using starch as indicator, 2.5 ml of Hypo was used for starch end point. Calculate the strength of free chlorine (in ppm) in the given water sample.
Answer: B
The free chlorine liberates iodine from KI, and the iodine is titrated with N/50 hypo. The hypo used is 2.5 ml of 0.02 N, which is 0.05 milli-equivalents. One milli-equivalent of chlorine is 35.5 mg, so the chlorine in the 50 ml sample is
\[ 0.05 \times 35.5 = 1.775\ \text{mg} \]Per litre (1000 ml) this is 1.775 x 20 = 35.5 mg/L, that is 35.5 ppm. The options 0.0355 and 0.001 are factors of 1000 smaller, which would result from stating the answer in grams per litre or from mixing up the units, and 1 has no basis in the data.
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