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2. (10) An old stockpile of bleaching powder used to disinfect swimming pool wat

ID: 1029864 • Letter: 2

Question

2. (10) An old stockpile of bleaching powder used to disinfect swimming pool water is found in a warehouse; the active ingredient in the bleach is calcium hypochlorite, Ca(OCI)2. Environmental regulations require that the hypochlorite content of the bleach be measured before disposal. It is determined iodometrically as follows: a 2.257 g sample of the bleach is dissolved in water and brought to volume in a 1.000 L volumetric flask. A 50.0 mL aliquot is pipeted into a flask, and is treated with excess KI and hydrochloric acid. The hypochlorite in the bleach reacts with the iodide to form molecular iodine: OCI(aq) 2 I(aq) 2 H(aq)I2(aq) + CI(aq) HO. The I2 formed is then titrated to a starch endpoint with sodium thiosulfate 12(aq) + 2 S:032(aq) 21(aq) + S4062(aq) It requires 17.54 mL of 0.1145 M Na2S,Os solution to titrate the iodine formed. Calculate the hypochlorite content ofthe bleach powder as weight % Ca(OCI)2. (Similar samples ofthis type ofbleach from other sources, when/resh, typically are 67-70% calcium hypochlorite.)

Explanation / Answer

Hi Dear Friend, we will solve this problem by back titration method.

moles present in 17.54 mL of 0.1145M Na2S2O3 = 17.54 mL x 10-3 L x 0.1145M = 0.001998025 mol

1 mol I2 requires 2 mol S2O32-

hence 0.001998025 mol S2O32- = 0.001998025 / 2 = 0.0009990125 mol I2

1 mol I2 formed from 1 mol OCl- and 2 mol KI, means 0.0009990125 / 2 = 0.00049950625 mol OCl- present in 50.0

mL bleach solution

in 1 liter = 1000 ml, moles will be = 0.00049950625 mol x (1000 mL / 50.0 mL) = 0.009990125 moles

convert moles to mass of calcium hypochlorite by multiplying with molar mass of calcium hypochlorite

mass of calcium hypochlorite = 0.009990125 mol x 142.98 g/mol = 1.4283880725 grams

weight % = (mass of calcium hypochlorite / mass of bleach) x 100%

weight % = (1.4283880725 grams / 2.257 g) x 100% = 63.2870213779 %

weight % = 63.3 %

Hope this helped you!

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