Question

3. A student followed the procedure of this experiment to determe t pe commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask and titrated a 20-mL aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na2S2O3 solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g/mL. Refer to the post laboratory assignment calculations for h-j a) Calculate the number of moles of S2O,2 required for the titration. b) Calculate the number of moles of 12 produced in the titration mixtu re. Calculate the number of moles of OCI ion present in the diluted bleaching solution c) titrated. titratod.

Need help with C please

Answer 1

The molecular formula of Bleaching powder is Ca(OCl)Cl

Now the reaction between Ca(OCl)Cl ans Na2S2O3 is

Ca(OCl)Cl + Na₂S₂O₃ -----------> CaCl₂ + Na₂S₂O₆

That means 1 mole of Na₂S₂O₃ will titrate 1 mole of Ca(OCl)Cl

Now the given amount of Na₂S₂O₃ required = 0.1052 M and 35.46 ml

That means the moles of Na₂S₂O₃ required = concentration * volume

= 0.1052 M * 35.46 ml

= 0.1052 mol/ 1000 ml * 35.46 ml

= 0.0037 moles

Now this Na₂S₂O₃ is used for the titration of 20 ml of diluted aliquot. Thus the moles of Ca(OCl)Cl present in that 20 ml of diluted aliquot = 0.0037 moles

Again from 1 mol of Ca(OCl)Cl , we will get 1 mole OCl-

Thus moles of OCl- present in that 20 ml of diluted aliquot, which was titrated = 0.0037 moles