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Question: 3. When A 2.27 G Sample Of Liquid-Plumr Drain Cleaner Was Titrated With 0.1025 M HCI, TWO... 3. When a 2.27 g sample of Liquid-Plumr drain cleaner was titrated with 0.1025 M HCl, two endpoints were obtained. The first endpoint, shown as V, in the diagram, was at 8.43 mli the second endpoint, shown as V, in the diagram, was at 25.18 ml. Calculate the mass percent of both NaOH and NaOCl in the drain cleaner (Beware that you should not use the total volume V, to find the amount of NaOCI present, since the NaOH had already neutralised all of the acid added to reach the endpoint at V.) volume of HCI titrant (mL)

Answer 1

Total weight = 2.279 NaOH + HCl - Nael +H2O Firest end point (v.) = 8.43 ml at this end point all NaOH is newbalised by Hel. no of moles of Hell used = volumex Molarity = 8.43 x0.1025 ovo = 0,84075 mole Tour Imole of Naoh reacts with 1 mole of Hel nof moles of NaOH in cleaner =0.84075 mass of NaOH = 0.84075* Mdar mass & NaoH -084075 LO love = 0.034563 g andreaction Naou +2HCl - Nad +H2O tel second end popnt 2) = 25.18 me out of 25.18 me, 8.43 ml is abready used for neutralisation of NaOH volume of Hel used for 25.18-8:43 this reaction J = 16.75ml

Imole & Naol nof moles of reacts with 2 mees & Hol' He used= 16:15*0.1025 - mole 1.7169 1000 no f moles of Naod a nof moly q Hel in cleaner = 1.71169 2x tece = 0.856xlo mole mars of Naol = 0.856x10 xMolarmats = 0.8566103 X 74.5 = 0.0638 9 mass percent of NaOH = 0.03456 3x100 1.52% mass percent of Naoct 0.0638 x 100 = 2.814. 2.27