Question

4. a) A solution of potassium hypochlorite (KOCI) is prepared by dissolving 0.61 g of potassium hypochlorite in water in a 250 cmº volumetric flask and making up to the mark. What is the molarity of the solution? b) A 10 cm3 aliquot of the potassium hypochlorite solution is added to a conical flask containing an excess of potassium iodide thus generating molecular iodine (12) according to the following reaction: KOCI + 2HCl + 2K1 + 3KCl + H2O + 12 The iodine is then titrated against a solution of sodium thiosulfate, Na2S2O3, of unknown concentration. Sodium thiosulfate and iodine react as: 2Na2S2O3 + 12 + 2Nal + Na2S406 Given that the average titre of sodium thiosulfate is 11.2 cm3 calculate the concentration of the sodium thiosulfate solution clearly showing all your workings.

Answer 1

LR 256 volume z Pot. Hypochlorite (1) cnyiven mass of Porebrium hypchlovite = 0.619 volume = 250cm = 2000 - 0.230 molarity = mass 3. Polassium hypchlonte x . rotar mass A) Pot. Hypochlorite more more x strost = 0.0 269 NO (6) Griven, locus aliquot of potassium hypochlon'te į added whose molarity = 0.0264 M . rounde = 10cm = 0.= .011 volume of sodium thiosulphate = 11.2em' = 16:24 = .01122 1000 now, uring (Movi ( m2 Y2) 0,50 0.0269 x 0.01L = 0.0112 x My Mg = _0'0269x0.01 -0.0240 M 00112