Question

Q 1(a) [13.33 Marks] Levels of dissolved oxygen in water can be determined by the Winkler method in the Leaving Certificate Chemistry course as follows: Under alkaline conditions manganese(II) sulphate produces a white precipitate of manganese(II) hydroxide: Mn2+(aq) + 2OH-(aq) → Mn(OH)2(s) This reacts with the dissolved oxygen to produce a brown precipitate. 2Mn(OH)2(s) + O2(aq) → 2MnO(OH)2(s) Addition of concentrated H2SO4 releases free iodine from KI. MnO(OH)2(s) + 4H+(aq) + 21-(aq) + Mn2+(aq) + 12(aq) + 3H2O(1) The free iodine is then titrated with standard sodium thiosulfate using starch as indicator. 12(aq) + 2S2O32-(aq) → 21:(aq) + S4062-(aq) (i) State how you would prepare the following solutions: 25 cm3 of 2.2M Mn(II) solution using solid MnSO4.4H2O 100 cm3 of 0.005M sodium thiosulphate solution from stock 0.1M solution Assuming the titre value of 13.05 cm3 of 0.005 M sodium thiosulphate was obtained from the experiment, determine the concentration of dissolved oxygen in a 50.0 cm3 sample of water. Express the dissolved oxygen concentration in moles/litre, g/l and also as ppm. Explain how starch is used as an indicator in this titration.

Answer 1

Molarity of Mo (T) solution 2.2M Volume to Solution 25 cm3. as mL Molar mass of Mn54. MO - 151 g/mol Molanity = Given mass x 1000 Mony Mass VML) x 1000 2.2 mol ? litre W 151. g mor! 25 litre w = 1 2.2 mol x IS19 x 25 litse litse moi 1000 W = 8305 g 8-305 g of Solid Masou. Hois dissolved into 25 cm3 to foom 2:27 Mn (1) solution 0 M, V, = M₂ (By using dilution formula) 100 crom X 0.005M = OIM XV2 V₂ = 5 cm² 5 cms of OIM stock Solution is taken in 100 cm Standard Flask and water is added upto monk in Standard Flask to make this Solution. % Number of milimoles = mdlarity x V (in cm3) milimdes of N9, S2 = 0.005 M X 13.05 cm3 - 0.06525 milimo I @q) + 25,03€ (89— 21°09) + Syros ad S. Mo 2. = 0.0326 milimol 006525 milimo le 2 Mo O OH), + 4H+ + 21 is @9 (aq) Molt (09) + I, 09) + 3H00) Problema mnoon, = 0.06525 milimele 2 Mn(OH), (2 +ą aq) — 2M COH), 6) Mon - Fryslens me = 0.06525 milimele 2 X2 Dissolved oxygen concentration - No.of milimole of Volume in cm² 2 0.06525 Milimol 4 x 500 ml = 0.000326 mol/line Dissolved oxygen concentration - 3.26 x64 mol / litse Dissolved Oxygen concentration - 3.26x10" mol x Ma litre = not x 329 3-26 X 10 litre 0001 · 104-32 X10-9 g litse Dissolved oxygen concentration - 1.043 x 109 /litse. Density of water- Ig/mL I litre - 1000 ml - 1000 Dissolved oxygen ppm concentration - 1.043x102x 10 looog : 10:43 ppm Dissolved oxygen Concentration = 10:43 ppm IL + Starch Complex Violet Colour Starch form violet Colour complex with iodine so it give information about end point during the titration.