Question

1. The Molar Solubility of Ca(10), in Pure Water Temperature of the saturated solution of calcium iodate: Volume (or mass) of saturated calcium iodate solution titrated: 25 °C 10 ml (or g) Dat Trial 1 Trial 2 Trial 3 Volume of Na S o titrant Final buret reading Initial buret reading ml mL mL mL Net volume of Na So 26.621 mL 30.019 mL Calculated concentration of 10 Standardized 0.05 M sodium thiosulfate (Na,8,0,) solution (10,51(E) - TIL 10 mols.0.2 x molso (16) For each trial.calculate the concentration of IO ; show a sample calculation in the space below. Average IO concentration M

Answer 1

solution: Reactions (i) 5 + 647 + I 03 > 3120+ 31, * (i) 1 + 252032 - 21+ Sy 06? from the stichioning of reaction in mules of Iz = 1/2 X males of SiO3- from the Stoichiometry of reaction (i moles of 103 = x makes of Iz . = 1/x x moles of Seas? Males of IO3 = 1/x moles of 52832 onscanner

molasity of NA 2.5, 0g = 0.05 m Table Trail-3 Tocil-2 Tail- 1 volume of Na2 5203 used [26.621 ml 1 30.019 mb molesy Naz Sz03 13:3310 X16311.5009 moles of 10 0.2218x103 10.2502 X10-3 I volume of Tož taken / 10mL I for titration concentration of I o 10.02218M 0.02502m 7 10ml average concentration of Iog formula used in Table (All calculations) 0 Anales of Naz5103 = molesity X Volume (2) Тcu? О = 0.05 mx 26.621 mL XL - 10oome ^ = 1:3310 X to 3 mal. ☺ gmales of 105 = 1 x males of Na2SiO3 Ren = 4x 1.3310 x103 = 0.22.18 X1033 mal © concentration of 107 = moles - Volume (2) Trail = 0-2218 X10-3 lo me da Scanned with -X16oonL = 0.022 18 m ad Cari stanner

PERVJUJU Mly solubility to Calcium idelade in 0.01001 KIO 0-0 м nutesi - 6 кб 03 - = 0.01 M Conventors kion of 103" foom K103 Determination of [103 Table - 2 ] total concorrection Teybol Volume of Nar. S01 dreef 1 26.84 me Trail 2 23.81 ml 1.1905 X10-3 Andles of Na. 25203 1342 X 103 0.1994 X 1003 n: 1 Тci some 0.223 7 X163 volume of To7 taken , lome pore fortifrution 0.0198 M concentration of 1057 0.0211 M average concentration same as in case of water i.e. All calculations are previous calculations.. Total 107 concentration = 0.0211M 107 concentration come from K LO3 = 0.0lm come from a (103), = 0.0211-0.07 concentration 13 - Ca(103)2 Scanned with . CamScanner = 0.0111 m 2+ + 21 05 S (0:01+25) i

WRITER 2% = 0.0211 -0.01 2 = 0.0111 8 = 000555 (mat) Triolar pelstility of Cal101), in BPO 0.011 Kio, is 0.00555 M. Molar & lubility of Ca (203)2 in Pune Water 2 = $x[103] = 1X0.0236 M S = 0.0118 M . composision of solubilities The sulubelity of Ca (103), in o.olm KIO in less than the pure water due to commen Ion effect. And it follow the Le chatelier's principle. Ca( IC3), = 210; + Ca2+ for this equilibrium , it concentration of Iož increases, the equilibrium shift in back wash direction and hence solubility of ca (103)2. decreases. Ksp in oom KIOZ Ksp = [ca?"] [105] = 8(0:01+28) 2 = 0.005554 (0.01 + 2X0.00555)? = 0.60555x(0.0211)2 = 2:47x166 I ksp = 2 478106 1 Shanned with CamScanner