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

Solutionlet w first calculate the No. of moles of Na, S202 - solution - Molarity of Na2SO3 = 0.05 m This means 5:0 1000 ml of solution contains Nag,03 - 0-05 moles .: 1 ml of solution contains Na, S2O3 = 0.05 moles 1000 to for trial 1 : Volume of Na, S, O2 = 26.621 me :: 26:621 ml of solution contains Nag S, 02 - DOS y 26:621 1000 = 0.00133 moles For brail 2 : Volume of Nq,8202 : 30.019 ml .: 30.0141 ml of solution contains Na, S202 = 0.05 x 30.019 1000 = 0.0015 moles Consider the reaction Na2S2O3 = 2 Nat + S202 It is clear from the reartion that one mole of Naz S203 gives one mole of $20 - ions :: For trial 1: No.ofas,62 ions : No.of moles of Na2S2O3 = 0.00133 molus for trial 2: No. of moles of 520 = 0.0015 moles Consider another reaction between iodate and thiosulfate ions 10, + 65,03 + 64 G 1 + 354062" + 3460 1. It is clear from the above reaction 6 moles of S₂0 reacts with one mole of IO ion. . I molo of S₂0² on reacts with Imoles of 10g ions. CScanned with CamScanner

toial 1: Noi of moles of 10, ion = 6* No. of moles of $20,1 ions - 6x0.00133 = 0.00798 moles For trial 2: No. of moles of 10 ions = 6x0.0015 = 0.009 moles Now, volume caleni calcium ìodate used = 10 ml = 0.012 For trial 1: Molosity of Tojions = No.of moles of 10ions - Volume of solution 0.00798 0.01 0:01 S .: (10,-) = 0.798 M For trial 2 : Molarity of 100+ ions = 0.099 = 0.9 M [10]= 0.9M Average (103) = 0.798 +0.9 = 0:49M 2 Now, we know that Molar Solubility is given by (10,*) = 2s. i s = [103] = 0.849 = 0.4245 2 2 Cleomes with cánce [s= 04245

Finally, the solubility product constant is given by Kop=16a??](10,*)= $(25) = 452 * Ksp = 482 = 4x (0.42452 Ksp = 0.12 0.72 CScanned with CamScanner