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

2. The Molar Solubility of Calcium lodate in 0.0100 M Potassium lodate Temperature of the saturated solution of calcium iodate: 25°C Volume (or mass) of saturated calcium iodate solution titrated: 10 ml (or g) Dat a Trial 1 Trial 2 Trial 3 mL mL Volume of Na S. titrant Final buret reading Initial buret reading Net volume of Na S EE ml mL 26.84 mL 23.81 mL mL Calculated concentration of 10 M For each trial, calculate the concentration of 10 ; show a sample calculation in the space below. (10,-|(m- L., O x mols,o, xumolto (16) 6 mol S,0,- Average I concentration M

Subtract the concentration of IO ion that came from the KIO from the average value of the total IO concentration to get the iodate ion concentration that came from dissolved Ca(IO). Total 10 3 concentration M IO3 concentration from KIO; —_ M 10 concentration from dissolved Ca(IO) Calculate the molar solubility s of Ca(IO), in 0.0100 M KIO solution. [10,-) = 0.01 + 2 (12) Molar solubility mol/L Compare the molar solubility of Ca(IO) in 0.0100 MKIO solution with the molar solubility in pure water, determined in Part 1. Is the change consistent with Le Châtelier’s principle? Calculate the solubility product constant, K , for a saturated solution of Ca(IO) in 0.0100 M KIO (see Equation (13)]. K = [Ca2+][IO, -12 = s(0.01 + 2s)2 (13) K =