Group 2 iodates such as Ca(IO3)2 and Ba(IO3)2 are sparingly soluble compounds. The Ksp of Ba(IO3)2 i — Solubility Equilibria and Thermodynamics Chemistry Question
Problem Context
Group 2 iodates such as Ca(IO3)2 and Ba(IO3)2 are sparingly soluble compounds.
The Ksp of Ba(IO3)2 is measured by the following experiment: An aqueous solution is saturated with Ba(IO3)2 at 298 K and a 10.00 mL aliquot of this solution (free of any solid) is added to a solution containing an excess of KI in 1 M aqueous HCl, causing the solution to turn yellow. This yellow solution is then titrated with 0.0150 M Na2S2O3 solution until the color disappears; 8.00 mL of the thiosulfate solution is required to reach this endpoint.
Write balanced net ionic equations for:
i. the reaction of the saturated Ba(IO3)2 solution with the acidified KI solution;
Model Answer
IO3 –(aq) + 8 I–(aq) + 6 H+(aq) → 3 I3 –(aq) + 3 H2O(l)
Write balanced net ionic equations for:
ii. the reaction of the yellow solution with the sodium thiosulfate solution.
Model Answer
I3 –(aq) + 2 S2O3 2-(aq) → 3 I–(aq) + S4O6 2-(aq)
Calculate the Ksp of Ba(IO3)2 at 298 K.
Model Answer
(8.00 × 10^-3 L titrant) × (0.0150 mol L^-1) = 1.20 × 10^-4 mol S2O3 2-
(1.20 × 10^-4 mol S2O3 2-) × (1 mol IO3 –/6 mol S2O3 2-) = 2.00 × 10^-5 mol IO3 –
So the concentration of IO3 – in the saturated Ba(IO3)2 solution is (2.00 × 10^-5 mol)/(0.01000 L) = 2.00 × 10^-3 M
[Ba2+] = 0.5[IO3 –] = 1.00 × 10^-3 M
Ksp = [Ba2+][IO3 –]^2 = (1.00 × 10^-3)(2.00 × 10^-3)^2 = 4.00 × 10^-9
The Ksp of Ca(IO3)2 can be measured by similar experiments. Below is shown a plot of ln(Ksp) of Ca(IO3)2 as a function of 1/T.
Calculate ∆Hº and ∆Sº for the dissolution of Ca(IO3)2 in the range of T = 273 to 317 K.
Model Answer
The figure given is a van't Hoff plot for the dissolution reaction of Ca(IO3)2, since Ksp = Keq for the dissolution reaction. Since ln(Keq) = –(∆Gº/RT) = (–∆Hº/R)(1/T) + (∆Sº/R), the slope of the plot is –∆Hº/R and the intercept is ∆Sº/R.
∆Hº = –(8.314 J mol^-1 K^-1)(–11400 K) = 94.8 kJ mol^-1
∆Sº = (8.314 J mol^-1 K^-1)(24.3) = 202 J mol^-1 K^-1
The ln(Ksp) vs. 1/T plot has a break in the curve because the stable crystalline form of calcium iodate changes in the temperature range shown. At some temperatures, the stable form is anhydrous Ca(IO3)2, while at other temperatures the hexahydrate Ca(IO3)2•6 H2O is the stable form. Assign which form is stable at which temperatures, and justify your assignment.
Model Answer
Dissolution of calcium iodate in the 317-363 K range takes place with a much smaller ∆Hº (2.7 kJ mol^-1) and much smaller (more negative) ∆Sº (–88.38 J mol^-1 K^-1) than in the 273-317 K range discussed in (c). The hexahydrate will certainly dissolve with a more positive ∆Sº, since the six lattice waters will have many more degrees of freedom in bulk water than they do in the crystal lattice. Thus Ca(IO3)2•6 H2O must correspond to the stable phase at low temperature (273-317 K) and the anhydrous Ca(IO3)2 must be stable in the 317-363 K range.