[12%] Potassium iodate (KIO3, M = 214.0) dissolves to a limited extent in water: KIO3(s) ⇌ K+(aq) + — Thermodynamics and Solubility Chemistry Question
Problem Context
[12%] Potassium iodate (KIO3, M = 214.0) dissolves to a limited extent in water:
KIO3(s) ⇌ K+(aq) + IO3–(aq)
To 100.67 g water in a well-insulated container were added small portions of solid KIO3. After each addition, the mixture was stirred extensively until the temperature stabilized. The temperatures, measured as a function of the added mass of KIO3, are shown as solid dots below. A best-fit line through the data points below 6 g KIO3 added is given for your convenience. (You may assume that the specific heat capacity and density of the solution are the same as that of pure water at all times during the experiment and may neglect the heat capacity of the insulated container.)
Calculate Ksp for KIO3 (that is, Keq for the dissolution reaction given above).
Model Answer
The KIO3 stops dissolving when the temperature stops changing, i.e. at the intersection of the solid and dashed lines. The position of this intersection can be found by reading the graph, or by setting the equations of the two lines equal to one another:
19.76 – 0.600x = 15.75
x = 6.68 g KIO3 to make a saturated solution
6.68 g KIO3/(214.0 g mol-1) = 0.0312 mol KIO3
[K+] = [IO3–] = 0.0312 mol/(0.10067 L + 0.00668 L) = 0.291 M
In the saturated solution, the dissolution reaction is at equilibrium:
Keq = Ksp of KIO3 = [K+][IO3–] = [0.291]^2 = 0.0846
Calculate ∆Hº for the dissolution reaction of KIO3.
Model Answer
qsolution = mCp∆T = (107.39 g solution)(4.184 J g-1 ºC-1)(15.75 ºC – 19.76 ºC)
qsolution = –1.80 kJ
∆Hºrxn = qrxn/(mol reacted) = –qsolution/(0.0312 mol) = +57.8 kJ mol-1
Calculate ∆Sº for the dissolution reaction of KIO3.
Model Answer
The temperature is not strictly constant in the experiment, but it is reasonable to use the average temperature of the experiment, 18 ºC = 291 K.
∆Gº = –RTln(Keq) = –(8.314 J mol-1 K-1)(291 K)ln(0.0846)
∆Gº291 K = 5.98 kJ mol-1
∆Gº = ∆Hº – T∆Sº
5980 J mol-1 = 57800 J mol-1 – (291 K)(∆Sº)
∆Sº = +178 J mol-1 K-1
One might expect dissolution reactions of ionic solids to invariably take place with large positive ∆Sº values, since an ordered solid is forming mobile ions in solution. In fact, ∆Sº values for such reactions are often small in magnitude and are frequently negative. Explain why.
Model Answer
The increase in entropy on dissolving the highly constrained solid is counterbalanced by the decrease in degrees of freedom of the solvent. Ions in aqueous solution are surrounded by a solvent shell of water molecules that are strongly oriented by the charges on the ions and so have much smaller entropies than bulk water. This contributes a negative term to the overall change in entropy of the dissolution reaction. (In the case of KIO3, the relatively large size and small charges on the ions means that this negative term is modest in magnitude; the overall entropy change of this dissolution is indeed rather positive.)