Consider the dissolving of lead iodide: PbI2 (s) →← Pb2+ (aq) + 2 I- (aq) The solubility of lead iodide (like many salts) increases at higher temperatures:
At 20°C, the molar solubility is 1.40 × 10-3 M
At 80°C, the molar solubility is 7.00 × 10-3 M
From this information, calculate ΔH° and ΔS° for the above reaction. You may assume that ΔH° andΔS° remain constant over this temperature range.