Final answer:
Among the listed compounds, PbCO3 (lead(II) carbonate) has a solubility greater than that calculated from its solubility product due to hydrolysis of the carbonate ion, which produces hydroxide ions and increases pH, leading to increased dissolution. The correct option is c.
Step-by-step explanation:
The student has asked which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO3, CuI, PbCO3, PbCl2, Tl2S, KClO4.
To determine which compounds' solubilities are increased due to anion hydrolysis, we need to identify the anions that can undergo hydrolysis in water and produce hydroxide ions, which raise the pH and lead to additional dissolution of the compound.
Among the options given, PbCO3 (lead(II) carbonate) is the compound that experiences increased solubility due to the hydrolysis of its anion. The carbonate ion (CO32-) can react with water to form bicarbonate (HCO3-) and hydroxide (OH-) ions. This hydrolysis raises the pH of the solution, which can lead to resolubilization of PbCO3 to a greater extent than predicted by its Ksp alone. The correct option is c.