Final answer:
The compound with a solubility greater in a 0.01-M solution of HClO4 than in pure water is CaCO3, due to the formation of soluble H2CO3 upon reaction with H+ ions. Additionally, CaF2 would have greater solubility due to the formation of HF, which reduces the concentration of fluoride ions in the solution.
Step-by-step explanation:
When considering the solubility of certain compounds in a 0.01-M solution of HClO4 compared to their solubility in pure water, one must take into account the common ion effect and the potential for complex formation. HClO4 is a strong acid and will dissociate completely in solution to form ClO4- and H+ ions.
- CuCl will not have increased solubility in the presence of HClO4 as it does not contain overlapping ions or form complexes with these ions.
- CaCO3 is largely insoluble in water, but in the presence of extra H+ ions from HClO4, the equilibrium can shift to form H2CO3, thereby increasing its solubility.
- MnS solubility would not be substantially increased by HClO4, as no common ions are shared and no favorable reaction with the acid occurs.
- PbBr2 solubility should not be affected by the addition of HClO4 under common conditions as no complexing or common ions are involved.
- CaF2 solubility can increase because HF can form by reaction of fluoride ions with HClO4, reducing the fluoride ion concentration and increasing solubility due to Le Chatelier's principle.
The most notable increase in solubility in a 0.01-M solution of HClO4 over pure water would be seen in compounds that can react with H+ to form soluble products—primarily CaCO3— and those that can undergo complexation or other reactions that reduce the concentration of one of the ions in the solid (such as CaF2 forming HF).