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Chapter 17: Problem 97

\(\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)\) has a larger \(K_{\mathrm{sp}}\) value than that of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{\mathrm{sp}}=1.4 \times 10^{-5}\right)\). Does it fol- low that \(\mathrm{CaSO}_{4}\) also has greater solubility \((\mathrm{g} / \mathrm{L}) ?\)

Short Answer

Expert verified
No, it does not follow that \(CaSO_{4}\) has greater solubility (in g/L) than \(Ag_{2}SO_{4}\) just because its \(K_{sp}\) value is higher.

Step by step solution

01

Identify solubility product constants

The \(K_{sp}\) value of \(CaSO_{4}\) is given as \(2.4 \times 10^{-5}\) and of \(Ag_{2}SO_{4}\) as \(1.4 \times 10^{-5}\). This indicates that at equilibrium, the product of the concentration of the ions in solution is greater for \(CaSO_{4}\) than for \(Ag_{2}SO_{4}\).
02

Identify the equilibrium reactions

The salts dissolve in water dissociating into ions: \(CaSO_{4} \rightarrow Ca^{2+} + SO^{2-}_{4}\), contributing 1+1=2 particles for each formula unit dissolved, whereas \(Ag_{2}SO_{4} \rightarrow 2Ag^{+} + SO^{2-}_{4}\), contributing 2+1=3 particles per formula unit dissolved.
03

Compare the number of ions

From the dissolution reactions, you can observe that \(Ag_{2}SO_{4}\) breaks up into more ions in solutions than \(CaSO_{4}\). This factor can affect the solubility (in g/L) of the salts because it influences the mass of the salt that can be dissolved in the solution.
04

Conclusion

Given that \(Ag_{2}SO_{4}\) produces more ions when it dissolves and since the solubility depends on the number of moles of ions per liter of solution, it is not certain that \(CaSO_{4}\) has greater solubility just because it has a bigger \(K_{sp}\). Additional information, such as molar mass and temperature, would be needed to accurate determine which of the two salts is more soluble (in g/L).

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