Which silver salt has the highest molar solubility in water?

Question

BlackTom AI · Accepted Answer

To determine which silver salt has the highest molar solubility in water, we must compare the solubility s for each salt based on its Ksp and dissolution equation.
Option 1: Ag2CrO4 (Ksp = 2.6 × 10^-12). For a salt that dissociates as Ag2CrO4 ⇌ 2 Ag+ + CrO4^2-, Ksp = [Ag+]^2 [CrO4^2-] = (2s)^2 (s) = 4 s^3. Therefore s = (Ksp/4)^(1/3) = (2.6 × 10^-12 / 4)^(1/3) ≈ (6.5 × 10^-13)^(1/3) ≈ 8.7 × 10^-5 M. This is relatively small due to the cubic root of a very small number.
Option 2: AgBr (Ksp = 5.0 × 10^-13). For a 1:1 salt, Ksp = s^2, so s = sqrt(Ksp) = sqrt(5.0 × 10^-13) ≈ 7.1 × 10^-7 M. The square root of a small Ksp yields a very small solubility, indicating low dissolution.
Option 3: AgCl (Ksp = 1.8 × 10^-10). Also a 1:1 salt, so s = sqrt(Ksp) = sqrt(1.8 × 10^-10) ≈ 1.34 × 10^-5 M. This is larger than the previous two options but still far smaller than the cubic-root result for Ag2CrO4.
Option 4: AgI (Ksp = 8.3 × 10^-17). Again 1:1, so s = sqrt(Ksp) = sqrt(8.3 × 10^-17) ≈ 9.1 × 10^-9 M. This is extremely insoluble due to the very small Ksp value.
Comparing the calculated solubilities: approximately 8.7 × 10^-5 M (Ag2CrO4) > 1.34 × 10^-5 M (AgCl) > 7.1 × 10^-7 M (AgBr) > 9.1 × 10^-9 M (AgI). The largest molar solubility corresponds to the first option, which reflects the cubic-root dependence for the sparingly soluble salt Ag2CrO4 versus the square-root dependences for the 1:1 salts. In other words, the interplay between stoichiometry (2:1 for Ag2CrO4) and the magnitude of Ksp drives s, and the calculation shows Ag2CrO4 yields the highest s among the listed salts.

Which silver salt has the highest molar solubility in water?Single choice

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