Question

Under what circumstances can we compare the solubilities of two salts by directly comparing the values of their solubility products?

Short answer

We can directly compare the solubilities of two salts by comparing their solubility products (Ksp) only if they have the same stoichiometry and charges on their ions - such as a 1:1, 2:2, or 3:3 ratio and equal charges (+1 and -1, +2 and -2, etc.). In these cases, larger Ksp indicates higher solubility. For salts with different stoichiometry and/or charges, we need to convert the solubility products into solubility values (molar solubility) and compare those values instead.

Step-by-step solution

Understand solubility and solubility product

Solubility refers to the maximum amount of a solute that can be dissolved in a given solvent at a specific temperature. The solubility product, denoted as Ksp, is a measure of the extent to which a salt can dissolve in a solvent (usually water). Ksp is determined by multiplying the concentrations of the ions raised to their stoichiometric coefficients in the balanced chemical equation. For example, for a general salt AB with a balanced chemical equation: A(s) <-> A^(p+) (aq) + B^(q-) (aq), the solubility product Ksp would be \(K_{sp} = [A^{p+}]^p [B^{q-}]^q\).

Comparing solubility products for salts with similar stoichiometry and charges

Two salts can be directly compared through their solubility products if they share the same stoichiometry and the charges on their ions are also the same. In simple words, the salts should have a 1:1, 2:2, or 3:3 ratio (where the coefficients in the balanced chemical equation are the same for both salts) and the charges on their ions should also be equal (such as +1 and -1, +2 and -2, etc.). In these cases, a larger solubility product indicates a higher solubility, as the salts dissociate into the same number and types of ions. For example, we can directly compare the solubilities of AgCl (Ksp = \(1.77 × 10^{-10}\)) and KCl (Ksp ≈ 143.2) because both salts have a 1:1 stoichiometry and the same charges on their ions (+1 and -1). In this case, since Ksp(KCl) > Ksp(AgCl), KCl is more soluble than AgCl.

Comparing solubility products for salts with different stoichiometry and/or charges

When the salts being compared have different stoichiometry and/or charges on their ions, a direct comparison of their solubility products is not meaningful. In these cases, we need to convert the solubility products into solubility values (typically, molar solubility) and then compare them. For example, the solubility of calcium fluoride (CaF2, Ksp = \(4 × 10^{-11}\)) cannot be directly compared with the solubility of silver chloride (AgCl, Ksp = \(1.77 × 10^{-10}\)) because the stoichiometry and charges are different (2:2 and 1:1, respectively). Under this circumstance, we should calculate the molar solubility of each salt and then make the comparison. In conclusion, we can compare the solubilities of two salts by directly comparing their solubility products if they have the same stoichiometry and the charges on their ions are also the same. In cases where stoichiometry and/or charges are different, we need to convert the solubility products into solubility values and compare those values instead.