Solubility Product

Insert the unknown solubility s for each ion concentration, weighted by its stoichiometric coefficient, and solve for s. For a 1:1 salt like AgCl, $K_{L} = s\cdot{}s = s^{2}$, so $s = \sqrt{K_{L}}$: with $K_{L} =\frac{ 1.7\times{}10^{-10} mol^{2}}{L^{2}}$ you get $s =\frac{ 1.3\times{}10^{-5} mol}{L}$. For an AB₂ salt like Mg(OH)₂ each formula unit releases two hydroxide ions: $K_{L} = s\cdot{}(2s)^{2} = 4s^{3}$, so $s = \sqrt[3]{(\frac{K_{L}}{4})}$. The factor in front of s must be raised to the power as well, which is the most common calculation error. Finally you can convert s with the molar mass into a mass solubility in grams per litre.

Compare the ion product Q with the solubility product K_L. Q is formed exactly like K_L, but with the concentrations actually present instead of the equilibrium values. If Q is smaller than K_L, the solution is unsaturated and more salt can dissolve. If Q equals K_L, the solution is saturated and in equilibrium with the solid. If Q exceeds the solubility product, the solution is supersaturated and the salt precipitates until Q has dropped back to K_L. When combining two solutions you must account for dilution first, because the volumes add up and lower the individual concentrations.

Because the law of mass action rigorously works with activities, and the activity of a pure solid is by definition 1. The solid has a fixed density and thus a constant "concentration" that does not change during dissolution; it is therefore absorbed into the constant. What remains is only the product of the ion concentrations in solution. This has a practical consequence: it does not matter whether one gram or one kilogram of undissolved salt sits at the bottom, the saturation concentrations above it are identical. All that matters is that some solid is present at all, so the equilibrium between dissolving and precipitating can exist.

A common ion lowers the solubility. If you add table salt to a saturated AgCl solution, [Cl⁻] rises sharply. Since the product $[Ag^{+}]\cdot{}[Cl^{-}]$ must not exceed K_L, [Ag⁺] has to fall accordingly: AgCl precipitates. Numerically, at $[Cl^{-}] =\frac{ 0.1 mol}{L}$ only $[Ag^{+}] =\frac{ K_{L}}{[Cl^{-}]} =\frac{ 1.7\times{}10^{-10}}{0.1} =\frac{ 1.7\times{}10^{-9} mol}{L}$ remains, roughly ten thousand times less than in pure water. This is Le Chatelier's principle applied to the dissolution equilibrium, and it is used deliberately in analytical chemistry to make precipitations as complete as possible, for example in gravimetric determinations.

Only if the salts have the same formula type. For two 1:1 salts like AgCl and AgBr, the smaller K_L really does mean the lower solubility. As soon as the stoichiometry differs, however, the K_L values carry different units and powers: an AB₂ salt with K_L in $\frac{mol^{3}}{L^{3}}$ cannot be compared directly with a 1:1 salt in $\frac{mol^{2}}{L^{2}}$. In that case you must calculate the molar solubility s for both and compare those values. An example: with $5.6\times{}10^{-12}$, Mg(OH)₂ has a numerically smaller K_L than some 1:1 salts, but because of the relation $s = \sqrt[3]{(\frac{K_{L}}{4})}$ it is not necessarily less soluble.