Solubility Equilibria and Ksp

Chalk, calcium carbonate, looks totally insoluble, but drop some in water and a tiny amount still dissolves until the solution is saturated. At that point solid CaCO3 is in equilibrium with the ions it releases: CaCO3(s) <→ Ca2+(aq) + CO3 2-(aq). Like any equilibrium, this one gets a constant. Because the solid's concentration never changes, it drops out, leaving Ksp = [Ca2+][CO3 2-]. That is the solubility product: the product of ion concentrations, each raised to its coefficient, when the solid is dissolving into a saturated solution.

To turn Ksp into an actual solubility, you build an ICE table starting from pure solid and water. If every mole of solid that dissolves gives one mole of each ion, both concentrations become s, and Ksp = s^2. For AgCl, Ksp = 1.8e-10, so s = √(1.8e-10) = 1.3e-5 M. That s is the molar solubility, the maximum moles of salt per liter that can dissolve. The same setup also answers whether a precipitate will form when two solutions mix: compute the reaction quotient Q using the actual concentrations right after mixing, and compare it to Ksp. If Q > Ksp, the solution is oversaturated and solid forms; if Q < Ksp, everything stays dissolved.

Ksp problems are really just ICE tables wearing a different costume, which is why this topic sits right after writing K expressions and finding equilibrium concentrations. The twist is that solubility equilibria start from a solid, not a mix of dissolved species, so the table always starts with 0 for both ions and builds up by s. That same logic also sets up the next step, the common ion effect, where adding extra A+ or X- ions from another source shifts the equilibrium left and lowers how much solid dissolves.