Solubility describes how much of a substance can dissolve in a solvent before the solution becomes saturated. For ionic solids in water, dissolving means the crystal separates into positive and negative ions. The solubility product constant, Ksp, gives a precise way to describe the equilibrium between undissolved solid and dissolved ions.
This matters in chemistry because precipitation, water hardness, mineral formation, and many lab separations depend on whether ions stay dissolved or form a solid.
Understanding Solubility and Ksp
The size of Ksp cannot be compared between different salts without care. Each salt breaks apart in its own ratio, and that ratio controls the powers used in the equilibrium expression. A salt that releases two fluoride ions for every calcium ion has a squared fluoride concentration in its expression.
This makes the calculation more sensitive to fluoride levels. A very small Ksp usually points to low solubility, but it does not directly tell the number of moles that dissolve per liter unless the ion ratio is considered. Students should begin every problem by writing the balanced dissolving reaction in words or symbols, then use its coefficients to connect the ion concentrations.
Molar solubility is often represented by one unknown amount. Suppose one mole of a solid produces one mole of positive ion and two moles of negative ion. If the molar solubility is s, the positive ion concentration is s and the negative ion concentration is two times s.
Substituting those values into the Ksp expression gives an equation that can be solved for s. This method works only when the water starts without those ions.
If one ion is already present, its starting concentration must be included. The dissolved amount may then be much smaller than the value found for pure water.
Qsp is useful before equilibrium has been reached. It is calculated from the ion concentrations present at a particular moment, such as just after two solutions are mixed. Comparing Qsp with Ksp predicts the direction of change.
When Qsp is smaller, more solid can dissolve if solid is available. When Qsp equals Ksp, the system is at equilibrium. When Qsp is larger, ions leave the solution and form a precipitate until the concentrations return to the equilibrium condition.
Mixing changes volume, so concentrations must be found using total moles divided by the new total volume. Forgetting this dilution step is a common source of wrong predictions.
The common ion effect appears in many real settings. Hard water can form scale inside pipes or kettles when dissolved ions reach conditions that favor a solid mineral. In a laboratory, chemists use controlled additions of ions to separate mixtures.
One ion may form a precipitate with a reagent while another stays dissolved. This is called selective precipitation. The order of precipitation depends on both Ksp values and the actual ion concentrations, not on Ksp alone.
Temperature, acidity, and complex ions can change solubility too. For school problems, check whether these extra conditions are given. If they are not, use the stated equilibrium model and keep units consistent throughout the calculation.
Key Facts
- For AgCl(s) ⇌ Ag+(aq) + Cl-(aq), Ksp = [Ag+][Cl-].
- Pure solids are not included in Ksp expressions because their activity is treated as constant.
- For CaF2(s) ⇌ Ca2+(aq) + 2F-(aq), Ksp = [Ca2+][F-]^2.
- Molar solubility is the number of moles of solid that dissolve per liter of solution at equilibrium.
- A precipitate forms when the ion product Qsp is greater than Ksp.
- Adding a common ion usually lowers solubility because equilibrium shifts toward the solid.
Vocabulary
- Solubility
- Solubility is the maximum amount of a substance that can dissolve in a given amount of solvent at a specific temperature.
- Saturated solution
- A saturated solution contains the maximum dissolved solute possible while remaining in equilibrium with undissolved solute.
- Ksp
- Ksp is the equilibrium constant for the dissolving of a slightly soluble ionic compound.
- Ion product
- The ion product, Qsp, is the current product of ion concentrations before or during equilibrium comparison.
- Common ion effect
- The common ion effect is the decrease in solubility caused by adding an ion already present in the dissolving equilibrium.
Common Mistakes to Avoid
- Including the solid in the Ksp expression is wrong because pure solids do not appear in equilibrium constant expressions.
- Forgetting coefficients become exponents is wrong because a salt such as CaF2 gives Ksp = [Ca2+][F-]^2, not [Ca2+][F-].
- Assuming larger Ksp always means larger solubility is wrong because different formulas produce different ion ratios, so Ksp values must be compared carefully.
- Ignoring the common ion effect is wrong because added ions change equilibrium concentrations and can greatly reduce how much solid dissolves.
Practice Questions
- 1 The Ksp of AgCl is 1.8 x 10^-10 at 25°C. What is the molar solubility of AgCl in pure water?
- 2 For CaF2, Ksp = 3.9 x 10^-11. If the molar solubility is s, write the Ksp expression in terms of s and solve for s.
- 3 A solution contains dissolved Pb2+ and I- ions, and a small amount of solid PbI2 is visible at the bottom. Explain what must be true about the rates of dissolving and crystallizing at equilibrium.