Many ionic compounds dissolve only slightly in water, so a saturated solution contains both dissolved ions and undissolved solid at the same time. The solubility product, Ksp, is the equilibrium constant that describes how many ions are present in that saturated solution. It matters because Ksp helps predict whether a precipitate forms and how much of a solid can dissolve.
Molar solubility connects this equilibrium idea to a measurable concentration in mol/L.
Understanding Chemistry: Ksp and Molar Solubility
The main challenge in these problems is keeping track of the ions made by each formula unit. Molar solubility describes the amount of solid formula units that enter one liter of solution. It does not automatically equal the concentration of every ion.
A salt with equal numbers of positive and negative ions gives equal ion concentrations in pure water. A salt such as calcium fluoride gives one calcium ion for each two fluoride ions. If s moles per liter dissolve, calcium has concentration s, while fluoride has concentration two s.
The equilibrium expression must use those actual concentrations. The powers in the expression come from the ion ratio, so they can make the final calculation very different from a simple square root.
An initial, change, equilibrium table helps organize the calculation. In pure water, the dissolved ion concentrations begin near zero. Dissolving the solid creates changes based on its formula.
If the water already contains one of the ions, that initial concentration must be included. Students often forget this step and treat every situation as pure water. Sometimes a large starting concentration makes the added amount from dissolution so small that it can be estimated as negligible.
That shortcut is useful only when a check shows the added amount is truly small compared with the starting amount. Otherwise, the full equation is needed. A calculator can solve the resulting quadratic or higher order equation, but the physically meaningful answer must be positive.
Molar solubility can be converted into grams per liter by multiplying moles per liter by the compound's molar mass in grams per mole. This gives a more familiar mass value for laboratory work. It also explains why two substances can have similar molar solubilities but very different masses dissolved.
A heavy compound contributes more grams for the same number of moles. Keep the temperature stated in a problem, because solubility data usually apply only at a particular temperature.
Real solutions can contain many dissolved ions, which can alter ion behavior slightly. Introductory calculations usually treat concentrations as ideal, but this assumption becomes less accurate in concentrated solutions.
Precipitation predictions require careful attention to mixing and dilution. When two solutions are combined, the volume increases, so each ion concentration must be recalculated using the total volume before forming the ion product. Comparing that product with Ksp tells whether the mixture can remain clear at equilibrium.
This idea appears in qualitative analysis, where chemists separate metal ions by adding a reagent that precipitates one ion before another. It matters in water treatment and in hard water deposits inside pipes. A calculated tendency to precipitate does not always mean a solid appears instantly.
Crystal formation can be slow if no suitable starting surface or seed crystal is present. Equilibrium predicts the final favored condition, while kinetics affects how quickly the visible change occurs.
Key Facts
- For AgCl(s) ⇌ Ag+(aq) + Cl-(aq), Ksp = [Ag+][Cl-].
- Pure solids are not included in the Ksp expression because their activity is constant.
- Molar solubility, s, is the number of moles of solid that dissolve per liter of saturated solution.
- For CaF2(s) ⇌ Ca2+(aq) + 2F-(aq), Ksp = [Ca2+][F-]^2 = s(2s)^2 = 4s^3.
- Ion product Q = product of ion concentrations before equilibrium; if Q > Ksp, precipitation occurs.
- A common ion lowers molar solubility because Le Châtelier's principle shifts the dissolution equilibrium toward solid.
Vocabulary
- Ksp
- The solubility product constant is the equilibrium constant for a slightly soluble ionic solid dissolving into its ions.
- Molar solubility
- Molar solubility is the concentration of dissolved solid in a saturated solution, usually measured in mol/L.
- Saturated solution
- A saturated solution contains the maximum amount of dissolved solute possible at a given temperature while undissolved solid remains.
- Common-ion effect
- The common-ion effect is the decrease in solubility caused by adding an ion already present in the dissolution equilibrium.
- Ion product
- The ion product, Q, is the current product of dissolved ion concentrations used to predict whether precipitation will occur.
Common Mistakes to Avoid
- Including the solid in the Ksp expression is wrong because pure solids do not appear in equilibrium constant expressions.
- Forgetting stoichiometric coefficients is wrong because ions may form in unequal amounts, such as [F-] = 2s for CaF2.
- Assuming Ksp equals molar solubility is wrong because Ksp depends on the ion concentrations raised to powers from the balanced equation.
- Thinking a common ion changes Ksp is wrong because Ksp depends only on temperature, while the common ion changes the equilibrium concentrations and solubility.
Practice Questions
- 1 AgCl has Ksp = 1.8 x 10^-10 at 25°C. For AgCl(s) ⇌ Ag+(aq) + Cl-(aq), calculate the molar solubility in pure water.
- 2 The molar solubility of CaF2 in pure water is 2.1 x 10^-4 M. For CaF2(s) ⇌ Ca2+(aq) + 2F-(aq), calculate Ksp.
- 3 Explain why AgCl is less soluble in 0.10 M NaCl than in pure water, and state whether Ksp changes.