Molar solubility is the number of moles of a salt that dissolve per liter of solution at saturation (units of M). On the AP Chem exam, you calculate it from Ksp using an ICE table, where the relationship depends on the stoichiometry of the dissolution equation (EK 7.11.A.2).
Molar solubility is how much of a salt actually dissolves, measured in moles per liter, before the solution hits saturation. It usually gets the symbol s in your ICE table. If CaF₂ has a molar solubility of s, then a saturated solution contains s mol/L of Ca²⁺ and 2s mol/L of F⁻, because every formula unit that dissolves releases one calcium ion and two fluoride ions.
Here's the move the AP exam tests over and over. Dissolution is an equilibrium (EK 7.11.A.1), so molar solubility and Ksp are two sides of the same coin, linked by stoichiometry. For AgCl (1:1 salt), Ksp = s². For CaF₂ (1:2 salt), Ksp = (s)(2s)² = 4s³. For a salt like M₃X₂, Ksp = (3s)³(2s)² = 108s⁵. That coefficient-to-exponent translation is exactly what EK 7.11.A.2 means when it says the relationship between solubility and Ksp "depends on the stoichiometry of the dissolution reaction." It also runs in reverse. If you measure the molar solubility of a saturated solution, you can calculate Ksp (EK 7.11.A.4).
Molar solubility lives in Unit 7 (Equilibrium), specifically Topics 7.11 and 7.12. Learning objective 7.11.A asks you to calculate the solubility of a salt from its Ksp, and 7.12.A asks you to do the same thing when a common ion is already floating around in solution. The common-ion effect (EK 7.12.A.1) is pure Le Châtelier. If the solution already contains one of the salt's ions, the dissolution equilibrium shifts back toward the solid, so molar solubility drops below its pure-water value. This concept is also one of the cleanest bridges in the course, connecting the qualitative solubility rules from Unit 4 to quantitative equilibrium math. EK 7.11.A.3 makes that link explicit, since Ksp values greater than 1 correspond to the salts you memorized as "soluble."
Keep studying AP Chemistry Unit 7
Solubility Product Constant, Ksp (Unit 7)
Ksp is the equilibrium constant for dissolution; molar solubility is the answer you extract from it. They're related but not interchangeable, and the conversion between them always goes through stoichiometry. Comparing raw Ksp values to rank solubility only works when the salts have the same ion ratio.
Common Ion Effect (Unit 7)
Drop a salt into a solution that already contains one of its ions and its molar solubility shrinks. Le Châtelier explains why (the equilibrium shifts toward the solid), and the Ksp expression lets you calculate exactly how much. This is the Topic 7.12 twist on the basic 7.11 calculation.
Precipitation Reactions and Solubility Rules (Unit 4)
The solubility rules from Topic 4.7 are the qualitative version of this whole story. Unit 7 puts numbers on them. "Insoluble" really means a tiny molar solubility and a Ksp far below 1, while genuinely soluble salts have Ksp values greater than 1 (EK 7.11.A.3).
Stoichiometric Coefficients (Units 4 & 7)
Coefficients in the dissolution equation become multipliers in the ICE table and exponents in the Ksp expression. Getting 2s versus s wrong is the single most common error in these problems, so the balanced equation is step one every time.
Molar solubility shows up in three main flavors. First, the straight calculation, like finding the molar solubility of Ag₂CrO₄ in pure water from its Ksp of 1.1 × 10⁻¹², where you have to set Ksp = (2s)²(s) = 4s³. Second, the symbolic version, where a multiple-choice stem gives you a generic formula like M₃X₂ and asks which expression correctly relates s to Ksp. No numbers, just stoichiometry. Third, the common-ion version, where ions are already present, such as finding the molar solubility of Cu(IO₃)₂ in a solution that already contains Cu²⁺ and IO₃⁻, or finding the solubility of Cu(OH)₂ in a buffer at pH 10 (the buffer fixes [OH⁻], which simplifies the math). On free-response questions, expect to write the dissolution equation, build an ICE table, solve for s, and then justify a solubility change qualitatively using Le Châtelier when a common ion is added. Show the Ksp expression explicitly; that step earns points.
Ksp is an equilibrium constant with no real concentration meaning on its own. Molar solubility is an actual concentration, the mol/L of salt that dissolves at saturation. You convert between them using the stoichiometry of the dissolution equation, so a salt with a smaller Ksp can actually have a higher molar solubility if its formula releases more ions. That's why "compare the Ksp values" only works for salts with matching ion ratios.
Molar solubility (s) is the moles of salt that dissolve per liter of solution at saturation, and it has units of molarity.
To find s from Ksp, write the balanced dissolution equation, set up an ICE table, and let the coefficients multiply s (so a 1:2 salt gives Ksp = 4s³).
The relationship between molar solubility and Ksp depends entirely on stoichiometry, so you can't compare solubilities of different salt types just by comparing Ksp values.
A common ion already in solution lowers molar solubility, which you can explain with Le Châtelier or calculate by plugging the existing ion concentration into the Ksp expression.
The process runs both ways, so a measured molar solubility lets you calculate Ksp, and a known Ksp lets you predict solubility.
Ksp values greater than 1 correspond to the soluble salts from the Unit 4 solubility rules, tying the qualitative rules to quantitative equilibrium.
It's the number of moles of a salt that dissolve per liter of solution before saturation, usually labeled s in an ICE table. You calculate it from Ksp using the stoichiometry of the dissolution equation, which is learning objective 7.11.A.
No. Ksp is an equilibrium constant, while molar solubility is an actual concentration in mol/L. For AgCl they're related by Ksp = s², but for CaF₂ it's Ksp = 4s³, so the conversion changes with the salt's formula.
No, and this is a classic trap. Ksp comparisons only predict solubility rankings for salts with the same ion ratio (like comparing two 1:1 salts). A 1:2 salt with a smaller Ksp can still dissolve more than a 1:1 salt, because the math relating s to Ksp is different.
It lowers it. If the solution already contains one of the salt's ions, the dissolution equilibrium shifts toward the solid (Le Châtelier), so less dissolves. Quantitatively, you plug the existing ion concentration into the Ksp expression instead of starting from zero, per EK 7.12.A.1.
Every formula unit of CaF₂ that dissolves releases one Ca²⁺ and two F⁻ ions. So if s mol/L of the salt dissolves, [F⁻] = 2s, and that 2s gets squared in the Ksp expression, giving Ksp = (s)(2s)² = 4s³.
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