7.11 Introduction to Solubility Equilibria

Solubility equilibria treat the dissolving of a slightly soluble salt as a reversible reaction with its own equilibrium constant, $K_{sp}$, called the solubility product. If you can write a balanced dissolution equation and set up an ICE table, you can calculate molar solubility from $K_{sp}$ or calculate $K_{sp}$ from measured solubility. For AP Chemistry, leave the solid out of the $K_{sp}$ expression.

Ksp AP Chem

In AP Chem, Ksp is the equilibrium constant for a slightly soluble ionic compound dissolving in water. Write the balanced dissolution equation, leave the solid out of the equilibrium expression, and raise each ion concentration to its coefficient.

For Topic 7.11, the important move is connecting Ksp to molar solubility. Let x represent the amount of solid that dissolves, use the balanced equation to express ion concentrations in terms of x, then solve the Ksp expression. To predict precipitation, compare Qsp to Ksp.

Why This Matters for the AP Chemistry Exam

Solubility equilibria connect everything you learned about equilibrium to the specific case of salts dissolving in water. The exam expects you to pick the right relationship for a problem, which here means writing a correct Ksp expression from a balanced equation and using stoichiometry to link Ksp and solubility.

This topic rewards careful setup. Because the math depends entirely on the dissolution stoichiometry, a 1:1 salt and a 2:1 salt give different relationships between Ksp and solubility even if their Ksp values look similar. You will use these skills again in the common-ion effect and in how pH affects solubility, so getting comfortable with the basics now pays off later in the course.

Key Takeaways

• The dissolving of a salt is a reversible process, so it reaches equilibrium and has an equilibrium constant called Ksp.
• Write the Ksp expression from the balanced dissolution equation, and leave the solid out because solids do not appear in equilibrium expressions.
• The link between Ksp and molar solubility depends on the dissolution stoichiometry, so always start from the balanced equation.
• A Ksp value greater than 1 corresponds to a soluble salt, which matches the solubility rules you already know.
• Convert solubility from g/L to mol/L before plugging into a Ksp calculation.
• Compare Qsp to Ksp to predict whether a solution will precipitate, dissolve more solid, or sit at equilibrium.

What Is Solubility Equilibria?

You spent earlier topics deciding whether a solute can dissolve in a solvent, almost always water. Now that you understand equilibrium, you can apply those ideas to compounds you once labeled "insoluble."

In reality, everything dissolves to some extent. The question is to what extent. NaCl dissolves much more easily than a substance like PbI2, which we usually call insoluble. But you can still write the dissolution reaction for PbI2:

PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)

You might think this reaction simply does not happen because PbI2 is "insoluble." Instead, the reaction just has a tiny K value. For lead iodide, that value is 4.41 x 10^-9. When the equilibrium is a dissolution, we call this constant Ksp, where sp stands for solubility product.

It is called the solubility product because finding Ksp means multiplying the ion concentrations raised to their stoichiometric coefficients. For lead iodide that is [Pb2+][I-]^2. The solid is left out because solids do not appear in equilibrium expressions.

Ksp values let you compare how soluble different salts are. For salts with the same ion ratio (like 1:1 salts such as AgCl and AgBr), the one with the larger Ksp is more soluble. For example, if AgCl has Ksp = 1.8 × 10^-10 and AgBr has Ksp = 5.0 × 10^-13, then AgCl is more soluble than AgBr.

For very soluble compounds such as NaCl or KOH, Ksp is well above 1, which reflects how far the dissolution goes forward. This ties back to the solubility rules you learned earlier. Salts that count as "soluble" by those rules have Ksp values greater than 1, while "insoluble" salts (like carbonates, sulfides, and most metal hydroxides) have Ksp values much less than 1. When Ksp is large (greater than 1), you can often treat the reaction as going mostly in one direction because most or all of the solute dissolves. You cannot treat the reaction that way when Ksp is very small, as with lead iodide.

Calculating Ksp

Solubility usually shows up two ways: grams per liter and moles per liter (molarity). When you calculate Ksp or find solubility, convert from grams per liter to molarity first.

Here is a sample problem for converting and finding the solubility product:

Start by writing the dissolution of lead chromate:

PbCrO4(s) ⇌ Pb2+(aq) + CrO42-(aq)

Next, convert from g/L to mol/L:

4.5 * 10^-5 g/L * 1mol/323.2g = 1.4 * 10^-7 mol/L.

Finally, plug into the solubility product expression:

Ksp = [1.4 * 10^-7][1.4 * 10^-7] = 1.9 * 10^-14.

If your equation has coefficients, adjust the equilibrium concentrations before solving for Ksp.

Calculating Molar Solubility

Calculating molar solubility works just like solving for equilibrium concentrations. Keep in mind that the solid reactant is not included in the equilibrium expression. The (R)ICE table is your tool here. Apply these ideas to a practice problem:

Calculate the molar solubility of CaCO3 (Ksp = 3.36 x 10^-9).

Write the dissolution equation for CaCO3:

CaCO3 (s) ⇌ Ca2+ (aq) + CO32- (aq)

Fill out an ICE table and solve for concentrations:

Plug into Ksp and solve: x^2 = 3.36 x 10^(-9) ⇒ x = 5.8*10^(-5).

Here x is the molar solubility in mol/L, because it equals the amount of CaCO3 that dissolved per liter. For a salt with different stoichiometry, the setup changes. For PbI2(s) ⇌ Pb2+(aq) + 2I-(aq), the iodide concentration is 2x, so Ksp = (x)(2x)^2 = 4x^3. This is why you always start from the balanced equation.

Relating Qsp and Ksp

Just like regular equilibrium, you can compare the reaction quotient to Ksp to predict whether a precipitate will form. The reaction quotient for a dissolution is often written as Qsp.

• If Qsp > Ksp, the solution has more dissolved ions than equilibrium allows, so solid will precipitate out.
• If Qsp < Ksp, the solution is unsaturated, so more solid can dissolve.
• If Qsp = Ksp, the solution is saturated and sitting at equilibrium.

How to Use This on the AP Chemistry Exam

Problem Solving

• Write the balanced dissolution equation first. Every later step depends on getting the stoichiometry right.
• Build the Ksp expression by multiplying ion concentrations raised to their coefficients, and leave the solid out.
• Use an ICE table when going from Ksp to molar solubility. Let x be the molar solubility, then express each ion in terms of x using the coefficients.
• Convert g/L to mol/L before using any solubility value in a Ksp calculation.
• Watch your units. Molar solubility comes out in mol/L.

Common Trap

• Do not assume each ion concentration equals x. For a 2:1 or 3:1 salt, one ion is 2x or 3x, which changes the Ksp expression.
• Do not include the solid in the Ksp expression or treat it as a concentration.
• Do not compare Ksp values directly across salts with different ion ratios to rank solubility. That shortcut only works when the stoichiometry matches.

Common Misconceptions

• "Insoluble salts do not dissolve at all." They dissolve a tiny bit. They just have very small Ksp values.
• "A bigger Ksp always means more soluble." That is only reliable when you compare salts with the same ion ratio. Different stoichiometry can flip the ranking.
• "Ksp and molar solubility are the same number." They are related but not equal. The relationship depends on the dissolution stoichiometry.
• "The solid belongs in the Ksp expression." Pure solids and pure liquids never appear in equilibrium expressions.
• "Qsp works differently from Q." It follows the same logic as any reaction quotient. Compare it to Ksp the same way you compare Q to K.

Related AP Chemistry Guides

• 7.1 Introduction to Equilibrium
• Unit 7 Overview: Equilibrium
• 7.2 Direction of Reversible Reactions
• 7.3 Reaction Quotient and Equilibrium Constant
• 7.4 Calculating the Equilibrium Constant
• 7.5 Magnitude of the Equilibrium Constant