Key Concepts of Solubility Product Constants

Why This Matters

Solubility product constants ($K_{sp}$) sit at the intersection of several major themes you'll be tested on: equilibrium principles, stoichiometry, Le Chatelier's principle, and acid-base chemistry. When you understand $K_{sp}$, you're not just memorizing another constant—you're applying equilibrium thinking to real-world scenarios like predicting whether a kidney stone will form, how water treatment plants remove heavy metals, or why antacids work the way they do.

The AP exam loves $K_{sp}$ problems because they test whether you can set up equilibrium expressions, use ICE tables, and apply Le Chatelier's principle to new situations. You're being tested on your ability to connect mathematical relationships to chemical behavior. Don't just memorize formulas—know why adding a common ion decreases solubility, how pH shifts dissolution equilibria, and when precipitation will occur based on comparing $Q$ to $K_{sp}$.

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Foundations: What $K_{sp}$ Actually Represents

Before you can manipulate $K_{sp}$ values, you need to understand what they're measuring. $K_{sp}$ is simply an equilibrium constant applied to the dissolution of sparingly soluble ionic compounds—it tells you how far the dissolution reaction proceeds before reaching equilibrium.

Definition of $K_{sp}$

• $K_{sp}$ is an equilibrium constant—it quantifies the extent to which a sparingly soluble ionic compound dissolves in water at a specific temperature
• The expression includes only dissolved ions—solid reactants don't appear in the expression because their concentration is constant
• Higher $K_{sp}$ means greater solubility—a compound with $K_{sp} = 10^{-5}$ dissolves more than one with $K_{sp} = 10^{-12}$

Writing $K_{sp}$ Expressions

• Derive from the balanced dissolution equation—for $\text{AB}_2 \rightarrow \text{A}^{2+} + 2\text{B}^-$, the expression is $K_{sp} = [\text{A}^{2+}][\text{B}^-]^2$
• Stoichiometric coefficients become exponents—this follows directly from equilibrium expression rules you learned earlier
• Products only, no reactants—pure solids have an activity of 1 and are omitted from the expression

Relationship Between $K_{sp}$ and Solubility

• Solubility (S) is the molar concentration of dissolved compound at equilibrium—it's what you can actually measure in the lab
• $K_{sp}$ and S are mathematically related but not identical—for a 1:1 salt, $K_{sp} = S^2$; for a 1:2 salt, $K_{sp} = 4S^3$
• You can convert between them using stoichiometry—this is essential for comparing solubilities of compounds with different formulas

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Calculations: From Data to Constants and Back

These calculation skills appear repeatedly on exams. Master the setup, and the math follows naturally.

Calculating $K_{sp}$ from Solubility Data

• Start with solubility in mol/L—convert from g/L if necessary using molar mass
• Use stoichiometry to find ion concentrations—if $S = 0.001$ M for $\text{PbCl}_2$, then $[\text{Pb}^{2+}] = S$ and $[\text{Cl}^-] = 2S$
• Substitute into the $K_{sp}$ expression—for $\text{PbCl}_2$: $K_{sp} = (S)(2S)^2 = 4S^3$

Comparing Solubilities Using $K_{sp}$

• Direct comparison works only for same-type salts—you can compare $K_{sp}$ values directly for two 1:1 salts like AgCl and AgBr
• Different stoichiometries require calculation—a 1:2 salt with $K_{sp} = 10^{-10}$ may be more soluble than a 1:1 salt with $K_{sp} = 10^{-8}$
• Always solve for S to compare unlike compounds—this is a common exam trap that catches students who just compare $K_{sp}$ values

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Predicting Behavior: Will It Precipitate?

This is where $K_{sp}$ becomes predictive. By comparing the ion product (Q) to $K_{sp}$, you can determine whether a solution is saturated, unsaturated, or supersaturated.

Predicting Precipitation Using $Q$ vs. $K_{sp}$

• Calculate the ion product $Q$ using actual concentrations—use the same form as the $K_{sp}$ expression but with current (not equilibrium) values
• Compare $Q$ to $K_{sp}$ to predict behavior—if $Q > K_{sp}$, precipitation occurs; if $Q < K_{sp}$, more solid can dissolve
• $Q = K_{sp}$ means the solution is exactly saturated—the system is at equilibrium with no net change

Selective Precipitation of Ions

• Add a reagent that precipitates one ion preferentially—the ion forming the compound with the lowest $K_{sp}$ precipitates first
• Separation depends on $K_{sp}$ differences—larger differences between $K_{sp}$ values give cleaner separations
• Used extensively in qualitative analysis—this technique identifies metal ions by systematically precipitating them in groups

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Factors That Shift Solubility Equilibria

Le Chatelier's principle governs how solubility responds to changes in conditions. Understanding these shifts is crucial for both multiple choice and free response questions.

Common Ion Effect

• Adding a common ion decreases solubility—Le Chatelier's principle shifts equilibrium toward the solid (left)
• The effect is quantitative and predictable—you can calculate the new, lower solubility using the $K_{sp}$ expression with the added ion concentration
• Applications include water softening and drug formulation—controlling solubility through common ions has real industrial significance

pH Effects on Solubility

• Basic anions become protonated in acidic solutions—this removes the anion from equilibrium, shifting dissolution to the right
• Salts of weak acids dissolve better at low pH—examples include carbonates, sulfides, and hydroxides
• Salts of strong acid anions are pH-independent—chlorides and nitrates don't react with $\text{H}^+$, so their solubility is unaffected by pH

Temperature Effects on $K_{sp}$

• $K_{sp}$ is temperature-dependent like all equilibrium constants—the value listed in tables applies only at that specific temperature (usually 25°C)
• Most salts have endothermic dissolution—$K_{sp}$ increases with temperature, meaning greater solubility when heated
• Some salts show decreased solubility when heated—calcium sulfate and some other compounds have exothermic dissolution, reversing the typical trend

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Quick Reference Table

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Self-Check Questions

1. Why can't you directly compare $K_{sp}$ values to determine which compound is more soluble when the compounds have different stoichiometries (e.g., AgCl vs. $\text{Ag}_2\text{CrO}_4$)?

2. A solution contains both $\text{Cl}^-$ and $\text{CrO}_4^{2-}$ ions. If you slowly add $\text{Ag}^+$, which precipitate forms first, and how would you determine when the second ion begins to precipitate?

3. Compare and contrast how the common ion effect and pH changes both affect the solubility of $\text{CaCO}_3$. Which mechanism is at play when you add $\text{Na}_2\text{CO}_3$? When you add HCl?

4. If $Q < K_{sp}$ for a solution, what does this tell you about the solution's saturation state, and what would happen if you added more solid solute?

5. An FRQ asks why $\text{Fe(OH)}_3$ dissolves in acidic solution but $\text{AgCl}$ does not. What concept explains this difference, and how would you structure your response?