Solubility Product Ksp Worked Examples Cheat Sheet

A printable reference covering Ksp expressions, molar solubility, ion product Q, common ion effects, and precipitation predictions for grades 11-12.

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Solubility product, written as $K_{sp}$ , helps students predict how much of a sparingly soluble ionic compound dissolves in water. This cheat sheet focuses on worked example patterns, including writing $K_{sp}$ expressions, finding molar solubility, and deciding whether a precipitate forms. Students need these tools for equilibrium problems where solids, ions, and saturated solutions appear together. The goal is to make each setup clear before doing the calculation. The key idea is that a slightly soluble salt reaches equilibrium between its solid form and its dissolved ions. For a salt $A_mB_n(s)$ , the expression is $K_{sp} = [A^{n+}]^m[B^{m-}]^n$ , and the solid is not included. The ion product $Q$ is compared with $K_{sp}$ to predict precipitation: if $Q > K_{sp}$ , a precipitate forms. Common ions reduce solubility because they shift the dissolution equilibrium toward the solid.

Key Facts

For $A_mB_n(s) \rightleftharpoons mA^{n+}(aq) + nB^{m-}(aq)$ , the solubility product is $K_{sp} = [A^{n+}]^m[B^{m-}]^n$ .
Pure solids and liquids are not included in $K_{sp}$ expressions because their activities are treated as constant.
For $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ , if the molar solubility is $s$ , then $K_{sp} = s^2$ .
For $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$ , if the molar solubility is $s$ , then $K_{sp} = s(2s)^2 = 4s^3$ .
The ion product has the same form as $K_{sp}$ , so for $PbI_2$ , $Q = [Pb^{2+}][I^-]^2$ .
If $Q < K_{sp}$ , the solution is unsaturated and more solid can dissolve.
If $Q = K_{sp}$ , the solution is saturated and the system is at solubility equilibrium.
If $Q > K_{sp}$ , the solution is supersaturated and precipitation is predicted until $Q$ falls to $K_{sp}$ .

Vocabulary

Solubility product: The equilibrium constant $K_{sp}$ for the dissolving of a sparingly soluble ionic solid into its ions.
Molar solubility: The number of moles of a solute that dissolve per liter of solution, usually represented by $s$ in $\mathrm{mol\,L^{-1}}$ .
Saturated solution: A solution in which dissolved ions are in equilibrium with undissolved solid at a given temperature.
Ion product: The value $Q$ calculated from current ion concentrations using the same exponent pattern as the $K_{sp}$ expression.
Common ion effect: The decrease in solubility caused by adding an ion already present in the dissolution equilibrium.
Precipitate: An insoluble or slightly soluble solid that forms when ion concentrations exceed the solubility limit.

Common Mistakes to Avoid

Including the solid in the $K_{sp}$ expression is wrong because $K_{sp}$ only uses dissolved ion concentrations, not the amount of solid present.
Forgetting coefficients become exponents is wrong because $CaF_2$ gives $K_{sp} = [Ca^{2+}][F^-]^2$ , not $K_{sp} = [Ca^{2+}][F^-]$ .
Using $s$ for every ion concentration without stoichiometry is wrong because $CaF_2$ produces $[Ca^{2+}] = s$ and $[F^-] = 2s$ .
Comparing $Q$ and $K_{sp}$ with different expressions is wrong because $Q$ must be built using the same ion powers as $K_{sp}$ .
Ignoring a common ion is wrong because an initial concentration such as $[Cl^-] = 0.10\,\mathrm{M}$ can greatly reduce the solubility of $AgCl$ .

Practice Questions

1 Write the $K_{sp}$ expression for $Ba_3(PO_4)_2(s) \rightleftharpoons 3Ba^{2+}(aq) + 2PO_4^{3-}(aq)$ .
2 The molar solubility of $AgCl$ is $1.3 \times 10^{-5}\,\mathrm{M}$ . Calculate $K_{sp}$ using $K_{sp} = s^2$ .
3 For $CaF_2$ , $K_{sp} = 3.9 \times 10^{-11}$ . Find the molar solubility in pure water using $K_{sp} = 4s^3$ .
4 A solution contains extra $Cl^-$ ions before $AgCl$ dissolves. Explain why the molar solubility of $AgCl$ is lower than in pure water.

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