DAT General Chemistry: Ksp and Molar Solubility

DAT High-Yield Concept

Solubility Product Constant ( $K_{\text{sp}}$ ): The equilibrium constant for a sparingly soluble ionic compound dissolving in water. Pure solids are always omitted from equilibrium expressions.
Stoichiometric Exponents: Coefficients in the balanced dissolution equation become exponents in the $K_{\text{sp}}$ expression AND multiplier coefficients when substituting molar solubility ( $x$ ).
Molar Solubility ( $x$ ): The maximum moles of ionic solute that dissolve per liter of solution to reach saturation ( $\pu{mol/L}$ ).
Comparing Solubilities: Direct comparison of $K_{\text{sp}}$ values to determine relative solubility is valid ONLY for salts that yield the same number of ions per formula unit.

The Solubility Product Constant ( $K_{\text{sp}}$ )

When a sparingly soluble ionic compound is added to water, it establishes a dynamic equilibrium between the undissolved solid and its dissociated aqueous ions in a saturated solution.

Concept Focus: Saturated Solution Equilibrium

In a saturated solution at a fixed temperature, dissolution and precipitation occur at equal rates. Because the concentration of pure solid salt remains constant, solids are excluded from the equilibrium expression.

For a general sparingly soluble salt dissolving in aqueous solution:

\ce{M_aX_b(s) <=> a M^{b+}(aq) + b X^{a-}(aq)}

The solubility product constant expression is written as:

K_{\text{sp}} = [\ce{M^{b+}}]^a [\ce{X^{a-}}]^b

Worked Example: Writing $K_{\text{sp}}$ Expressions

Problem: Write the balanced dissolution equilibrium and $K_{\text{sp}}$ expression for calcium phosphate, $\ce{Ca3(PO4)2}$ .

Step-by-Step Solution:

Write the balanced dissociation reaction: $\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO4^{3-}(aq)}$
Formulate the $K_{\text{sp}}$ expression by placing stoichiometric coefficients as exponents: $K_{\text{sp}} = [\ce{Ca^{2+}}]^3 [\ce{PO4^{3-}}]^2$

DAT Exam Trap: Do not compare raw $K_{\text{sp}}$ values to rank solubility between compounds with different ion stoichiometries (e.g., comparing $\ce{AgCl}$ to $\ce{CaF2}$ ). You must solve for molar solubility ( $x$ ) first.

Calculating Molar Solubility from $K_{\text{sp}}$

Reaction Reference

From Ksp to molar solubility, one formula type at a time — A solubility product is an ordinary equilibrium constant for a solid dissolving, so the pure solid gets no term and Ksp is just the product of the dissolved ion concentrations, each raised to its coefficient. Molar solubility s is a different quantity, the moles of salt that dissolve per litre, and converting between the two is where the stoichiometry has to be handled carefully: silver chloride gives s for both ions, while calcium fluoride gives 2s for fluoride and the 2 appears both as a multiplier and as an exponent to make 4s cubed. Two consequences are worth carrying: adding more solid to an already saturated solution dissolves nothing extra, and the ion concentration a question asks for is often 2s rather than s.

References

Sources returned by Google Search grounding during lesson verification.

17.2: Molar Solubility and Ksp - Chemistry LibreTexts — chem.libretexts.org
18.1: Solubility Product Constant, Ksp - Chemistry LibreTexts — chem.libretexts.org

Ksp and Molar Solubility

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