MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5a Solubility Solubility Product

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This deck focuses on 5a Solubility Solubility Product, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Flashcard 1: Which condition predicts precipitation: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?

Answer: Precipitation when $Q_{sp} > K_{sp}$. Exceeding Ksp drives the reverse reaction, forming precipitate to reduce ion concentrations to equilibrium levels.

Flashcard 2: Which condition indicates a saturated solution: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?

Answer: Saturated when $Q_{sp} = K_{sp}$. Equality with Ksp signifies equilibrium, where net dissolution or precipitation does not occur.

Flashcard 3: Which condition indicates an unsaturated solution: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?

Answer: Unsaturated when $Q_{sp} < K_{sp}$. Below Ksp, the forward dissolution reaction is favored to increase ion concentrations toward equilibrium.

Flashcard 4: Identify the relationship between $K_{sp}$ magnitude and solubility for salts with the same stoichiometry.

Answer: Larger $K_{sp}$ implies higher molar solubility (same stoichiometry). Higher Ksp reflects greater equilibrium ion concentrations, indicating increased solubility for comparable ionic compositions.

Flashcard 5: State the molar solubility relation for $AB(s) \rightleftharpoons A^+ + B^-$ in pure water.

Answer: $K_{sp} = s^2$. Dissolving s mol/L produces equal [A+] and [B-], yielding the squared relationship in pure water.

Flashcard 6: State the molar solubility relation for $A_2B(s) \rightleftharpoons 2A^+ + B^{2-}$ in pure water.

Answer: $K_{sp} = 4s^3$. Stoichiometry gives [A+] = 2s and [B^{2-}] = s, resulting in the cubic expression at equilibrium.

Flashcard 7: What is the definition of solubility for a solute in a given solvent at fixed temperature?

Answer: Maximum dissolved solute concentration at equilibrium (given $T$). Solubility quantifies the equilibrium concentration of solute when the solution is saturated with undissolved solid at specified temperature.

Flashcard 8: What is the definition of a saturated solution at a specified temperature?

Answer: Solution at equilibrium with undissolved solute present. Saturation occurs when dissolution and precipitation rates balance in the presence of excess undissolved solute.

Flashcard 9: What is the definition of an unsaturated solution at a specified temperature?

Answer: Contains less solute than the solubility limit at that $T$. Unsaturated solutions have ion products below $K_{sp}$, allowing more solute to dissolve to reach equilibrium.

Flashcard 10: What is the definition of a supersaturated solution?

Answer: Contains more dissolved solute than equilibrium solubility. Supersaturation exceeds equilibrium solubility, creating an unstable state prone to precipitation upon disturbance.

Flashcard 11: State the equilibrium expression for $K_{sp}$ of $A_mB_n(s) \rightleftharpoons mA^{n+} + nB^{m-}$.

Answer: $K_{sp} = [A^{n+}]^m[B^{m-}]^n$. Ksp is the equilibrium constant for sparingly soluble salt dissolution, using ion concentrations raised to stoichiometric coefficients.

Flashcard 12: State the molar solubility relation for $AB_2(s) \rightleftharpoons A^{2+} + 2B^-$ in pure water.

Answer: $K_{sp} = 4s^3$. Stoichiometry yields [A^{2+}] = s and [B-] = 2s, leading to the cubic form in pure water.

Flashcard 13: State the molar solubility relation for $A_3B_2(s) \rightleftharpoons 3A^{2+} + 2B^{3-}$ in pure water.

Answer: $K_{sp} = 108s^5$. Equilibrium concentrations are [A^{2+}] = 3s and [B^{3-}] = 2s, producing the fifth-power relationship.

Flashcard 14: Find $K_{sp}$ in terms of $s$ for $M(OH)_2(s) \rightleftharpoons M^{2+} + 2OH^-$ in pure water.

Answer: $K_{sp} = 4s^3$. Analogous to AB_2 salts, with [M^{2+}] = s and [OH^-] = 2s in equilibrium expression.

Flashcard 15: Calculate $s$ for an $AB$ salt in pure water if $K_{sp} = 1.0 \times 10^{-8}$.

Answer: $s = 1.0 \times 10^{-4}\,\text{M}$. Square root of Ksp gives molar solubility for 1:1 stoichiometry in pure water.

Flashcard 16: Calculate $s$ for an $A_2B$ salt in pure water if $K_{sp} = 4.0 \times 10^{-12}$.

Answer: $s = 1.0 \times 10^{-4}\,\text{M}$. Cube root of Ksp/4 yields s for 2:1 stoichiometry in pure water.

Flashcard 17: Identify the common-ion effect on solubility of a sparingly soluble salt.

Answer: A common ion decreases molar solubility. Le Chatelier's principle dictates that excess common ion shifts equilibrium toward undissolved solid.

Flashcard 18: For $AB(s)$ in $0.10\,\text{M}$ $A^+$, what approximation gives solubility $s$ in terms of $K_{sp}$?

Answer: $s \approx \frac{K_{sp}}{[A^+]}$. Valid when s is negligible compared to initial [A^+], simplifying the quadratic equation.

Flashcard 19: For $AB(s)$ with $K_{sp} = 1.0 \times 10^{-10}$ in $0.10\,\text{M}$ $A^+$, estimate $s$.

Answer: $s \approx 1.0 \times 10^{-9}\,\text{M}$. Approximation applies as s << 0.10 M, dividing Ksp by initial concentration.

Flashcard 20: Which change increases solubility of salts with basic anions (for example, $CO_3^{2-}$): adding acid or adding base?

Answer: Adding acid increases solubility (consumes the basic anion). Protonation reduces anion concentration, shifting dissolution equilibrium to increase solubility.

Flashcard 21: Identify the net effect of complex-ion formation on the solubility of a metal salt (for example, $AgCl$ in $NH_3$).

Answer: Complex formation increases solubility (reduces free metal ion). Ligand binding lowers free ion concentration, driving more dissolution by Le Chatelier's principle.

Flashcard 22: Which species are omitted from a $K_{sp}$ expression: pure solids, pure liquids, or aqueous ions?

Answer: Pure solids and pure liquids are omitted. Heterogeneous equilibrium constants incorporate constant activities of pure solids and liquids as unity.

Flashcard 23: Which salt is less soluble: one with $K_{sp} = 1.0 \times 10^{-12}$ or one with $K_{sp} = 1.0 \times 10^{-8}$ (same stoichiometry)?

Answer: The salt with $K_{sp} = 1.0 \times 10^{-12}$ is less soluble. Smaller Ksp corresponds to lower equilibrium concentrations, indicating reduced solubility for identical stoichiometries.

Flashcard 24: What is the ion product $Q_{sp}$ for $A_mB_n(s) \rightleftharpoons mA^{n+} + nB^{m-}$?

Answer: $Q_{sp} = [A^{n+}]^m[B^{m-}]^n$ (current concentrations). Qsp assesses deviation from equilibrium by substituting current ion concentrations into the Ksp expression.