Solubility and Solubility Product (5A) - MCAT Chemical and Physical Foundations of Biological Systems
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Which condition predicts precipitation: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?
Which condition predicts precipitation: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?
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Precipitation when $Q_{sp} > K_{sp}$. Exceeding Ksp drives the reverse reaction, forming precipitate to reduce ion concentrations to equilibrium levels.
Precipitation when $Q_{sp} > K_{sp}$. Exceeding Ksp drives the reverse reaction, forming precipitate to reduce ion concentrations to equilibrium levels.
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Which condition indicates a saturated solution: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?
Which condition indicates a saturated solution: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?
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Saturated when $Q_{sp} = K_{sp}$. Equality with Ksp signifies equilibrium, where net dissolution or precipitation does not occur.
Saturated when $Q_{sp} = K_{sp}$. Equality with Ksp signifies equilibrium, where net dissolution or precipitation does not occur.
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Which condition indicates an unsaturated solution: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?
Which condition indicates an unsaturated solution: $Q_{sp} > K_{sp}$, $Q_{sp} = K_{sp}$, or $Q_{sp} < K_{sp}$?
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Unsaturated when $Q_{sp} < K_{sp}$. Below Ksp, the forward dissolution reaction is favored to increase ion concentrations toward equilibrium.
Unsaturated when $Q_{sp} < K_{sp}$. Below Ksp, the forward dissolution reaction is favored to increase ion concentrations toward equilibrium.
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Identify the relationship between $K_{sp}$ magnitude and solubility for salts with the same stoichiometry.
Identify the relationship between $K_{sp}$ magnitude and solubility for salts with the same stoichiometry.
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Larger $K_{sp}$ implies higher molar solubility (same stoichiometry). Higher Ksp reflects greater equilibrium ion concentrations, indicating increased solubility for comparable ionic compositions.
Larger $K_{sp}$ implies higher molar solubility (same stoichiometry). Higher Ksp reflects greater equilibrium ion concentrations, indicating increased solubility for comparable ionic compositions.
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State the molar solubility relation for $AB(s) \rightleftharpoons A^+ + B^-$ in pure water.
State the molar solubility relation for $AB(s) \rightleftharpoons A^+ + B^-$ in pure water.
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$K_{sp} = s^2$. Dissolving s mol/L produces equal [A+] and [B-], yielding the squared relationship in pure water.
$K_{sp} = s^2$. Dissolving s mol/L produces equal [A+] and [B-], yielding the squared relationship in pure water.
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State the molar solubility relation for $A_2B(s) \rightleftharpoons 2A^+ + B^{2-}$ in pure water.
State the molar solubility relation for $A_2B(s) \rightleftharpoons 2A^+ + B^{2-}$ in pure water.
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$K_{sp} = 4s^3$. Stoichiometry gives [A+] = 2s and [B^{2-}] = s, resulting in the cubic expression at equilibrium.
$K_{sp} = 4s^3$. Stoichiometry gives [A+] = 2s and [B^{2-}] = s, resulting in the cubic expression at equilibrium.
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What is the definition of solubility for a solute in a given solvent at fixed temperature?
What is the definition of solubility for a solute in a given solvent at fixed temperature?
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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.
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.
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What is the definition of a saturated solution at a specified temperature?
What is the definition of a saturated solution at a specified temperature?
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Solution at equilibrium with undissolved solute present. Saturation occurs when dissolution and precipitation rates balance in the presence of excess undissolved solute.
Solution at equilibrium with undissolved solute present. Saturation occurs when dissolution and precipitation rates balance in the presence of excess undissolved solute.
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What is the definition of an unsaturated solution at a specified temperature?
What is the definition of an unsaturated solution at a specified temperature?
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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.
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.
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What is the definition of a supersaturated solution?
What is the definition of a supersaturated solution?
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Contains more dissolved solute than equilibrium solubility. Supersaturation exceeds equilibrium solubility, creating an unstable state prone to precipitation upon disturbance.
Contains more dissolved solute than equilibrium solubility. Supersaturation exceeds equilibrium solubility, creating an unstable state prone to precipitation upon disturbance.
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State the equilibrium expression for $K_{sp}$ of $A_mB_n(s) \rightleftharpoons mA^{n+} + nB^{m-}$.
State the equilibrium expression for $K_{sp}$ of $A_mB_n(s) \rightleftharpoons mA^{n+} + nB^{m-}$.
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$K_{sp} = [A^{n+}]^m[B^{m-}]^n$. Ksp is the equilibrium constant for sparingly soluble salt dissolution, using ion concentrations raised to stoichiometric coefficients.
$K_{sp} = [A^{n+}]^m[B^{m-}]^n$. Ksp is the equilibrium constant for sparingly soluble salt dissolution, using ion concentrations raised to stoichiometric coefficients.
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State the molar solubility relation for $AB_2(s) \rightleftharpoons A^{2+} + 2B^-$ in pure water.
State the molar solubility relation for $AB_2(s) \rightleftharpoons A^{2+} + 2B^-$ in pure water.
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$K_{sp} = 4s^3$. Stoichiometry yields [A^{2+}] = s and [B-] = 2s, leading to the cubic form in pure water.
$K_{sp} = 4s^3$. Stoichiometry yields [A^{2+}] = s and [B-] = 2s, leading to the cubic form in pure water.
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State the molar solubility relation for $A_3B_2(s) \rightleftharpoons 3A^{2+} + 2B^{3-}$ in pure water.
State the molar solubility relation for $A_3B_2(s) \rightleftharpoons 3A^{2+} + 2B^{3-}$ in pure water.
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$K_{sp} = 108s^5$. Equilibrium concentrations are [A^{2+}] = 3s and [B^{3-}] = 2s, producing the fifth-power relationship.
$K_{sp} = 108s^5$. Equilibrium concentrations are [A^{2+}] = 3s and [B^{3-}] = 2s, producing the fifth-power relationship.
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Find $K_{sp}$ in terms of $s$ for $M(OH)_2(s) \rightleftharpoons M^{2+} + 2OH^-$ in pure water.
Find $K_{sp}$ in terms of $s$ for $M(OH)_2(s) \rightleftharpoons M^{2+} + 2OH^-$ in pure water.
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$K_{sp} = 4s^3$. Analogous to AB_2 salts, with [M^{2+}] = s and [OH^-] = 2s in equilibrium expression.
$K_{sp} = 4s^3$. Analogous to AB_2 salts, with [M^{2+}] = s and [OH^-] = 2s in equilibrium expression.
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Calculate $s$ for an $AB$ salt in pure water if $K_{sp} = 1.0 \times 10^{-8}$.
Calculate $s$ for an $AB$ salt in pure water if $K_{sp} = 1.0 \times 10^{-8}$.
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$s = 1.0 \times 10^{-4},\text{M}$. Square root of Ksp gives molar solubility for 1:1 stoichiometry in pure water.
$s = 1.0 \times 10^{-4},\text{M}$. Square root of Ksp gives molar solubility for 1:1 stoichiometry in pure water.
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Calculate $s$ for an $A_2B$ salt in pure water if $K_{sp} = 4.0 \times 10^{-12}$.
Calculate $s$ for an $A_2B$ salt in pure water if $K_{sp} = 4.0 \times 10^{-12}$.
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$s = 1.0 \times 10^{-4},\text{M}$. Cube root of Ksp/4 yields s for 2:1 stoichiometry in pure water.
$s = 1.0 \times 10^{-4},\text{M}$. Cube root of Ksp/4 yields s for 2:1 stoichiometry in pure water.
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Identify the common-ion effect on solubility of a sparingly soluble salt.
Identify the common-ion effect on solubility of a sparingly soluble salt.
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A common ion decreases molar solubility. Le Chatelier's principle dictates that excess common ion shifts equilibrium toward undissolved solid.
A common ion decreases molar solubility. Le Chatelier's principle dictates that excess common ion shifts equilibrium toward undissolved solid.
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For $AB(s)$ in $0.10,\text{M}$ $A^+$, what approximation gives solubility $s$ in terms of $K_{sp}$?
For $AB(s)$ in $0.10,\text{M}$ $A^+$, what approximation gives solubility $s$ in terms of $K_{sp}$?
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$s \approx \frac{K_{sp}}{[A^+]}$. Valid when s is negligible compared to initial [A^+], simplifying the quadratic equation.
$s \approx \frac{K_{sp}}{[A^+]}$. Valid when s is negligible compared to initial [A^+], simplifying the quadratic equation.
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For $AB(s)$ with $K_{sp} = 1.0 \times 10^{-10}$ in $0.10,\text{M}$ $A^+$, estimate $s$.
For $AB(s)$ with $K_{sp} = 1.0 \times 10^{-10}$ in $0.10,\text{M}$ $A^+$, estimate $s$.
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$s \approx 1.0 \times 10^{-9},\text{M}$. Approximation applies as s << 0.10 M, dividing Ksp by initial concentration.
$s \approx 1.0 \times 10^{-9},\text{M}$. Approximation applies as s << 0.10 M, dividing Ksp by initial concentration.
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Which change increases solubility of salts with basic anions (for example, $CO_3^{2-}$): adding acid or adding base?
Which change increases solubility of salts with basic anions (for example, $CO_3^{2-}$): adding acid or adding base?
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Adding acid increases solubility (consumes the basic anion). Protonation reduces anion concentration, shifting dissolution equilibrium to increase solubility.
Adding acid increases solubility (consumes the basic anion). Protonation reduces anion concentration, shifting dissolution equilibrium to increase solubility.
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Identify the net effect of complex-ion formation on the solubility of a metal salt (for example, $AgCl$ in $NH_3$).
Identify the net effect of complex-ion formation on the solubility of a metal salt (for example, $AgCl$ in $NH_3$).
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Complex formation increases solubility (reduces free metal ion). Ligand binding lowers free ion concentration, driving more dissolution by Le Chatelier's principle.
Complex formation increases solubility (reduces free metal ion). Ligand binding lowers free ion concentration, driving more dissolution by Le Chatelier's principle.
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Which species are omitted from a $K_{sp}$ expression: pure solids, pure liquids, or aqueous ions?
Which species are omitted from a $K_{sp}$ expression: pure solids, pure liquids, or aqueous ions?
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Pure solids and pure liquids are omitted. Heterogeneous equilibrium constants incorporate constant activities of pure solids and liquids as unity.
Pure solids and pure liquids are omitted. Heterogeneous equilibrium constants incorporate constant activities of pure solids and liquids as unity.
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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)?
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)?
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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.
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.
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What is the ion product $Q_{sp}$ for $A_mB_n(s) \rightleftharpoons mA^{n+} + nB^{m-}$?
What is the ion product $Q_{sp}$ for $A_mB_n(s) \rightleftharpoons mA^{n+} + nB^{m-}$?
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$Q_{sp} = [A^{n+}]^m[B^{m-}]^n$ (current concentrations). Qsp assesses deviation from equilibrium by substituting current ion concentrations into the Ksp expression.
$Q_{sp} = [A^{n+}]^m[B^{m-}]^n$ (current concentrations). Qsp assesses deviation from equilibrium by substituting current ion concentrations into the Ksp expression.
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