Solubility and Solubility Product (5A)
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MCAT Chemical and Physical Foundations of Biological Systems › Solubility and Solubility Product (5A)
In a blood-chemistry simulation at 37°C, calcium oxalate precipitation is monitored: $\text{CaC}_2\text{O}_4(s) \rightleftharpoons \text{Ca}^{2+} + \text{C}_2\text{O}4^{2-}$ with $K{sp}=2.3\times10^{-9}$. A sample has $\text{Ca}^{2+} = 1.2\times10^{-3},\text{M}$ and $\text{C}_2\text{O}_4^{2-} = 5.0\times10^{-6},\text{M}$. If a clinician administers CaCl$2$ that increases free $\text{Ca}^{2+}$ while oxalate remains constant, which shift in equilibrium is most consistent with maintaining $K{sp}$?
Shift right; increased $[\text{Ca}^{2+}]$ increases solubility so that $Q$ returns to $K_{sp}$
No shift; $K_{sp}$ changes to accommodate the higher $[\text{Ca}^{2+}]$ at constant temperature
Shift left; increased $[\text{Ca}^{2+}]$ promotes precipitation, decreasing dissolved oxalate
Shift right; precipitation occurs until $[\text{Ca}^{2+}]$ decreases, which increases $Q$ back to $K_{sp}$
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle states that Ksp = [Ca²⁺][C₂O₄²⁻] must remain constant at fixed temperature. When CaCl₂ is administered, it increases [Ca²⁺] in the blood while [C₂O₄²⁻] initially remains constant, causing Q to exceed Ksp temporarily. To restore equilibrium and maintain the constant Ksp, the system must shift left, promoting precipitation of CaC₂O₄(s) which removes both Ca²⁺ and C₂O₄²⁻ from solution until Q again equals Ksp. This shift left decreases the dissolved oxalate concentration as solid calcium oxalate forms. Choice B incorrectly suggests the equilibrium shifts right, which would further increase Q above Ksp rather than restoring equilibrium. A useful principle is that when one ion concentration increases in a saturated solution, the other ion concentration must decrease proportionally to maintain constant Ksp, typically through precipitation.
A researcher studying biomineralization prepares a saturated solution of calcium carbonate at 25°C: $\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+} + \text{CO}3^{2-}$ with constant $K{sp}$. They then bubble CO$_2$ into the solution, which increases dissolved CO$_2$ and shifts $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HCO}_3^- + \text{H}^+$, reducing free $\text{CO}_3^{2-}$ via protonation (conditions otherwise unchanged). Which statement best explains the resulting effect on CaCO$3$ solubility in terms of $K{sp}$?
Solubility increases because $K_{sp}$ becomes larger when CO$_2$ is added, independent of temperature
Solubility is unchanged because CO$_2$ affects only acid–base equilibria, not solubility equilibria
Solubility increases because consuming $\text{CO}_3^{2-}$ allows more CaCO$_3$ to dissolve while maintaining $[\text{Ca}^{2+}][\text{CO}3^{2-}] = K{sp}$
Solubility decreases because lowering $[\text{CO}3^{2-}]$ forces $[\text{Ca}^{2+}]$ to decrease to keep $K{sp}$ constant
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle states that Ksp = [Ca²⁺][CO₃²⁻] remains constant at fixed temperature, but the individual ion concentrations can change if one ion is consumed by another reaction. When CO₂ is bubbled through the solution, it forms carbonic acid which protonates CO₃²⁻ ions to form HCO₃⁻, effectively removing CO₃²⁻ from the dissolution equilibrium. To maintain constant Ksp as [CO₃²⁻] decreases, more CaCO₃ must dissolve to increase both [Ca²⁺] and [CO₃²⁻], with the net effect being increased solubility of CaCO₃. Choice A incorrectly suggests both ion concentrations must decrease together, missing that dissolution increases both simultaneously. A useful principle is that removing one ion from a dissolution equilibrium (through complexation, protonation, or precipitation in another reaction) always increases the solubility of the original salt as the system shifts to restore Ksp.
A dialysis-fluid study examines whether adding chloride affects dissolution of calcium chloride deposits modeled as $\text{CaCl}2(s) \rightleftharpoons \text{Ca}^{2+} + 2\text{Cl}^-$. (Assume a hypothetical sparingly soluble form with a defined $K{sp}$ at fixed temperature.) If the dialysate is supplemented with NaCl, increasing $\text{Cl}^-$ while leaving temperature unchanged, which statement best predicts the direction of change in dissolved $\text{Ca}^{2+}$ at equilibrium?
Dissolved $[\text{Ca}^{2+}]$ decreases because added $\text{Cl}^-$ is a common ion that shifts the dissolution equilibrium left
Dissolved $[\text{Ca}^{2+}]$ decreases because NaCl lowers temperature via endothermic dissolution, reducing $K_{sp}$
Dissolved $[\text{Ca}^{2+}]$ is unchanged because only the amount of solid controls equilibrium concentrations
Dissolved $[\text{Ca}^{2+}]$ increases because adding a spectator electrolyte always increases solubility
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle combined with the common ion effect predicts that adding Cl⁻ ions from NaCl will decrease the solubility of CaCl₂. At equilibrium, Ksp = [Ca²⁺][Cl⁻]² must be satisfied, so when [Cl⁻] increases due to added NaCl, [Ca²⁺] must decrease proportionally to maintain the constant Ksp value. This represents a shift of the dissolution equilibrium to the left, favoring the solid phase and reducing dissolved calcium concentration. Choice A incorrectly suggests spectator electrolytes increase solubility, when in fact Cl⁻ is not a spectator but a common ion participating in the equilibrium. The key principle for identifying common ion effects is recognizing when an added salt shares an ion with the sparingly soluble salt; this shared ion always decreases solubility by shifting equilibrium toward precipitation.
A lab prepares a saturated solution of lead(II) iodide at 25°C for an electrode calibration: $\text{PbI}2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^-$. The measured $K{sp}$ is $7.0\times10^{-9}$. The technician accidentally reports the iodide concentration in the saturated solution as $1.0\times10^{-3},\text{mM}$ instead of $1.0\times10^{-3},\text{M}$. Without doing a full calculation, which statement best describes the impact of this unit error on the implied ion product $Q$ computed from the reported values?
It would not change $Q$ because $Q$ is dimensionless and unit choices cancel
It would make the computed $Q$ much larger than the true $Q$, potentially suggesting precipitation when the solution is actually saturated
It would make the computed $Q$ much smaller than the true $Q$, potentially suggesting undersaturation when the solution is actually saturated
It would change $K_{sp}$ rather than $Q$, because $K_{sp}$ depends on the units used for concentration
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle requires consistent units when calculating Q to compare with Ksp. If iodide concentration is mistakenly reported as 1.0×10⁻³ mM instead of 1.0×10⁻³ M, this represents a 1000-fold error since 1 mM = 10⁻³ M, making the actual value 1.0×10⁻⁶ M. When computing Q = [Pb²⁺][I⁻]², using the erroneous smaller concentration value would make Q appear much smaller than its true value. This could lead to incorrectly concluding the solution is undersaturated (Q < Ksp) when it might actually be saturated or supersaturated. Choice B reverses the direction of the error, while choice C incorrectly claims units don't matter for dimensionless quantities. The critical lesson is that concentration units must be consistent (typically M) when calculating Q or Ksp, as unit errors can lead to incorrect predictions about precipitation or dissolution.
In a nephrology study modeling kidney stone risk, researchers prepared a saturated solution of calcium fluoride at 25°C: $\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq)$. In pure water, the measured molar solubility was $s = 2.0\times10^{-4},\text{M}$. A second trial used the same solid mass but the solvent initially contained $1.0\times10^{-3},\text{M}$ NaF (fully dissociated), approximating elevated urinary fluoride. Assuming activity coefficients ~1 and constant temperature, which shift in equilibrium is most consistent with adding the common ion $\text{F}^-$?
Shift right; dissolution increases until $[\text{Ca}^{2+}] = [\text{F}^-]$ to satisfy electroneutrality
Shift right; $[\text{Ca}^{2+}]$ increases because adding NaF increases ionic strength and dissolves more $\text{CaF}_2$
Shift left; dissolution decreases because added $\text{F}^-$ drives precipitation to keep $K_{sp}$ constant
No shift; $K_{sp}$ depends only on temperature, so added $\text{F}^-$ cannot change equilibrium concentrations
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle states that for a sparingly soluble salt at equilibrium, the product of ion concentrations raised to their stoichiometric powers equals a constant Ksp at a given temperature. In this scenario, adding NaF introduces F⁻ ions to the solution, creating a common ion effect where the shared F⁻ ion from both CaF₂ and NaF affects the equilibrium. According to Le Chatelier's principle, adding F⁻ ions increases the concentration of a product, causing the equilibrium to shift left to maintain the constant Ksp value, which decreases the dissolution of CaF₂ and reduces [Ca²⁺]. Choice A incorrectly suggests that ionic strength increases solubility, when in fact the common ion effect dominates and decreases solubility. A useful check for common ion problems is to remember that adding a common ion always decreases the solubility of a sparingly soluble salt, as the system must maintain Ksp = [Ca²⁺][F⁻]² constant.
To model mineral deposition in bile, a researcher compares two sparingly soluble salts at 25°C: $\text{CaCO}3(s) \rightleftharpoons \text{Ca}^{2+}+\text{CO}3^{2-}$ with $K{sp}=3.3\times10^{-9}$ and $\text{CaF}2(s) \rightleftharpoons \text{Ca}^{2+}+2\text{F}^-$ with $K{sp}=1.5\times10^{-10}$. Each salt is placed separately into pure water until saturation. Based on $K{sp}$ and stoichiometry (no calculations required), which statement best compares their molar solubilities?
CaCO$_3$ must be more soluble because it produces fewer ions per formula unit than CaF$_2$
CaCO$_3$ must be less soluble because it forms a 1:1 ion pair, which always lowers solubility
CaF$2$ must be more soluble because its $K{sp}$ is smaller, indicating weaker ionic interactions
A direct comparison requires considering both $K_{sp}$ magnitude and dissolution stoichiometry; smaller $K_{sp}$ does not automatically mean lower molar solubility
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle shows that comparing molar solubilities requires considering both Ksp values and dissolution stoichiometry. For CaCO₃ with Ksp = 3.3×10⁻⁹, if molar solubility is s, then Ksp = s², giving s = √(3.3×10⁻⁹) ≈ 5.7×10⁻⁵ M. For CaF₂ with Ksp = 1.5×10⁻¹⁰, if molar solubility is s, then Ksp = s(2s)² = 4s³, giving s = ∛(Ksp/4) ≈ 3.3×10⁻⁴ M. Despite having a smaller Ksp, CaF₂ actually has higher molar solubility due to its different stoichiometry. Choice A incorrectly assumes smaller Ksp always means lower solubility, ignoring stoichiometric differences. The key insight is that Ksp values cannot be directly compared across different stoichiometries; the relationship between Ksp and molar solubility depends on how many ions are produced per formula unit dissolved.
A pharmacology team prepares an oral suspension containing slightly soluble magnesium hydroxide: $\text{Mg(OH)}2(s) \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-$. At 25°C, $K{sp}=5.6\times10^{-12}$. The formulation is modified by adding NaOH to raise the initial $\text{OH}^-$ to $1.0\times10^{-3},\text{M}$ before adding solid $\text{Mg(OH)}_2$. Assuming ideal behavior, which outcome is most consistent with the common-ion effect on the dissolved $\text{Mg}^{2+}$ at equilibrium?
$[\text{Mg}^{2+}]$ is unchanged because $K_{sp}$ fixes the solubility regardless of added ions
$[\text{Mg}^{2+}]$ decreases because added $\text{OH}^-$ shifts the equilibrium left, lowering solubility
$[\text{Mg}^{2+}]$ increases until $[\text{Mg}^{2+}] = 2[\text{OH}^-]$ to match stoichiometry
$[\text{Mg}^{2+}]$ increases because adding base shifts dissolution right to consume $\text{OH}^-$
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle combined with the common ion effect predicts that adding a common ion (OH⁻) to a saturated solution decreases the solubility of the sparingly soluble salt. When NaOH is added, it increases [OH⁻] in solution, and since Ksp = [Mg²⁺][OH⁻]² must remain constant at fixed temperature, [Mg²⁺] must decrease proportionally to maintain this product. The equilibrium shifts left, favoring precipitation of Mg(OH)₂ and reducing the dissolved magnesium concentration. Choice A incorrectly suggests the equilibrium shifts right to consume OH⁻, but this would increase the Ksp expression value, which cannot happen at constant temperature. To verify common ion effects, remember that adding any ion already present in the dissolution equilibrium always decreases the solubility of the original salt by shifting equilibrium toward the solid phase.
A physiology lab examines calcium phosphate precipitation in serum. At 37°C, the relevant equilibrium is $\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}4^{3-}$ with $K{sp}=2.0\times10^{-33}$. A serum sample has free $\text{Ca}^{2+} = 1.0\times10^{-3},\text{M}$ and $\text{PO}_4^{3-} = 1.0\times10^{-6},\text{M}$. Assuming ideal behavior, which statement best predicts whether precipitation is thermodynamically favored at these ion concentrations?
No precipitation is favored because solids do not participate in equilibrium calculations, so $Q$ cannot be compared to $K_{sp}$
No precipitation is favored because $Q$ is less than $K_{sp}$ at these concentrations
Precipitation is favored because $Q = [\text{Ca}^{2+}]^3[\text{PO}4^{3-}]^2$ exceeds $K{sp}$
Precipitation is favored because $K_{sp}$ increases with higher ion concentrations in serum
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle requires comparing the ion product Q to Ksp to determine if precipitation occurs: if Q > Ksp, the solution is supersaturated and precipitation is favored. For Ca₃(PO₄)₂, Q = [Ca²⁺]³[PO₄³⁻]² = (1.0×10⁻³)³(1.0×10⁻⁶)² = 1.0×10⁻²¹, which is much greater than Ksp = 2.0×10⁻³³. Since Q > Ksp, the system will shift toward precipitation to reduce ion concentrations until Q equals Ksp at equilibrium. Choice C incorrectly claims Q < Ksp without performing the calculation, while choice B misunderstands that Ksp is a constant at fixed temperature. When checking precipitation problems, always calculate Q using the actual ion concentrations and stoichiometric coefficients, then compare to Ksp: precipitation occurs when Q > Ksp.
A biochemistry lab uses $\text{SrSO}_4$ as a nonradioactive proxy for calcium sulfate scaling: $\text{SrSO}_4(s) \rightleftharpoons \text{Sr}^{2+} + \text{SO}4^{2-}$ with $K{sp}=3.2\times 10^{-7}$ at 25°C. A solution initially has $\text{Sr}^{2+}=4.0\times 10^{-4}\ \text{M}$ and $\text{SO}_4^{2-}=4.0\times 10^{-4}\ \text{M}$. The team adds $\text{Na}_2\text{SO}_4$ to increase sulfate while keeping volume constant. Which shift in equilibrium is most consistent with this intervention?
Shift right; increasing sulfate lowers $Q$ and promotes dissolution
Shift left; increasing sulfate raises $Q$ and promotes precipitation
Shift right; increasing sulfate increases $K_{sp}$ by mass action
No shift; the added sulfate is buffered and cannot affect $Q$
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle requires first calculating Q to determine the system's initial state: Q = [Sr²⁺][SO₄²⁻] = (4.0 × 10⁻⁴)(4.0 × 10⁻⁴) = 1.6 × 10⁻⁷, which is less than Ksp = 3.2 × 10⁻⁷, indicating undersaturation. Adding Na₂SO₄ increases [SO₄²⁻], raising Q toward and potentially above Ksp, which shifts the equilibrium left toward precipitation. The correct answer (B) properly identifies that increasing sulfate concentration raises Q and promotes precipitation when Q exceeds Ksp. Choice A incorrectly suggests that adding a product ion would promote dissolution. When adding ions to an undersaturated solution, check whether the new Q exceeds Ksp to predict precipitation.
In a study of kidney stone prevention, researchers modeled calcium phosphate precipitation by preparing a buffered aqueous solution at 37°C containing $1.0\ \text{mM}$ $\text{Ca}^{2+}$ and $0.60\ \text{mM}$ $\text{PO}_4^{3-}$ (assume $\text{PO}_4^{3-}$ is the relevant phosphate species at this pH). For $\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}4^{3-}$, $K{sp}=2.0\times 10^{-29}$ at 37°C. The team then adds $\text{CaCl}_2$ to raise free $\text{Ca}^{2+}$ without changing volume appreciably. Which shift in equilibrium is most consistent with adding the common ion $\text{Ca}^{2+}$ under these conditions?
Shift right; increased $\text{Ca}^{2+}$ drives dissolution to restore $K_{sp}$
No shift; $K_{sp}$ fixes ion concentrations so added $\text{Ca}^{2+}$ remains in solution
Shift right; added $\text{Ca}^{2+}$ lowers $Q$ below $K_{sp}$ by dilution
Shift left; increased $\text{Ca}^{2+}$ promotes precipitation of $\text{Ca}_3(\text{PO}_4)_2(s)$
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle states that at equilibrium, the product of ion concentrations raised to their stoichiometric coefficients equals Ksp, and adding a common ion shifts the equilibrium to reduce its concentration. In this scenario, adding Ca²⁺ to a solution already containing calcium and phosphate ions increases the concentration of a common ion in the dissolution equilibrium. The correct answer (B) follows because increasing [Ca²⁺] raises the reaction quotient Q above Ksp, driving the equilibrium left toward precipitation to restore equilibrium. Choice A incorrectly suggests dissolution would increase when adding more product, violating Le Chatelier's principle. To verify common ion effects, remember that adding any ion already present in the equilibrium always shifts the reaction away from that ion's side, promoting precipitation for dissolution equilibria.