Solubility and Solubility Product (5A)
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MCAT Chemical and Physical Foundations of Biological Systems › Solubility and Solubility Product (5A)
To minimize free fluoride in a mouth-rinse, a chemist considers adding CaCl$_2$ to a solution in equilibrium with $\text{CaF}_2(s)$: $\text{CaF}2(s) \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-$. At 25°C, $K{sp}=3.9\times10^{-11}$. Which shift in equilibrium is most consistent with adding the common ion $\text{Ca}^{2+}$?
Shift right; more $\text{CaF}_2$ dissolves and $[\text{F}^-]$ increases.
Shift right; added $\text{Ca}^{2+}$ must be consumed, increasing solubility.
Shift left; precipitation is promoted and dissolved $[\text{F}^-]$ decreases.
No shift; $\text{Ca}^{2+}$ does not appear in the dissolution reaction.
Explanation
This question tests solubility and solubility product concepts (5A). The common ion effect predicts that adding Ca²⁺ to a saturated CaF₂ solution shifts the equilibrium CaF₂(s) ⇌ Ca²⁺ + 2F⁻ to the left. Since Ca²⁺ is a product in this dissolution equilibrium, increasing its concentration drives the reaction toward solid formation according to Le Chatelier's principle. This shift causes more CaF₂ to precipitate and reduces the dissolved [F⁻] concentration. The correct answer B accurately describes this shift left with precipitation promoted and decreased [F⁻]. Answer D incorrectly suggests that added Ca²⁺ must be consumed by shifting right, but this would increase dissolution rather than decrease it. When a common ion is added to any sparingly soluble salt equilibrium, the shift is always toward precipitation (left).
In a controlled precipitation experiment at 25°C, a team maintains a suspension of $\text{CaCO}_3(s)$: $\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+} + \text{CO}3^{2-}$ with $K{sp}=3.3\times10^{-9}$. They bubble CO$_2$ and observe that free $\text{CO}3^{2-}$ decreases (assume only this change is relevant). With $\text{Ca}^{2+}$ initially unchanged, which statement best predicts the equilibrium response based on $Q$ vs $K{sp}$?
No change occurs because $K_{sp}$ is constant and forces $Q$ to remain unchanged without dissolution.
More $\text{CaCO}_3$ precipitates because lowering $[\text{CO}_3^{2-}]$ increases $Q$.
More $\text{CaCO}_3$ dissolves because adding CO$_2$ adds a common ion to the dissolution reaction.
More $\text{CaCO}_3$ dissolves because lowering $[\text{CO}3^{2-}]$ lowers $Q$ below $K{sp}$.
Explanation
This question tests solubility and solubility product concepts (5A). The reaction quotient Q equals [Ca²⁺][CO₃²⁻] for the dissolution equilibrium, and the system responds to maintain Q = Ksp at equilibrium. When CO₂ is bubbled through the solution, it reacts with carbonate ions (CO₃²⁻ + CO₂ + H₂O → 2HCO₃⁻), decreasing the free [CO₃²⁻] concentration. With [Ca²⁺] initially unchanged and [CO₃²⁻] decreased, Q becomes less than Ksp, creating a driving force for more CaCO₃ to dissolve until Q again equals Ksp. Choice A incorrectly suggests precipitation would occur when Q < Ksp, while choice C fails to recognize that the system must respond to restore equilibrium when Q ≠ Ksp. Choice D misidentifies CO₂ as a common ion when it actually removes carbonate ions from solution. To verify equilibrium shifts in solubility problems, calculate whether Q < Ksp (dissolution) or Q > Ksp (precipitation) after the perturbation.
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 decreases because lowering $[\text{CO}3^{2-}]$ forces $[\text{Ca}^{2+}]$ to decrease to keep $K{sp}$ constant
Solubility is unchanged because CO$_2$ affects only acid–base equilibria, not solubility equilibria
Solubility increases because $K_{sp}$ becomes larger when CO$_2$ is added, independent of temperature
Solubility increases because consuming $\text{CO}_3^{2-}$ allows more CaCO$_3$ to dissolve while maintaining $[\text{Ca}^{2+}][\text{CO}3^{2-}] = K{sp}$
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 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 make the computed $Q$ much larger than the true $Q$, potentially suggesting precipitation when the solution is actually saturated
It would change $K_{sp}$ rather than $Q$, because $K_{sp}$ depends on the units used for concentration
It would not change $Q$ because $Q$ is dimensionless and unit choices cancel
It would make the computed $Q$ much smaller than the true $Q$, potentially suggesting undersaturation when the solution is actually saturated
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.
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?
$[\text{Mg}^{2+}]$ increases because adding base shifts dissolution right to consume $\text{OH}^-$
$[\text{Mg}^{2+}]$ is unchanged because $K_{sp}$ fixes the solubility regardless of added ions
$[\text{Mg}^{2+}]$ increases until $[\text{Mg}^{2+}] = 2[\text{OH}^-]$ to match stoichiometry
$[\text{Mg}^{2+}]$ decreases because added $\text{OH}^-$ shifts the equilibrium left, lowering solubility
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 $K_{sp}$
Precipitation is favored because $Q = [\text{Ca}^{2+}]^3[\text{PO}4^{3-}]^2$ exceeds $K{sp}$
No precipitation is favored because $Q$ is less than $K_{sp}$ at these concentrations
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.
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 left; increased $\text{Ca}^{2+}$ promotes precipitation of $\text{Ca}_3(\text{PO}_4)_2(s)$
Shift right; added $\text{Ca}^{2+}$ lowers $Q$ below $K_{sp}$ by dilution
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.
To study lead exposure, an environmental health team models precipitation of lead(II) iodide: $\text{PbI}2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^-$. At 25°C, $K{sp}=8.5\times 10^{-9}$. In a water sample, $\text{Pb}^{2+}=1.0\times 10^{-5}\ \text{M}$ and $\text{I}^-=5.0\times 10^{-3}\ \text{M}$. Which statement best predicts whether $\text{PbI}_2(s)$ will form?
Precipitation is disfavored because $Q$ should be computed as $[\text{Pb}^{2+}]^2[\text{I}^-]$
Precipitation is favored because $Q=[\text{Pb}^{2+}][\text{I}^-]^2$ is greater than $K_{sp}$
Precipitation is disfavored because $K_{sp}$ increases automatically when $[\text{I}^-]$ increases
No precipitation occurs because iodide is a spectator ion and does not affect solubility
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle requires calculating Q with proper stoichiometry: for PbI₂, Q = [Pb²⁺][I⁻]² because the equilibrium shows one Pb²⁺ and two I⁻ ions. Calculating Q = (1.0 × 10⁻⁵)(5.0 × 10⁻³)² = (1.0 × 10⁻⁵)(2.5 × 10⁻⁵) = 2.5 × 10⁻¹⁰, which exceeds Ksp = 8.5 × 10⁻⁹. The correct answer (A) properly calculates Q and concludes precipitation occurs because Q > Ksp. Choice B incorrectly reverses the stoichiometric coefficients in the Q expression. Always match the powers in Q to the coefficients in the balanced dissolution equation, then compare to Ksp.
A researcher compares two formulations containing $\text{CaCO}_3(s)$ in aqueous buffer at 25°C: $\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+} + \text{CO}3^{2-}$ with $K{sp}=3.3\times 10^{-9}$. Formulation 1 contains added $\text{CaCl}_2$ (raising $\text{Ca}^{2+}$), while Formulation 2 contains added $\text{Na}_2\text{CO}_3$ (raising $\text{CO}_3^{2-}$). Assuming equal total ionic strength and no complexation, which statement best describes the relative solubility of $\text{CaCO}_3$ in the two formulations?
Solubility is unchanged because $K_{sp}$ fixes solubility independent of added ions
Solubility is lower in both formulations because each adds a common ion to the dissolution equilibrium
Solubility is lower only in Formulation 1 because $\text{Ca}^{2+}$ appears first in the $K_{sp}$ expression
Solubility is higher in both formulations because adding either ion increases the driving force to dissolve
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product principle combined with the common ion effect states that adding any ion present in the dissolution equilibrium decreases the solubility of the solid. Formulation 1 adds Ca²⁺ (a common ion), while Formulation 2 adds CO₃²⁻ (also a common ion), both of which appear in the CaCO₃ dissolution equilibrium. The correct answer (B) recognizes that both formulations contain a common ion, so both will have lower CaCO₃ solubility compared to pure water. Choice C incorrectly suggests the order of ions in the Ksp expression matters, when both ions have equal importance in shifting equilibrium. The common ion effect applies equally whether you add the cation or anion from the dissolution equilibrium.
A microbiology team studies barium toxicity and uses sulfate to precipitate barium from aqueous media at 25°C. The equilibrium is $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+} + \text{SO}4^{2-}$ with $K{sp}=1.1\times10^{-10}$. A culture medium is saturated with BaSO4 and then Na2SO4 is added (no volume change). Which shift in equilibrium is most consistent with adding the common ion $\text{SO}_4^{2-}$?
Shift right, dissolving more BaSO4(s) to keep $[\text{Ba}^{2+}]$ constant
Shift left only if pH decreases, because sulfate is a base and controls $K_{sp}$ directly
Shift left, reducing dissolved $\text{Ba}^{2+}$ as more BaSO4(s) forms
No shift, because $K_{sp}$ is constant and therefore concentrations cannot change
Explanation
This question tests solubility and solubility product concepts (5A). The solubility product expression for BaSO₄ is Ksp = [Ba²⁺][SO₄²⁻], which remains constant at constant temperature. Adding Na₂SO₄ increases [SO₄²⁻], causing the ion product Q to temporarily exceed Ksp. The system responds by shifting left to form more solid BaSO₄, thereby reducing the concentration of dissolved Ba²⁺ until equilibrium is restored. Choice A incorrectly predicts a rightward shift that would dissolve more solid, which would further increase Q rather than restore equilibrium. The common ion effect is particularly important in medical contexts, as it explains why barium sulfate can be safely used as a radiopaque contrast agent—the low solubility is further reduced in the presence of sulfate ions, minimizing toxic barium absorption.