Solubility and Solubility Product (5A) - Chem & Physics

1. A formulation scientist compares $\text{BaSO}_4$ and $\text{CaSO}_4$ as insoluble fillers. At 25°C: $K_{sp}(\text{BaSO}_4)=1.1\times10^{-10}$ and $K_{sp}(\text{CaSO}_4)=2.4\times10^{-5}$ for $\text{MSO}_4(s) \rightleftharpoons \text{M}^{2+} + \text{SO}_4^{2-}$. Which statement best explains the solubility difference observed in water?
   Answer: $\text{CaSO}_4$ is more soluble because its larger $K_{sp}$ indicates greater ion concentrations at equilibrium. This question tests solubility and solubility product concepts (5A). For salts with identical stoichiometry (both MSO₄ dissociating to M²⁺ + SO₄²⁻), the salt with the larger Ksp has greater solubility. BaSO₄ has Ksp = 1.1×10⁻¹⁰ while CaSO₄ has Ksp = 2.4×10⁻⁵, making CaSO₄ about 200,000 times more soluble. The much larger Ksp for CaSO₄ indicates that its equilibrium ion concentrations are much higher than those of BaSO₄. The correct answer B states that CaSO₄ is more soluble because its larger Ksp indicates greater ion concentrations at equilibrium. Answer A incorrectly claims BaSO₄ is more soluble despite having the smaller Ksp. When comparing salts of the same type, always remember that Ksp directly reflects solubility: larger Ksp means more dissolved ions at equilibrium.
2. 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?
   Answer: It would make the computed $Q$ much smaller than the true $Q$, potentially suggesting undersaturation when the solution is actually saturated 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.
3. A physiology lab examines whether calcium oxalate can precipitate in serum. At 37°C, for $\text{CaC}_2\text{O}_4(s) \rightleftharpoons \text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-}$, $K_{sp}=2.3\times 10^{-9}$. A sample is estimated to have free $[\text{Ca}^{2+}]=1.2\ \text{mM}$ and free $[\text{C}_2\text{O}_4^{2-}]=0.50\ \text{mM}$. Assuming ideal behavior and constant temperature, which statement best predicts what will happen in this sample?
   Answer: Precipitation is favored because $Q=[\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}]$ exceeds $K_{sp}$ This question tests solubility and solubility product concepts (5A). The solubility product principle requires comparing the reaction quotient Q to Ksp: when Q > Ksp, precipitation occurs to restore equilibrium. For this calcium oxalate system, Q = [Ca²⁺][C₂O₄²⁻] = (1.2 × 10⁻³)(0.50 × 10⁻³) = 6.0 × 10⁻⁷, which exceeds Ksp = 2.3 × 10⁻⁹. The correct answer (A) properly identifies that precipitation is favored because Q > Ksp. Choice D incorrectly suggests that millimolar concentrations need special conversion beyond simple multiplication, when in fact Q is correctly calculated as the product of the molar concentrations. When evaluating precipitation, always calculate Q using actual ion concentrations and compare directly to Ksp—precipitation occurs when Q > Ksp.
4. In a cell-culture medium at 25°C, investigators consider adding phosphate buffer. They model calcium phosphate precipitation as $\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-}$ with $K_{sp}=2.0\times10^{-33}$. If $[\text{Ca}^{2+}]$ is held constant and $[\text{PO}_4^{3-}]$ is increased, which shift in equilibrium is most consistent with adding the common ion $\text{PO}_4^{3-}$?
   Answer: Shift left; precipitation is promoted and free $[\text{Ca}^{2+}]$ decreases. This question tests solubility and solubility product concepts (5A). The common ion effect predicts that adding PO₄³⁻ to a system at equilibrium with Ca₃(PO₄)₂(s) shifts the reaction Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺ + 2PO₄³⁻ to the left. Since PO₄³⁻ is a product in this equilibrium, increasing its concentration drives precipitation according to Le Chatelier's principle. This shift causes more solid to form and reduces the free [Ca²⁺] concentration, even though [Ca²⁺] was stated to be held constant (likely meaning the total calcium, not free calcium). The correct answer B accurately describes the shift left with precipitation promoted and decreased free [Ca²⁺]. Answer A incorrectly suggests dissolution to consume phosphate, but phosphate is a product that drives precipitation when added. Remember that adding any product ion to a dissolution equilibrium always favors precipitation.
5. 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?
   Answer: A direct comparison requires considering both $K_{sp}$ magnitude and dissolution stoichiometry; smaller $K_{sp}$ does not automatically mean lower molar solubility 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.
6. A researcher compares two salts as potential controlled-release sources of $\text{Ag}^+$ in an antimicrobial dressing: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ with $K_{sp}=1.8\times10^{-10}$ and $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+ + \text{Br}^-$ with $K_{sp}=5.0\times10^{-13}$ at 25°C. In pure water, which statement best explains the solubility difference observed?
   Answer: $\text{AgCl}$ is more soluble because its larger $K_{sp}$ indicates a greater extent of dissolution. This question tests solubility and solubility product concepts (5A). For 1:1 salts like AgCl and AgBr with the same stoichiometry, the salt with the larger Ksp value has greater solubility in pure water. Since both salts produce equal numbers of ions per formula unit (Ag⁺ + X⁻), we can directly compare Ksp values: AgCl has Ksp = 1.8×10⁻¹⁰ while AgBr has Ksp = 5.0×10⁻¹³. The larger Ksp for AgCl indicates higher equilibrium ion concentrations and thus greater solubility. The correct answer B states that AgCl is more soluble because its larger Ksp indicates greater dissolution. Answer A incorrectly claims that a smaller Ksp leads to more dissolution, which contradicts the fundamental meaning of Ksp. When comparing salts with identical stoichiometry, always remember that larger Ksp means greater solubility.
7. To test the common-ion effect in a physiological context, investigators examine $\text{Mg(OH)}_2$: $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-$. At 25°C, $K_{sp}=5.6\times10^{-12}$. A saturated suspension is adjusted by adding NaOH so that $[\text{OH}^-]$ increases. Which shift in equilibrium is most consistent with adding the common ion $\text{OH}^-$?
   Answer: Shift left; $\text{Mg(OH)}_2$ precipitates and dissolved $[\text{Mg}^{2+}]$ decreases. This question tests solubility and solubility product concepts (5A). The common ion effect predicts that adding OH⁻ to a saturated Mg(OH)₂ solution shifts the equilibrium Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻ to the left. Since OH⁻ is a product in this equilibrium, increasing its concentration drives the reaction toward solid formation according to Le Chatelier's principle. This shift causes more Mg(OH)₂ to precipitate and reduces the dissolved [Mg²⁺] concentration. The correct answer B accurately describes this shift left with decreased [Mg²⁺]. Answer D incorrectly suggests a shift right to consume OH⁻, but this would increase dissolution rather than precipitation. Remember that for any sparingly soluble salt, adding a common ion always decreases solubility by shifting equilibrium toward the solid phase.
8. In a cell culture medium, researchers monitor formation of silver chloride deposits on electrodes: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$. At 25°C, $K_{sp}=1.8\times10^{-10}$. The medium initially contains $[\text{Cl}^-]=1.0\times10^{-2}\,\text{M}$ from NaCl. Assuming activity coefficients ~1, which statement best describes the equilibrium dissolved $[\text{Ag}^+]$ after excess solid AgCl is present?
   Answer: $[\text{Ag}^+]$ is lower than in pure water because the common ion $\text{Cl}^-$ suppresses AgCl dissolution This question tests solubility and solubility product concepts (5A). The solubility product principle states that at equilibrium, Ksp = [Ag⁺][Cl⁻] must be satisfied regardless of the source of ions in solution. With pre-existing [Cl⁻] = 1.0×10⁻² M from NaCl, the equilibrium [Ag⁺] can be calculated as [Ag⁺] = Ksp/[Cl⁻] = 1.8×10⁻¹⁰/1.0×10⁻² = 1.8×10⁻⁸ M, which is much lower than the Ag⁺ concentration in pure water. This demonstrates the common ion effect: the presence of Cl⁻ from another source suppresses AgCl dissolution to maintain the constant Ksp. Choice C incorrectly assumes [Ag⁺] must equal [Cl⁻], which is only true when dissolving AgCl in pure water with no other ion sources. A key principle for common ion problems is that the equilibrium concentrations must satisfy Ksp, but the individual ion concentrations need not be equal when other sources contribute ions.
9. In a serum-mimic buffer at 37°C (pH 7.40), a team studies precipitation risk of calcium phosphate, approximated as $\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-}$. They use $K_{sp}=2.0\times10^{-33}$ (activities ≈ concentrations). Initially, $[\text{Ca}^{2+}] = 1.2\times10^{-3}\,\text{M}$ and $[\text{PO}_4^{3-}] = 1.0\times10^{-6}\,\text{M}$. If dietary phosphate transiently increases $[\text{PO}_4^{3-}]$ by 10-fold with $[\text{Ca}^{2+}]$ unchanged, which statement best predicts the equilibrium response based on $Q$ vs $K_{sp}$?
   Answer: Precipitation becomes more likely because $Q$ increases and can exceed $K_{sp}$, lowering free $[\text{Ca}^{2+}]$. This question tests solubility and solubility product concepts (5A). The solubility product principle states that when Q (the ion product) exceeds Ksp, precipitation occurs to reduce ion concentrations until Q = Ksp. Initially, Q = [Ca²⁺]³[PO₄³⁻]² = (1.2×10⁻³)³(1.0×10⁻⁶)² = 1.73×10⁻¹⁸, which is much greater than Ksp = 2.0×10⁻³³. When [PO₄³⁻] increases 10-fold to 1.0×10⁻⁵ M, Q becomes (1.2×10⁻³)³(1.0×10⁻⁵)² = 1.73×10⁻¹⁶, which is even further above Ksp, making precipitation more likely. The correct answer B states that precipitation becomes more likely because Q increases and can exceed Ksp, lowering free [Ca²⁺]. Answer A incorrectly claims that increasing an anion always increases solubility, when actually increasing a common ion decreases solubility. To solve similar problems, calculate Q using the given concentrations and compare to Ksp: if Q > Ksp, precipitation occurs; if Q < Ksp, the solution is unsaturated.
10. A materials team prepares a saturated $\text{AgCl}$ solution at 25°C: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ with $K_{sp}=1.8\times10^{-10}$. They then add AgNO$_3$ to increase $[\text{Ag}^+]$ without changing volume significantly. Which shift in equilibrium is most consistent with adding the common ion $\text{Ag}^+$?
    Answer: Shift left; more $\text{AgCl}(s)$ forms and dissolved $[\text{Cl}^-]$ decreases. This question tests solubility and solubility product concepts (5A). The common ion effect predicts that adding Ag⁺ (from AgNO₃) to a saturated AgCl solution shifts the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left. Since Ag⁺ 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 AgCl to precipitate and reduces the dissolved [Cl⁻] concentration. The correct answer A accurately describes this shift left with more solid forming and decreased [Cl⁻]. Answer B incorrectly suggests dissolution to consume Ag⁺, but Ag⁺ is a product that drives precipitation when added, not a reactant consumed by dissolution. Remember that adding either product ion to a dissolution equilibrium always favors precipitation.
11. A researcher compares two hydroxides as antacid components at 25°C: $\text{Mg(OH)}_2$ with $K_{sp}=5.6\times10^{-12}$ and $\text{Ca(OH)}_2$ with $K_{sp}=5.5\times10^{-6}$ for $\text{M(OH)}_2(s) \rightleftharpoons \text{M}^{2+} + 2\text{OH}^-$. Which statement best explains the solubility difference observed in water?
    Answer: $\text{Ca(OH)}_2$ is more soluble because its larger $K_{sp}$ indicates more dissolved ions at equilibrium. This question tests solubility and solubility product concepts (5A). The solubility product constant (Ksp) represents the equilibrium constant for a sparingly soluble salt dissolving in water, where a larger Ksp indicates greater solubility at equilibrium. For compounds with the same stoichiometry like Mg(OH)₂ and Ca(OH)₂, both following M(OH)₂ ⇌ M²⁺ + 2OH⁻, the Ksp values can be directly compared to determine relative solubility. Since Ca(OH)₂ has a Ksp of 5.5×10⁻⁶ compared to Mg(OH)₂'s Ksp of 5.6×10⁻¹², Ca(OH)₂ is approximately 10⁶ times more soluble, producing more dissolved ions at equilibrium. Choice A incorrectly reverses the relationship between Ksp and solubility, while choices C and D introduce irrelevant factors (ion count and ionic radius) that don't directly determine solubility when Ksp values are given. To check similar problems, remember that for compounds with identical stoichiometry, the one with the larger Ksp is always more soluble.
12. 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?
    Answer: Dissolved $[\text{Ca}^{2+}]$ decreases because added $\text{Cl}^-$ is a common ion that shifts the dissolution equilibrium left 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.
13. To mimic gastrointestinal antacid chemistry, a researcher studies $\text{Mg(OH)}_2(s)$ in water at 25°C: $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-$. The solubility product is $K_{sp}=5.6\times 10^{-12}$. The solution is then dosed with NaOH, increasing $[\text{OH}^-]$ while keeping volume essentially constant. Which shift in equilibrium is most consistent with adding the common ion $\text{OH}^-$?
    Answer: Shift left; higher $[\text{OH}^-]$ decreases the solubility of $\text{Mg(OH)}_2$ This question tests solubility and solubility product concepts (5A). The solubility product principle combined with Le Chatelier's principle predicts that adding a common ion shifts the equilibrium to reduce that ion's concentration. In the Mg(OH)₂ dissolution equilibrium, OH⁻ is a product, so adding NaOH increases [OH⁻], which is a common ion. The correct answer (B) follows because increasing [OH⁻] raises Q above Ksp, shifting the equilibrium left toward solid formation and decreasing solubility. Choice A incorrectly claims that Ksp changes with concentration, when Ksp is only temperature-dependent. To analyze common ion effects, identify which ions appear in the equilibrium expression and remember that adding any of these ions always decreases solubility of the solid.
14. In a wastewater treatment test at 25°C, a chemist adds sulfide to precipitate zinc as $\text{ZnS}$: $\text{ZnS}(s) \rightleftharpoons \text{Zn}^{2+} + \text{S}^{2-}$ with $K_{sp}=2.0\times10^{-25}$. If $[\text{S}^{2-}]$ is increased while temperature is constant, which shift in equilibrium is most consistent with adding the common ion $\text{S}^{2-}$?
    Answer: Shift left; precipitation is promoted and dissolved $[\text{Zn}^{2+}]$ decreases. This question tests solubility and solubility product concepts (5A). The common ion effect predicts that adding S²⁻ to a system with ZnS(s) shifts the equilibrium ZnS(s) ⇌ Zn²⁺ + S²⁻ to the left. Since S²⁻ is a product in this dissolution equilibrium, increasing its concentration drives precipitation according to Le Chatelier's principle. This shift causes more ZnS to precipitate and reduces the dissolved [Zn²⁺] concentration. The correct answer A accurately describes this shift left with precipitation promoted and decreased [Zn²⁺]. Answer B incorrectly suggests dissolution to consume sulfide, but S²⁻ is a product of dissolution, not consumed by it. When treating wastewater with sulfide precipitation, adding excess S²⁻ ensures maximum removal of metal ions by driving equilibrium toward the solid phase.
15. A lab studies $\text{Fe(OH)}_3$ precipitation in a wound environment, modeling $\text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+} + 3\text{OH}^-$. At 25°C, $K_{sp}=2.8\times10^{-39}$. If the local pH increases (higher $[\text{OH}^-]$) while temperature is constant, which shift in equilibrium is most consistent with adding the common ion $\text{OH}^-$?
    Answer: Shift left; precipitation is promoted and free $[\text{Fe}^{3+}]$ decreases. This question tests solubility and solubility product concepts (5A). The common ion effect predicts that increasing [OH⁻] in a system at equilibrium with Fe(OH)₃(s) shifts the reaction Fe(OH)₃(s) ⇌ Fe³⁺ + 3OH⁻ to the left. Since OH⁻ is a product in this dissolution equilibrium, raising pH (increasing [OH⁻]) drives precipitation according to Le Chatelier's principle. This shift causes more Fe(OH)₃ solid to form and reduces the free [Fe³⁺] concentration. The correct answer A accurately describes this shift left with precipitation promoted and decreased [Fe³⁺]. Answer B incorrectly suggests dissolution to neutralize base, but OH⁻ is a product of dissolution, not consumed by it. For metal hydroxides, remember that increasing pH always decreases solubility by providing excess OH⁻ common ion.
16. 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?
    Answer: $[\text{Mg}^{2+}]$ decreases because added $\text{OH}^-$ shifts the equilibrium left, lowering solubility 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.
17. To probe drug–electrolyte interactions, researchers prepare a saturated solution of $\text{AgCl}(s)$ in water at 25°C: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-\,(K_{sp}=1.8\times10^{-10})$. They then add NaCl to raise $[\text{Cl}^-]$ while maintaining constant temperature and volume. Which shift in equilibrium is most consistent with adding the common ion $\text{Cl}^-$?
    Answer: Shift left, decreasing $[\text{Ag}^+]$ and promoting AgCl(s) formation This question tests solubility and solubility product concepts (5A). The solubility product for AgCl is Ksp = [Ag⁺][Cl⁻], which must remain constant at a given temperature. When NaCl is added to a saturated AgCl solution, the chloride ion concentration increases, temporarily making Q > Ksp. To restore equilibrium, the system shifts left according to Le Châtelier's principle, converting dissolved Ag⁺ and Cl⁻ ions into solid AgCl until Q again equals Ksp. Choice D incorrectly states that adding an electrolyte increases solubility through the common ion effect, when in fact the common ion effect always decreases solubility of the original salt. Students should remember that adding a common ion to a saturated solution always promotes precipitation, not dissolution, which is why hard water (high in Ca²⁺) readily forms scale deposits when combined with soap containing fatty acid anions.
18. 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 $K_{sp}$?
    Answer: More $\text{CaCO}_3$ dissolves because lowering $[\text{CO}_3^{2-}]$ lowers $Q$ below $K_{sp}$. 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.
19. A microbiology team uses $\text{AgCl}$ as a slow-release antimicrobial, but the growth medium contains chloride. At 25°C, $K_{sp}(\text{AgCl})=1.8\times10^{-10}$. Holding temperature constant, which statement best predicts how increasing $[\text{Cl}^-]$ in the medium affects dissolved $[\text{Ag}^+]$ at equilibrium with solid present?
    Answer: Dissolved $[\text{Ag}^+]$ decreases because added $\text{Cl}^-$ shifts dissolution equilibrium left. This question tests solubility and solubility product concepts (5A). For AgCl at equilibrium, Ksp = [Ag⁺][Cl⁻] must remain constant at a given temperature. When [Cl⁻] increases in the medium, the product [Ag⁺][Cl⁻] would exceed Ksp unless [Ag⁺] decreases proportionally. The system responds by shifting the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left, precipitating more AgCl and reducing [Ag⁺] until Ksp is satisfied. The correct answer A states that dissolved [Ag⁺] decreases because added Cl⁻ shifts dissolution equilibrium left. Answer C incorrectly claims [Ag⁺] is independent of [Cl⁻], but Ksp = [Ag⁺][Cl⁻] shows they are inversely related at equilibrium. This common ion effect is why AgCl antimicrobials are less effective in high-chloride environments.
20. A physiology lab models precipitation of $\text{Ca(OH)}_2$ in an alkaline microenvironment: $\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-$. At 25°C, $K_{sp}=5.5\times10^{-6}$. If pH rises (higher $[\text{OH}^-]$) while $[\text{Ca}^{2+}]$ is held constant, which statement best predicts the equilibrium response?
    Answer: Solubility decreases because added $\text{OH}^-$ is a common ion that shifts equilibrium left. This question tests solubility and solubility product concepts (5A). For Ca(OH)₂ dissolution equilibrium Ca(OH)₂(s) ⇌ Ca²⁺ + 2OH⁻, increasing pH means increasing [OH⁻], which is a common ion effect. Since OH⁻ is a product in this equilibrium, raising its concentration shifts the reaction left toward precipitation according to Le Chatelier's principle. This decreases the solubility of Ca(OH)₂ by promoting solid formation. The correct answer B states that solubility decreases because added OH⁻ is a common ion that shifts equilibrium left. Answer A incorrectly claims that base drives dissolution right, when actually OH⁻ is produced by dissolution, not consumed. For any sparingly soluble hydroxide, increasing pH (adding OH⁻) always decreases solubility through the common ion effect.

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