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

What this quiz covers

This quiz focuses on 5a Solubility Solubility Product, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Shift right; more $\text{CaF}_2$ dissolves and $[\text{F}^-]$ increases.
2. Shift left; precipitation is promoted and dissolved $[\text{F}^-]$ decreases. (correct answer)
3. No shift; $\text{Ca}^{2+}$ does not appear in the dissolution reaction.
4. Shift right; added $\text{Ca}^{2+}$ must be consumed, increasing solubility.

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).

Question 2

1. More $\text{CaCO}_3$ precipitates because lowering $[\text{CO}_3^{2-}]$ increases $Q$.
2. More $\text{CaCO}_3$ dissolves because lowering $[\text{CO}_3^{2-}]$ lowers $Q$ below $K_{sp}$. (correct answer)
3. No change occurs because $K_{sp}$ is constant and forces $Q$ to remain unchanged without dissolution.
4. More $\text{CaCO}_3$ dissolves because adding CO$_2$ adds a common ion to the dissolution reaction.

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.

Question 3

1. Solubility decreases because lowering $[\text{CO}_3^{2-}]$ forces $[\text{Ca}^{2+}]$ to decrease to keep $K_{sp}$ constant
2. Solubility increases because consuming $\text{CO}_3^{2-}$ allows more CaCO$_3$ to dissolve while maintaining $[\text{Ca}^{2+}][\text{CO}_3^{2-}] = K_{sp}$ (correct answer)
3. Solubility is unchanged because CO$_2$ affects only acid–base equilibria, not solubility equilibria
4. Solubility increases because $K_{sp}$ becomes larger when CO$_2$ is added, independent of temperature

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.

Question 4

1. It would make the computed $Q$ much smaller than the true $Q$, potentially suggesting undersaturation when the solution is actually saturated (correct answer)
2. It would make the computed $Q$ much larger than the true $Q$, potentially suggesting precipitation when the solution is actually saturated
3. It would not change $Q$ because $Q$ is dimensionless and unit choices cancel
4. 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.

Question 5

1. $[\text{Mg}^{2+}]$ increases because adding base shifts dissolution right to consume $\text{OH}^-$
2. $[\text{Mg}^{2+}]$ decreases because added $\text{OH}^-$ shifts the equilibrium left, lowering solubility (correct answer)
3. $[\text{Mg}^{2+}]$ is unchanged because $K_{sp}$ fixes the solubility regardless of added ions
4. $[\text{Mg}^{2+}]$ increases until $[\text{Mg}^{2+}] = 2[\text{OH}^-]$ to match stoichiometry

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.

Question 6

1. Precipitation is favored because $Q = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$ exceeds $K_{sp}$ (correct answer)
2. Precipitation is favored because $K_{sp}$ increases with higher ion concentrations in serum
3. No precipitation is favored because $Q$ is less than $K_{sp}$ at these concentrations
4. No precipitation is favored because solids do not participate in equilibrium calculations, so $Q$ cannot be compared to $K_{sp}$

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.

Question 7

1. Shift right; increased $\text{Ca}^{2+}$ drives dissolution to restore $K_{sp}$
2. Shift left; increased $\text{Ca}^{2+}$ promotes precipitation of $\text{Ca}_3(\text{PO}_4)_2(s)$ (correct answer)
3. No shift; $K_{sp}$ fixes ion concentrations so added $\text{Ca}^{2+}$ remains in solution
4. 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.

Question 8

1. Precipitation is favored because $Q=[\text{Pb}^{2+}][\text{I}^-]^2$ is greater than $K_{sp}$ (correct answer)
2. Precipitation is disfavored because $Q$ should be computed as $[\text{Pb}^{2+}]^2[\text{I}^-]$
3. No precipitation occurs because iodide is a spectator ion and does not affect solubility
4. Precipitation is disfavored because $K_{sp}$ increases automatically when $[\text{I}^-]$ increases

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.

Question 9

1. Solubility is higher in both formulations because adding either ion increases the driving force to dissolve
2. Solubility is lower in both formulations because each adds a common ion to the dissolution equilibrium (correct answer)
3. Solubility is lower only in Formulation 1 because $\text{Ca}^{2+}$ appears first in the $K_{sp}$ expression
4. Solubility is unchanged because $K_{sp}$ fixes solubility independent of added ions

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.

Question 10

1. Shift right, dissolving more BaSO4(s) to keep $[\text{Ba}^{2+}]$ constant
2. Shift left, reducing dissolved $\text{Ba}^{2+}$ as more BaSO4(s) forms (correct answer)
3. No shift, because $K_{sp}$ is constant and therefore concentrations cannot change
4. Shift left only if pH decreases, because sulfate is a base and controls $K_{sp}$ directly

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.

Question 11

1. Shift right, increasing $[\text{C}_2\text{O}_4^{2-}]$ until $Q$ again equals $K_{sp}$
2. Shift left, favoring formation of CaC2O4(s) and decreasing dissolved oxalate at equilibrium (correct answer)
3. No shift, because $K_{sp}$ depends only on temperature and not ion concentrations
4. Shift right, because adding $\text{Ca}^{2+}$ increases ionic strength and always increases solubility

Explanation: This question tests solubility and solubility product concepts (5A). The solubility product constant (Ksp) represents the equilibrium between a solid and its dissolved ions, where Ksp = [Ca²⁺][C₂O₄²⁻] remains constant at a given temperature. When Ca²⁺ is added to a saturated solution, the ion product Q temporarily exceeds Ksp, creating a non-equilibrium state. According to Le Châtelier's principle, the system responds by shifting left to form more solid CaC₂O₄, thereby reducing the concentration of dissolved oxalate ions until Q again equals Ksp. Choice A incorrectly suggests the equilibrium shifts right, which would further increase Q above Ksp rather than restore equilibrium. The common ion effect demonstrates that adding an ion already present in the equilibrium always decreases the solubility of the salt, making precipitation more likely in biological fluids when common ions are elevated.

Question 12

1. Shift right, increasing $[\text{CO}_3^{2-}]$ until $Q$ equals $K_{sp}$ again
2. Shift left, decreasing $[\text{CO}_3^{2-}]$ as more $\text{Ag}_2\text{CO}_3(s)$ forms (correct answer)
3. No shift, because adding $\text{Ag}^+$ changes $Q$ but cannot change the amount of solid present
4. Shift right, because increasing cation concentration always increases the solubility of the corresponding salt

Explanation: This question tests solubility and solubility product concepts (5A). The solubility product for Ag₂CO₃ is Ksp = [Ag⁺]²[CO₃²⁻], where silver concentration is squared due to the 2:1 stoichiometry. Adding AgNO₃ increases [Ag⁺], causing Q to exceed Ksp dramatically since Q increases with [Ag⁺]². The system responds by shifting left to precipitate more Ag₂CO₃ solid, thereby reducing the dissolved carbonate concentration until equilibrium is restored. Choice D incorrectly claims that increasing cation concentration always increases solubility, contradicting the fundamental principle of the common ion effect. Students should recognize that the common ion effect applies regardless of which ion is added, and the magnitude of the effect depends on the stoichiometric coefficient of the added ion in the dissolution equation.

Question 13

1. Shift right, increasing $[\text{PO}_4^{3-}]$ to offset the added $\text{Ca}^{2+}$
2. Shift left, decreasing dissolved phosphate as more solid forms (correct answer)
3. No shift, because $K_{sp}$ fixes $[\text{Ca}^{2+}]$ at a constant value regardless of additions
4. Shift right, because adding CaCl2 increases total ionic concentration and therefore increases solubility

Explanation: This question tests solubility and solubility product concepts (5A). For Ca₃(PO₄)₂, the solubility product is Ksp = [Ca²⁺]³[PO₄³⁻]², where calcium concentration is cubed due to the 3:2 stoichiometry. Adding CaCl₂ increases [Ca²⁺], causing Q to exceed Ksp dramatically since Q varies with [Ca²⁺]³. To restore equilibrium, the system shifts left according to Le Châtelier's principle, forming more solid calcium phosphate and reducing the dissolved phosphate concentration. Choice C incorrectly suggests that Ksp fixes ion concentrations at constant values, when actually Ksp fixes only the mathematical product of the ion concentrations raised to their stoichiometric powers. The common ion effect is particularly important in biological systems where calcium and phosphate concentrations must be carefully regulated to prevent pathological calcification.

Question 14

1. Shift right, because increasing $[\text{OH}^-]$ increases the solubility of $\text{Mg(OH)}_2$
2. Shift left, decreasing dissolved $[\text{Mg}^{2+}]$ as more solid forms (correct answer)
3. No shift, because $\text{OH}^-$ is not in the $K_{sp}$ expression for a sparingly soluble base
4. Shift right, because $K_{sp}$ increases when any ion is added to solution

Explanation: This question tests solubility and solubility product concepts (5A). For Mg(OH)₂, the solubility product is Ksp = [Mg²⁺][OH⁻]², where the hydroxide concentration is squared due to the stoichiometry. Adding NaOH increases [OH⁻], causing Q to exceed Ksp since Q varies with [OH⁻]². To restore equilibrium, the system shifts left, precipitating more Mg(OH)₂ solid and reducing the dissolved Mg²⁺ concentration. Choice C incorrectly claims OH⁻ is not in the Ksp expression, confusing the solid phase (which doesn't appear) with the dissolved ions (which do appear). Students should recognize that the common ion effect applies to any ion in the dissolution equation, and the effect is amplified when the common ion has a coefficient greater than one, as small changes in [OH⁻] produce large changes in Q.

Question 15

1. Shift left, lowering $[\text{Ca}^{2+}]$ as CaF2(s) precipitates (correct answer)
2. Shift right, increasing $[\text{Ca}^{2+}]$ because $[\text{F}^-]$ appears squared in $K_{sp}$
3. No shift, because adding NaF does not affect the CaF2 equilibrium when solid is present
4. Shift right, because precipitation can occur only when $Q<K_{sp}$

Explanation: This question tests solubility and solubility product concepts (5A). The solubility product for CaF₂ is Ksp = [Ca²⁺][F⁻]², where fluoride concentration is squared due to the 1:2 stoichiometry. Adding NaF increases [F⁻], causing Q to exceed Ksp dramatically since Q increases with the square of [F⁻]. The system responds by shifting left to form more solid CaF₂, thereby reducing the dissolved Ca²⁺ concentration until Q equals Ksp again. Choice D incorrectly states that precipitation occurs when Q < Ksp, when actually precipitation occurs when Q > Ksp (supersaturation). Students should remember that the common ion effect is particularly pronounced when the common ion has a coefficient greater than one in the balanced equation, making fluoride addition highly effective at removing calcium from solution.

Question 16

1. The solution remains unsaturated because increasing an anion always increases solubility.
2. Precipitation becomes more likely because $Q$ increases and can exceed $K_{sp}$, lowering free $[\text{Ca}^{2+}]$. (correct answer)
3. No change occurs because $K_{sp}$ depends only on temperature, so $Q$ is irrelevant.
4. Dissolution is favored because added $\text{PO}_4^{3-}$ shifts equilibrium right to consume it.

Explanation: 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.

Question 17

1. Precipitation is expected because $Q>K_{sp}$ under these conditions.
2. No precipitation is expected because $Q<K_{sp}$ under these conditions. (correct answer)
3. Precipitation is expected because $Q$ must always equal $K_{sp}$ in any mixture.
4. No precipitation is expected because increasing $[\text{Ca}^{2+}]$ increases solubility via complexation (not stated).

Explanation: This question tests solubility and solubility product concepts (5A). The ion product Q determines whether precipitation occurs by comparison with Ksp: if Q < Ksp, no precipitation occurs; if Q > Ksp, precipitation occurs. For CaC₂O₄, Q = [Ca²⁺][C₂O₄²⁻] = (2.0×10⁻³)(5.0×10⁻⁷) = 1.0×10⁻⁹. Since Q (1.0×10⁻⁹) < Ksp (2.3×10⁻⁹), the solution is unsaturated and no precipitation occurs. The correct answer B states that no precipitation is expected because Q < Ksp. Answer A incorrectly claims Q > Ksp, which would require Q to be larger than 2.3×10⁻⁹. To predict precipitation, always calculate Q using actual ion concentrations and compare numerically to Ksp - don't rely on qualitative assessments.

Question 18

1. $\text{BaSO}_4$ is more soluble because its $K_{sp}$ is smaller.
2. $\text{CaSO}_4$ is more soluble because its larger $K_{sp}$ indicates greater ion concentrations at equilibrium. (correct answer)
3. They have the same solubility because both produce two ions per formula unit.
4. $\text{BaSO}_4$ is more soluble because barium has a higher charge density.

Explanation: 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.

Question 19

1. Solubility increases because added base drives dissolution to the right.
2. Solubility decreases because added $\text{OH}^-$ is a common ion that shifts equilibrium left. (correct answer)
3. No change occurs because hydroxide is not included in the $K_{sp}$ expression.
4. Solubility increases because $K_{sp}$ increases as $[\text{OH}^-]$ increases.

Explanation: 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.

Question 20

1. Shift left; precipitation is promoted and dissolved $[\text{Zn}^{2+}]$ decreases. (correct answer)
2. Shift right; more $\text{ZnS}$ dissolves to consume sulfide.
3. No shift; $K_{sp}$ changes to keep $Q$ constant.
4. Shift right; added $\text{S}^{2-}$ increases $K_{sp}$.

Explanation: 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.