College Chemistry Quiz: Solubility
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SolubilityQuestion 1 of 19

A solution contains 0.10 M NaCl\text{NaCl} and 0.050 M AgNO3\text{AgNO}_3. Given that KspK_{sp} for AgCl\text{AgCl} is 1.8×10−101.8 \times 10^{-10}, what happens when these solutions are mixed?

No precipitation occurs because Q < Ksp for all possible combinations
AgCl precipitates because Q = 5.0 × 10⁻³, which exceeds Ksp significantly
AgNO₃ precipitates because its concentration exceeds the solubility limit
NaCl precipitates because Q > Ksp when calculated for sodium chloride
Both AgCl and NaCl precipitate because both Q values exceed their respective Ksp values
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College Chemistry Quiz

College Chemistry Quiz: Solubility

Practice Solubility in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solubility, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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.

All questions

Question 1

A solution contains 0.10 M NaCl\text{NaCl} and 0.050 M AgNO3\text{AgNO}_3. Given that KspK_{sp} for AgCl\text{AgCl} is 1.8×10−101.8 \times 10^{-10}, what happens when these solutions are mixed?

  1. No precipitation occurs because Q < Ksp for all possible combinations
  2. AgCl precipitates because Q = 5.0 × 10⁻³, which exceeds Ksp significantly (correct answer)
  3. AgNO₃ precipitates because its concentration exceeds the solubility limit
  4. NaCl precipitates because Q > Ksp when calculated for sodium chloride
  5. Both AgCl and NaCl precipitate because both Q values exceed their respective Ksp values
Explanation: When you encounter a precipitation problem, you need to calculate the reaction quotient (Q) and compare it to the solubility product constant (KspK_{sp}) to predict whether a precipitate will form. The key reaction here is: Ag++Cl−⇌AgCl(s)\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(s)} To find Q, you multiply the concentrations of the ions that could form the precipitate: Q=[Ag+][Cl−]=(0.050)(0.10)=5.0×10−3Q = [\text{Ag}^+][\text{Cl}^-] = (0.050)(0.10) = 5.0 \times 10^{-3} Since Q=5.0×10−3Q = 5.0 \times 10^{-3} and Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10}, we have Q>>KspQ >> K_{sp}. When Q exceeds KspK_{sp}, the solution is supersaturated and precipitation occurs to restore equilibrium. Looking at the wrong answers: A) is incorrect because Q does exceed KspK_{sp} significantly—by about 7 orders of magnitude. C) is wrong because AgNO3\text{AgNO}_3 is highly soluble and won't precipitate under these conditions. D) makes no sense because NaCl\text{NaCl} is also highly soluble, and you wouldn't use the KspK_{sp} of AgCl\text{AgCl} to predict NaCl\text{NaCl} precipitation. The answer is B—AgCl\text{AgCl} precipitates because Q greatly exceeds KspK_{sp}. Study tip: Always identify which ions can combine to form a low-solubility compound, then calculate Q using those specific ion concentrations. If Q>KspQ > K_{sp}, precipitation occurs; if Q<KspQ < K_{sp}, no precipitation happens.

Question 2

Which statement best explains why AgCl\text{AgCl} is more soluble in ammonia solution than in pure water?

  1. Ammonia increases the pH, which shifts the equilibrium toward greater AgCl dissolution
  2. Ag⁺ ions form stable complex ions with NH₃, removing free Ag⁺ and driving dissolution forward (correct answer)
  3. NH₃ molecules hydrogen bond with Cl⁻ ions, stabilizing them in solution
  4. Ammonia acts as a common ion that increases the solubility through Le Châtelier's principle
  5. The higher polarity of ammonia solution provides better solvation of both AgCl molecules
Explanation: This question tests your understanding of complex ion formation and its effect on solubility equilibria. When you encounter solubility problems involving ligands like ammonia, think about how metal ions can form coordination complexes. Silver chloride has low solubility in water due to the equilibrium: AgCl(s)⇌Ag++Cl−\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-. In ammonia solution, however, Ag+\text{Ag}^+ ions form stable complex ions with NH3\text{NH}_3 molecules: Ag++2NH3→[Ag(NH3)2]+\text{Ag}^+ + 2\text{NH}_3 \rightarrow \text{[Ag(NH}_3\text{)}_2\text{]}^+. This complexation removes free Ag+\text{Ag}^+ ions from solution, which by Le Châtelier's principle drives the dissolution equilibrium forward to replace them. More AgCl\text{AgCl} dissolves to maintain equilibrium, dramatically increasing overall solubility. Choice A incorrectly focuses on pH effects. While ammonia is basic, the increased solubility isn't primarily due to pH changes affecting the equilibrium position. Choice C misidentifies the interaction—NH3\text{NH}_3 doesn't significantly hydrogen bond with Cl−\text{Cl}^- ions, and this wouldn't explain the solubility increase. Choice D incorrectly applies the common ion effect. Ammonia isn't a "common ion" (it's not Ag+\text{Ag}^+ or Cl−\text{Cl}^-), and the common ion effect actually decreases solubility by shifting equilibrium toward the solid. Remember this pattern: when a sparingly soluble salt containing a metal cation is dissolved in a solution of a good ligand, look for complex ion formation as the driving force behind increased solubility. This is a key concept in analytical chemistry separations.

Question 3

Which factor would most significantly increase the solubility of CaCO3\text{CaCO}_3 in an aqueous solution?

  1. Increasing the temperature from 25°C to 75°C
  2. Adding a small amount of solid NaCl to increase ionic strength
  3. Decreasing the pH by adding HCl to protonate carbonate ions (correct answer)
  4. Adding Ca(NO₃)₂ to provide additional calcium ions for complexation
  5. Increasing the pressure to force more solid into solution
Explanation: When you encounter solubility questions involving ionic compounds, think about Le Châtelier's principle and how different factors affect the equilibrium between the solid and its dissolved ions. For CaCO3\text{CaCO}_3, the dissolution equilibrium is: CaCO3(s)⇌Ca2+(aq)+CO32−(aq)\text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) However, carbonate ions are basic and readily accept protons: CO32−+H+⇌HCO3−\text{CO}_3^{2-} + \text{H}^+ \rightleftharpoons \text{HCO}_3^- Option C is correct because adding HCl decreases pH, providing H+\text{H}^+ ions that protonate carbonate ions. This removes CO32−\text{CO}_3^{2-} from solution, driving the dissolution equilibrium forward according to Le Châtelier's principle, dramatically increasing solubility. Option A is incorrect because most ionic solids, including CaCO3\text{CaCO}_3, have very low temperature dependence for solubility. The increase would be minimal. Option B is wrong because increasing ionic strength through NaCl addition has only a small effect on solubility. While it can slightly increase solubility through activity coefficient changes, the effect is nowhere near as significant as pH changes. Option D represents a common misconception. Adding Ca(NO3)2\text{Ca(NO}_3)_2 increases Ca2+\text{Ca}^{2+} concentration, which actually decreases CaCO3\text{CaCO}_3 solubility due to the common ion effect, pushing the equilibrium toward the solid. Study tip: For sparingly soluble salts containing basic anions (like CO32−\text{CO}_3^{2-}, PO43−\text{PO}_4^{3-}, or S2−\text{S}^{2-}), decreasing pH almost always dramatically increases solubility by protonating the anion and removing it from the equilibrium.

Question 4

A solution contains equal molar concentrations of Cl−\text{Cl}^-, Br−\text{Br}^-, and I−\text{I}^- ions, each at 0.010 M. When AgNO3\text{AgNO}_3 solution is slowly added, which silver halide will precipitate first? (KspK_{sp} values: AgCl = 1.8×10−101.8 \times 10^{-10}, AgBr = 5.4×10−135.4 \times 10^{-13}, AgI = 8.5×10−178.5 \times 10^{-17})

  1. AgCl precipitates first because it has the highest Ksp value
  2. AgI precipitates first because it requires the lowest [Ag⁺] to reach saturation (correct answer)
  3. AgBr precipitates first because it has an intermediate Ksp value
  4. All three precipitate simultaneously because they have equal anion concentrations
  5. AgCl precipitates first because chloride ions are smallest and react fastest
Explanation: When you encounter precipitation problems with multiple competing equilibria, you need to determine which compound will reach its saturation point first as the precipitating ion is gradually added. To find which silver halide precipitates first, calculate the minimum [Ag+][\text{Ag}^+] needed to saturate each solution using the relationship Ksp=[Ag+][X−]K_{sp} = [\text{Ag}^+][\text{X}^-], where [X−]=0.010 M[\text{X}^-] = 0.010 \text{ M} for all halides. For AgCl: [Ag+]=1.8×10−100.010=1.8×10−8 M[\text{Ag}^+] = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \text{ M} For AgBr: [Ag+]=5.4×10−130.010=5.4×10−11 M[\text{Ag}^+] = \frac{5.4 \times 10^{-13}}{0.010} = 5.4 \times 10^{-11} \text{ M} For AgI: [Ag+]=8.5×10−170.010=8.5×10−15 M[\text{Ag}^+] = \frac{8.5 \times 10^{-17}}{0.010} = 8.5 \times 10^{-15} \text{ M} AgI requires the lowest [Ag+][\text{Ag}^+] concentration to reach saturation, so it precipitates first when AgNO3\text{AgNO}_3 is added dropwise. Answer B is correct. Answer A incorrectly suggests that higher KspK_{sp} values lead to earlier precipitation—actually, lower KspK_{sp} values indicate less soluble compounds that precipitate more readily. Answer C arbitrarily chooses the intermediate KspK_{sp} value without proper calculation. Answer D ignores the fact that different KspK_{sp} values mean different saturation points, even with equal anion concentrations. Study tip: In competitive precipitation problems, always calculate the minimum precipitating ion concentration needed for each compound. The one requiring the lowest concentration precipitates first, regardless of which has the highest or lowest KspK_{sp}.

Question 5

Which of the following best explains why CaF2\text{CaF}_2 is less soluble in CaCl2\text{CaCl}_2 solution than in pure water?

  1. Cl⁻ ions compete with F⁻ ions for solvation by water molecules
  2. The common Ca²⁺ ion shifts the equilibrium toward less CaF₂ dissolution (correct answer)
  3. CaCl₂ forms ion pairs with CaF₂, reducing the effective concentration
  4. The higher ionic strength decreases the activity coefficients of all ions
  5. Cl⁻ ions form stronger ionic bonds with Ca²⁺ than F⁻ ions do
Explanation: When you encounter solubility problems involving solutions that share a common ion, think about Le Châtelier's principle and equilibrium shifts. The dissolution of calcium fluoride follows this equilibrium: CaF2(s)⇌Ca2+(aq)+2F−(aq)\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) When CaF2\text{CaF}_2 dissolves in a CaCl2\text{CaCl}_2 solution instead of pure water, the solution already contains Ca2+\text{Ca}^{2+} ions from the dissolved CaCl2\text{CaCl}_2. This creates a "common ion effect." According to Le Châtelier's principle, adding more Ca2+\text{Ca}^{2+} ions shifts the equilibrium to the left, favoring the solid form and reducing CaF2\text{CaF}_2 solubility. This makes answer B correct. Let's examine why the other options are wrong: A is incorrect because Cl−\text{Cl}^- and F−\text{F}^- ions don't significantly compete for water solvation in a way that would affect solubility to this extent. C misrepresents the chemistry—CaCl2\text{CaCl}_2 and CaF2\text{CaF}_2 don't form meaningful ion pairs that would reduce effective concentrations in this context. D mentions activity coefficients and ionic strength effects, which are real phenomena but secondary compared to the dominant common ion effect occurring here. Study tip: Whenever you see solubility questions involving solutions with shared ions, immediately think "common ion effect" and Le Châtelier's principle. The shared ion will always reduce the solubility of the compound compared to pure water.

Question 6

Which compound would be expected to be most soluble in a polar solvent like water?

  1. I2\text{I}_2 (nonpolar covalent compound with large, polarizable atoms)
  2. CCl4\text{CCl}_4 (nonpolar covalent compound with symmetrical tetrahedral structure)
  3. KBr\text{KBr} (ionic compound with high lattice energy and small ion charges) (correct answer)
  4. C6H14\text{C}_6\text{H}_{14} (nonpolar hydrocarbon with only dispersion forces)
  5. SiO2\text{SiO}_2 (covalent network solid with strong directional bonds)
Explanation: When approaching solubility questions, remember the fundamental principle "like dissolves like" — polar solvents dissolve polar or ionic substances best, while nonpolar solvents dissolve nonpolar substances. Water is a highly polar solvent with the ability to form hydrogen bonds and surround ions through ion-dipole interactions. For a compound to dissolve well in water, it must be able to overcome its own intermolecular forces and interact favorably with water molecules. KBr\text{KBr} (choice C) is an ionic compound that dissociates into K+\text{K}^+ and Br−\text{Br}^- ions. Water molecules can surround these ions through ion-dipole interactions, where the partially negative oxygen atoms orient toward K+\text{K}^+ and the partially positive hydrogen atoms orient toward Br−\text{Br}^-. This process, called hydration, releases enough energy to overcome the lattice energy and make KBr\text{KBr} highly soluble. Choice A (I2\text{I}_2) is nonpolar and held together by weak London dispersion forces, but these interactions with water are much weaker than water's hydrogen bonding with itself. Choice B (CCl4\text{CCl}_4) has a symmetrical tetrahedral structure making it completely nonpolar despite having polar C-Cl bonds. Choice D (C6H14\text{C}_6\text{H}_{14}) is a hydrocarbon with only weak dispersion forces and no ability to interact meaningfully with polar water molecules. Study tip: When comparing solubility in polar solvents, rank compounds in this order of preference: ionic compounds > polar covalent compounds > nonpolar compounds. Ionic compounds almost always win due to the strength of ion-dipole interactions.

Question 7

A solution at equilibrium contains 1.2×10−31.2 \times 10^{-3} M Ca2+\text{Ca}^{2+} and 3.4×10−23.4 \times 10^{-2} M F−\text{F}^-. What is the reaction quotient Q if additional CaF2\text{CaF}_2 solid is added to this solution?

  1. Q = 0 because solid CaF₂ is not included in the expression
  2. Q = 1.4×10−61.4 \times 10^{-6} and the system is at equilibrium
  3. Q = 4.1×10−84.1 \times 10^{-8} and more CaF₂ will dissolve
  4. Q = 1.4×10−61.4 \times 10^{-6} and CaF₂ will precipitate if this exceeds Ksp (correct answer)
  5. Q cannot be calculated without knowing the amount of solid CaF₂ added
Explanation: When dealing with solubility equilibria, you need to understand how the reaction quotient Q relates to the equilibrium constant Ksp to predict what happens when conditions change. For the dissolution of CaF2\text{CaF}_2: CaF2(s)⇌Ca2+(aq)+2F−(aq)\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq), the reaction quotient is Q=[Ca2+][F−]2Q = [\text{Ca}^{2+}][\text{F}^-]^2. Note that the fluoride concentration is squared because of the stoichiometry. Calculate Q using the given concentrations: Q=(1.2×10−3)(3.4×10−2)2=(1.2×10−3)(1.156×10−3)=1.4×10−6Q = (1.2 \times 10^{-3})(3.4 \times 10^{-2})^2 = (1.2 \times 10^{-3})(1.156 \times 10^{-3}) = 1.4 \times 10^{-6} When additional solid CaF2\text{CaF}_2 is added, Q remains the same initially because the ion concentrations haven't changed yet—solids don't affect Q values. However, if this Q value exceeds the Ksp for CaF2\text{CaF}_2, precipitation will occur to restore equilibrium. Choice A is wrong because while solids aren't included in Q expressions, Q still has a meaningful value based on the dissolved ions. Choice B incorrectly assumes the system remains at equilibrium after adding solid. Choice C uses an incorrect calculation—likely forgetting to square the fluoride concentration—and draws the wrong conclusion about dissolution versus precipitation. The key insight is that adding solid to a solution already containing dissolved ions creates a situation where you must compare Q to Ksp to predict the direction of reaction. Always remember to include proper stoichiometric coefficients as exponents when calculating Q or K expressions.

Question 8

The solubility of oxygen gas in water decreases as temperature increases. Which statement best explains this observation?

  1. Higher temperature increases kinetic energy, making O₂ molecules escape solution more easily (correct answer)
  2. Gas dissolution is always exothermic, so higher temperature shifts equilibrium toward reactants
  3. Increased molecular motion at higher temperature disrupts hydrogen bonding between O₂ and H₂O
  4. Higher temperature decreases water density, providing less space for dissolved gas molecules
  5. Gas solubility follows Henry's law, which has an inverse temperature dependence
Explanation: When you encounter gas solubility problems involving temperature changes, think about the kinetic molecular theory and how thermal energy affects molecular behavior in different phases. Gas solubility in liquids follows Henry's Law and is influenced by temperature through molecular kinetics. As temperature increases, dissolved gas molecules gain kinetic energy, making them move faster and more likely to overcome the intermolecular forces keeping them in solution. This increased molecular motion provides enough energy for gas molecules to escape from the liquid phase back into the gas phase, reducing solubility. Choice A correctly identifies this mechanism - higher kinetic energy at elevated temperatures makes it easier for O₂ molecules to break free from solution and return to the gas phase above the liquid. Choice B incorrectly assumes gas dissolution is always exothermic. While many gas dissolution processes are exothermic, this isn't universal, and the question asks for the best explanation of the molecular mechanism, not just Le Chatelier's principle. Choice C misrepresents the intermolecular forces involved. Oxygen is nonpolar and doesn't form hydrogen bonds with water; instead, O₂ dissolves through weaker van der Waals forces. Choice D incorrectly suggests that decreased water density creates less space for gas. Actually, lower density typically means more space between molecules, and the solubility decrease isn't primarily due to volume changes. Remember: for gas-liquid solubility problems, focus on how temperature affects molecular kinetic energy and the ability of gas molecules to escape solution.

Question 9

Which of the following correctly predicts the relative solubilities of the alkaline earth metal hydroxides in water?

  1. Mg(OH)₂ > Ca(OH)₂ > Sr(OH)₂ > Ba(OH)₂ (decreases down the group)
  2. Ba(OH)₂ > Sr(OH)₂ > Ca(OH)₂ > Mg(OH)₂ (increases down the group) (correct answer)
  3. Ca(OH)₂ > Mg(OH)₂ > Sr(OH)₂ > Ba(OH)₂ (maximum in the middle)
  4. All have approximately equal solubility since they have the same charge ratios
  5. Sr(OH)₂ > Ba(OH)₂ > Mg(OH)₂ > Ca(OH)₂ (irregular pattern due to hydration)
Explanation: When analyzing trends in alkaline earth metal compounds, you need to consider how ionic size affects lattice energy and hydration energy. For hydroxides specifically, solubility increases as you move down Group 2 from magnesium to barium. This trend occurs because of competing energy factors. As the metal cation gets larger (Mg²⁺ < Ca²⁺ < Sr²⁺ < Ba²⁺), the lattice energy—the energy holding the ionic solid together—decreases more rapidly than the hydration energy. Larger cations have weaker electrostatic attractions to the hydroxide ions, making the crystal lattice easier to break apart. While the hydration energy also decreases for larger ions, the net effect favors increased solubility down the group. Option B correctly shows this trend: Ba(OH)₂ > Sr(OH)₂ > Ca(OH)₂ > Mg(OH)₂, with barium hydroxide being quite soluble and magnesium hydroxide being relatively insoluble. Option A reverses this trend, suggesting smaller cations form more soluble hydroxides—this contradicts experimental observations. Option C incorrectly places calcium hydroxide as most soluble, which isn't supported by the systematic decrease in lattice energy down the group. Option D ignores the significant differences in ionic radii; while all these compounds do have 2:1 charge ratios, the varying cation sizes create substantial solubility differences. Remember: For alkaline earth hydroxides, larger cations mean greater solubility. This contrasts with some other compound types, so always consider the specific anion when predicting Group 2 solubility trends.

Question 10

A solution contains 0.0200.020 M Pb2+\text{Pb}^{2+} and 0.0300.030 M Mg2+\text{Mg}^{2+}. If solid NaOH\text{NaOH} is gradually added, which hydroxide will precipitate first? (KspK_{sp}: Pb(OH)2=1.4×10−20\text{Pb(OH)}_2 = 1.4 \times 10^{-20}, Mg(OH)2=5.6×10−12\text{Mg(OH)}_2 = 5.6 \times 10^{-12})

  1. Pb(OH)₂ precipitates first because it has the lower Ksp value
  2. Mg(OH)₂ precipitates first because Mg²⁺ has higher concentration
  3. Both precipitate simultaneously since they have the same stoichiometry
  4. Pb(OH)₂ precipitates first because it requires lower [OH⁻] to reach saturation (correct answer)
  5. Mg(OH)₂ precipitates first because it has the higher Ksp value
Explanation: When you encounter competing precipitation problems, you need to determine which compound reaches its solubility limit first by calculating the required ion concentrations for saturation. To find which hydroxide precipitates first, calculate the [OH−][\text{OH}^-] needed for each compound to reach its KspK_{sp}. For both hydroxides, the equilibrium expression is Ksp=[M2+][OH−]2K_{sp} = [\text{M}^{2+}][\text{OH}^-]^2. For Pb(OH)2\text{Pb(OH)}_2: 1.4×10−20=(0.020)[OH−]21.4 \times 10^{-20} = (0.020)[\text{OH}^-]^2 Solving: [OH−]2=7.0×10−19[\text{OH}^-]^2 = 7.0 \times 10^{-19}, so [OH−]=2.6×10−10 M[\text{OH}^-] = 2.6 \times 10^{-10} \text{ M} For Mg(OH)2\text{Mg(OH)}_2: 5.6×10−12=(0.030)[OH−]25.6 \times 10^{-12} = (0.030)[\text{OH}^-]^2 Solving: [OH−]2=1.9×10−10[\text{OH}^-]^2 = 1.9 \times 10^{-10}, so [OH−]=1.4×10−5 M[\text{OH}^-] = 1.4 \times 10^{-5} \text{ M} Since Pb(OH)2\text{Pb(OH)}_2 requires much lower [OH−][\text{OH}^-] to precipitate, it forms first as you gradually add NaOH\text{NaOH}. Answer choice A is wrong because having a lower KspK_{sp} doesn't automatically mean precipitation occurs first—you must consider the actual ion concentrations. Choice B incorrectly assumes higher metal ion concentration determines precipitation order. Choice C is wrong because even though both have 1:2 stoichiometry, their different KspK_{sp} values and concentrations create different saturation requirements. Study tip: For selective precipitation problems, always calculate the required precipitating ion concentration for each compound rather than just comparing KspK_{sp} values directly. The compound needing the lowest concentration of the added ion precipitates first.

Question 11

If the KspK_{sp} of BaSO4\text{BaSO}_4 is 1.1×10−101.1 \times 10^{-10} at 25°C, what is the minimum concentration of SO42−\text{SO}_4^{2-} needed to precipitate BaSO4\text{BaSO}_4 from a solution containing 2.0×10−42.0 \times 10^{-4} M Ba2+\text{Ba}^{2+}?

  1. 5.5×10−75.5 \times 10^{-7} M (correct answer)
  2. 1.1×10−101.1 \times 10^{-10} M
  3. 2.2×10−142.2 \times 10^{-14} M
  4. 1.5×10−61.5 \times 10^{-6} M
  5. 4.7×10−44.7 \times 10^{-4} M
Explanation: When you encounter solubility and precipitation problems, you're dealing with equilibrium chemistry. The key insight is that precipitation occurs when the ion product exceeds the solubility product constant (KspK_{sp}). For BaSO4\text{BaSO}_4, the equilibrium expression is: Ksp=[Ba2+][SO42−]=1.1×10−10K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = 1.1 \times 10^{-10} At the exact point of precipitation, the ion product equals KspK_{sp}. You can solve for the minimum [SO42−][\text{SO}_4^{2-}] needed: 1.1×10−10=(2.0×10−4)[SO42−]1.1 \times 10^{-10} = (2.0 \times 10^{-4})[\text{SO}_4^{2-}] [SO42−]=1.1×10−102.0×10−4=5.5×10−7 M[\text{SO}_4^{2-}] = \frac{1.1 \times 10^{-10}}{2.0 \times 10^{-4}} = 5.5 \times 10^{-7} \text{ M} This confirms answer A is correct. Looking at the wrong answers: B (1.1×10−101.1 \times 10^{-10} M) represents the KspK_{sp} value itself—a common trap where students confuse the equilibrium constant with a concentration. C (2.2×10−142.2 \times 10^{-14} M) likely comes from incorrectly multiplying KspK_{sp} by the Ba2+\text{Ba}^{2+} concentration instead of dividing. D (1.5×10−61.5 \times 10^{-6} M) may result from calculation errors or using incorrect significant figures. Study tip: In precipitation problems, always set up the KspK_{sp} expression first, then substitute known concentrations. Remember that precipitation begins when the ion product equals KspK_{sp}—any higher concentration will cause precipitation to occur.

Question 12

Which factor does NOT significantly affect the solubility of an ionic compound in water?

  1. The lattice energy of the ionic solid, which affects energy required for dissolution
  2. The hydration energy of the constituent ions, which affects solvation stability
  3. The presence of common ions in solution, which shifts dissolution equilibrium
  4. The molecular weight of the compound, which determines mass per formula unit (correct answer)
  5. The temperature of the solution, which affects equilibrium position and kinetics
Explanation: When analyzing ionic compound solubility, you need to understand the thermodynamic and equilibrium factors that govern the dissolution process. Solubility depends on the balance between forces breaking apart the solid and forces stabilizing the dissolved ions. The molecular weight of the compound (D) has no direct impact on solubility. While molecular weight affects how much mass dissolves for a given molar concentration, it doesn't determine whether or how much of the compound will dissolve. For example, LiF (molecular weight 26 g/mol) is much less soluble than CsI (molecular weight 260 g/mol), despite being ten times lighter. Option A is incorrect because lattice energy absolutely affects solubility. Compounds with very high lattice energies (like MgO) require enormous energy input to break apart the ionic solid, making them poorly soluble. Lower lattice energy generally correlates with higher solubility. Option B is wrong because hydration energy is crucial for solubility. When ions dissolve, they become surrounded by water molecules, releasing hydration energy. Strong hydration (high hydration energy) helps compensate for the energy needed to break the lattice, promoting dissolution. Option C is incorrect due to the common ion effect. Adding ions already present in the dissolution equilibrium shifts the equilibrium toward the solid according to Le Châtelier's principle, decreasing solubility. This is why AgCl is less soluble in NaCl solution than in pure water. Remember: solubility depends on energetics (lattice vs. hydration energy) and equilibrium effects (common ions), not on the simple mass of the formula unit.

Question 13

A student observes that when solid NH4Cl\text{NH}_4\text{Cl} dissolves in water, the solution temperature decreases noticeably. What can be concluded about the dissolution process?

  1. The process is exothermic because heat is absorbed from the surroundings
  2. The process is endothermic because the system absorbs heat from the surroundings (correct answer)
  3. The lattice energy of NH₄Cl is less than the hydration energy of the ions
  4. The process involves no net energy change since the compound dissolves completely
  5. The temperature decrease is due to dilution rather than thermodynamic effects
Explanation: When you encounter a dissolution problem involving temperature changes, you're dealing with thermodynamics and the balance between energy required to break apart the solid and energy released when ions interact with water. The key observation here is that the solution temperature decreases when NH4Cl\text{NH}_4\text{Cl} dissolves. This tells you that heat is flowing from the surroundings (including the water) into the system to drive the dissolution process. When a process absorbs heat from its surroundings, it's defined as endothermic, and the surroundings get cooler as a result. Option B correctly identifies this as an endothermic process because the system absorbs heat from the surroundings, which is exactly what happens when the solution temperature drops. Option A contains a fundamental contradiction - it correctly notes that heat is absorbed from the surroundings but incorrectly labels this as exothermic. Exothermic processes release heat to the surroundings, making them feel warmer. Option C gets the energy relationship backwards. Since dissolution requires a net input of energy (endothermic), the lattice energy (energy needed to break apart the solid) must be greater than the hydration energy (energy released when ions interact with water). Option D is wrong because the temperature change directly indicates a net energy change is occurring. Complete dissolution doesn't mean the process is energetically neutral. Remember: temperature decrease during dissolution = endothermic process. The direction of heat flow (into or out of the system) determines whether a process is endothermic or exothermic, regardless of whether the substance actually dissolves.

Question 14

The solubility of BaF2\text{BaF}_2 decreases significantly when the solution temperature is lowered from 80°C to 20°C. What can be concluded about the dissolution process?

  1. The dissolution is exothermic because higher temperature decreases solubility
  2. The dissolution is endothermic because lower temperature decreases solubility (correct answer)
  3. The dissolution involves no heat change since BaF₂ is an ionic compound
  4. The dissolution is exothermic because ionic compounds always release heat when dissolving
  5. The temperature effect is due to changes in ionic strength rather than thermodynamics
Explanation: When you encounter questions about how temperature affects solubility, you're dealing with the thermodynamics of dissolution. The key insight is recognizing the relationship between temperature changes and the heat involved in the dissolving process. Since BaF2\text{BaF}_2 solubility decreases when temperature drops from 80°C to 20°C, this tells you that heat favors the dissolution process. When you remove heat (lower temperature), less solid dissolves, meaning the dissolution reaction must absorb heat from the surroundings to proceed. This makes it an endothermic process. Think of it this way: if a process needs heat to happen efficiently, removing heat will slow it down or reverse it. That's exactly what we see here—less BaF2\text{BaF}_2 dissolves at lower temperature because there's less thermal energy available to drive the endothermic dissolution. Choice A incorrectly states that higher temperature decreases solubility, which contradicts the given information. Choice C wrongly assumes ionic compounds can't involve heat changes during dissolution—they absolutely can, and most do. Choice D makes a false generalization that all ionic dissolutions are exothermic, which isn't true. The correct answer is B because lower temperature decreasing solubility indicates that heat is required for dissolution, making it endothermic. Study tip: Remember Le Châtelier's principle—if removing heat decreases solubility, then heat must be a "reactant" in the dissolution process, making it endothermic. Temperature-solubility relationships directly reveal whether dissolution absorbs or releases energy.

Question 15

A student prepares a saturated solution of AgCl\text{AgCl} in pure water at 25°C. If Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10} for AgCl\text{AgCl}, what is the molar solubility of AgCl\text{AgCl} in this solution?

  1. 1.3×10−51.3 \times 10^{-5} M (correct answer)
  2. 1.8×10−101.8 \times 10^{-10} M
  3. 3.6×10−103.6 \times 10^{-10} M
  4. 9.0×10−69.0 \times 10^{-6} M
  5. 3.2×10−203.2 \times 10^{-20} M
Explanation: When you encounter solubility equilibrium problems, you're dealing with the relationship between a compound's solubility product constant (KspK_{sp}) and its molar solubility in solution. The key is setting up the equilibrium expression correctly and recognizing the stoichiometric relationships. For AgCl\text{AgCl}, the dissolution equilibrium is: AgCl(s)⇌Ag(aq)++Cl(aq)−\text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} Let ss represent the molar solubility of AgCl\text{AgCl}. When ss moles of AgCl\text{AgCl} dissolve, they produce ss moles of Ag+\text{Ag}^+ and ss moles of Cl−\text{Cl}^-. The KspK_{sp} expression becomes: Ksp=[Ag+][Cl−]=(s)(s)=s2K_{sp} = [\text{Ag}^+][\text{Cl}^-] = (s)(s) = s^2 Solving for ss: s=Ksp=1.8×10−10=1.3×10−5s = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} M Answer A (1.3×10−51.3 \times 10^{-5} M) is correct—this is the calculated molar solubility. Answer B (1.8×10−101.8 \times 10^{-10} M) represents the KspK_{sp} value itself, not the solubility. This confuses the equilibrium constant with concentration. Answer C (3.6×10−103.6 \times 10^{-10} M) appears to double the KspK_{sp} value, possibly from incorrectly adding ion concentrations instead of multiplying them. Answer D (9.0×10−69.0 \times 10^{-6} M) might result from calculation errors or mishandling the square root. Remember: for compounds with 1:1 stoichiometry like AgCl\text{AgCl}, molar solubility equals Ksp\sqrt{K_{sp}}. Always write the balanced equation first to identify the correct stoichiometric relationships before setting up your KspK_{sp} expression.

Question 16

The solubility of CaF2\text{CaF}_2 in pure water is 2.1×10−42.1 \times 10^{-4} M at 25°C. What is the solubility of CaF2\text{CaF}_2 in a solution that already contains 0.10 M NaF\text{NaF}?

  1. 2.1×10−42.1 \times 10^{-4} M (unchanged due to different cation)
  2. 3.7×10−93.7 \times 10^{-9} M (significantly reduced by common ion effect) (correct answer)
  3. 4.2×10−44.2 \times 10^{-4} M (doubled due to increased ionic strength)
  4. 1.1×10−61.1 \times 10^{-6} M (moderately reduced by competing equilibrium)
  5. 8.4×10−58.4 \times 10^{-5} M (reduced by factor of 2.5 due to fluoride interference)
Explanation: When you encounter solubility problems involving a "common ion," you're dealing with the common ion effect—a key application of Le Châtelier's principle that significantly reduces solubility when a solution already contains one of the ions from the dissolving compound. First, let's find the KspK_{sp} using the pure water data. For CaF2⇌Ca2++2F−\text{CaF}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{F}^-, if solubility is 2.1×10−42.1 \times 10^{-4} M, then [Ca2+]=2.1×10−4[\text{Ca}^{2+}] = 2.1 \times 10^{-4} M and [F−]=2×2.1×10−4=4.2×10−4[\text{F}^-] = 2 \times 2.1 \times 10^{-4} = 4.2 \times 10^{-4} M. Therefore: Ksp=[Ca2+][F−]2=(2.1×10−4)(4.2×10−4)2=3.7×10−11K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = (2.1 \times 10^{-4})(4.2 \times 10^{-4})^2 = 3.7 \times 10^{-11}. Now in 0.10 M NaF solution, [F−]=0.10[\text{F}^-] = 0.10 M initially. Let ss = solubility of CaF2\text{CaF}_2. Then [Ca2+]=s[\text{Ca}^{2+}] = s and [F−]=0.10+2s≈0.10[\text{F}^-] = 0.10 + 2s \approx 0.10 (since ss will be very small). Using KspK_{sp}: 3.7×10−11=s(0.10)23.7 \times 10^{-11} = s(0.10)^2, so s=3.7×10−9s = 3.7 \times 10^{-9} M. Answer A incorrectly assumes different cations don't affect solubility—but the common fluoride ion is what matters. Answer C suggests increased ionic strength doubles solubility, which contradicts the common ion effect. Answer D shows some reduction but miscalculates the magnitude of the effect. Remember: when calculating common ion problems, always find KspK_{sp} from pure water data first, then apply it to the new conditions where the common ion concentration dominates the equilibrium.

Question 17

A chemist wants to separate AgCl\text{AgCl} from a mixture containing both AgCl\text{AgCl} and AgBr\text{AgBr} by selective precipitation. Which approach would be most effective?

  1. Add excess NaCl solution to precipitate AgCl preferentially due to common ion effect
  2. Add dilute ammonia solution to selectively dissolve AgCl through complex formation (correct answer)
  3. Heat the mixture to increase solubility differences between the two compounds
  4. Add excess AgNO₃ to precipitate both compounds completely for easier separation
  5. Use different pH conditions since AgCl and AgBr respond differently to acid
Explanation: When you encounter selective precipitation problems, you need to think about exploiting chemical differences between similar compounds—not just their precipitation behavior, but their ability to form complexes or undergo other reactions. The key insight here is that AgCl\text{AgCl} and AgBr\text{AgBr} have different abilities to dissolve in ammonia solution. AgCl\text{AgCl} readily forms a soluble complex with ammonia: AgCl+2NH3→[Ag(NH3)2]++Cl−\text{AgCl} + 2\text{NH}_3 \rightarrow [\text{Ag}(\text{NH}_3)_2]^+ + \text{Cl}^-. However, AgBr\text{AgBr} has a much lower tendency to form this complex due to its stronger ionic bonding, so it remains as a solid precipitate. This difference allows you to selectively dissolve the AgCl\text{AgCl} while leaving AgBr\text{AgBr} behind, making option B correct. Option A is flawed because both compounds contain Ag+\text{Ag}^+ ions, so the common ion effect would actually suppress dissolution of both compounds equally—it wouldn't create selectivity. Option C misunderstands the problem: while heating might change solubilities, both compounds would become more soluble together, making separation harder, not easier. Option D defeats the purpose entirely since adding AgNO3\text{AgNO}_3 would precipitate more of both chloride and bromide ions, making your mixture problem worse rather than achieving separation. Remember this pattern: when separating similar ionic compounds, look for differences in complex formation, acid-base behavior, or redox properties rather than just solubility differences. These secondary reactions often provide the selectivity that simple precipitation cannot.

Question 18

Based on the data shown in the table, which compound has the greatest molar solubility in water at 25°C?

  1. Compound A
  2. Compound B (correct answer)
  3. Compound C
  4. Compound D
  5. Compounds A and D have equal molar solubilities
Explanation: To find molar solubility, we calculate s from Ksp for each compound using their dissolution stoichiometry. Compound A (MX): s = √Ksp = √(4.0×10⁻⁹) = 6.3×10⁻⁵ M. Compound B (MX₂): Ksp = 4s³, so s = ∛(2.7×10⁻¹¹/4) = 3.0×10⁻⁴ M. Compound C (M₂X): Ksp = 4s³, so s = ∛(1.6×10⁻¹²/4) = 7.4×10⁻⁵ M. Compound D (MX): s = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. Compound B has the highest molar solubility despite not having the highest Ksp. Students often incorrectly compare Ksp values directly without considering stoichiometry.

Question 19

The data in the table shows the effect of different anions on the solubility of silver salts at 25°C. Which conclusion is best supported by this data?

  1. Solubility increases with increasing anion charge due to stronger electrostatic interactions
  2. Smaller anions form more soluble silver salts due to better lattice packing
  3. Solubility decreases as anion polarizability increases due to stronger covalent character (correct answer)
  4. The data shows no clear periodic trend in silver salt solubilities
  5. Solubility correlates with anion electronegativity through hydrogen bonding effects
Explanation: The data shows decreasing solubility in the order F⁻ > Cl⁻ > Br⁻ > I⁻, which correlates with increasing anion size and polarizability. Larger, more polarizable anions form bonds with Ag⁺ that have more covalent character, leading to lower solubility. Choice A is incorrect about charge effects. Choice B reverses the size-solubility relationship. Choice D denies the clear trend shown. Choice E incorrectly invokes hydrogen bonding with anions.