College Chemistry Quiz: Introduction To Solubility Equilibria
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Introduction To Solubility EquilibriaQuestion 1 of 20

The KspK_{sp} for calcium fluoride (CaF2CaF_2) is 3.9×10−113.9 \times 10^{-11} at 25°C. What is the molar solubility of CaF2CaF_2 in pure water at this temperature?

1.3×10−111.3 \times 10^{-11} M
2.1×10−42.1 \times 10^{-4} M
6.2×10−66.2 \times 10^{-6} M
1.2×10−51.2 \times 10^{-5} M
3.4×10−43.4 \times 10^{-4} M
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College Chemistry Quiz

College Chemistry Quiz: Introduction To Solubility Equilibria

Practice Introduction To Solubility Equilibria 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 Introduction To Solubility Equilibria, 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.

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Question 1

The KspK_{sp} for calcium fluoride (CaF2CaF_2) is 3.9×10−113.9 \times 10^{-11} at 25°C. What is the molar solubility of CaF2CaF_2 in pure water at this temperature?

  1. 1.3×10−111.3 \times 10^{-11} M
  2. 2.1×10−42.1 \times 10^{-4} M (correct answer)
  3. 6.2×10−66.2 \times 10^{-6} M
  4. 1.2×10−51.2 \times 10^{-5} M
  5. 3.4×10−43.4 \times 10^{-4} M
Explanation: When you encounter a solubility product (KspK_{sp}) problem, you're dealing with equilibrium expressions for slightly soluble ionic compounds. The key is setting up the dissolution equation and relating the concentrations of ions to the molar solubility. For calcium fluoride, the dissolution equilibrium is: CaF2(s)⇌Ca2+(aq)+2F−(aq)CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq) Let ss represent the molar solubility of CaF2CaF_2. When ss moles of CaF2CaF_2 dissolve per liter, you get ss moles of Ca2+Ca^{2+} ions and 2s2s moles of F−F^- ions (notice the 1:2 stoichiometry from the formula). The KspK_{sp} expression becomes: Ksp=[Ca2+][F−]2=(s)(2s)2=4s3K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3 Substituting the given KspK_{sp} value: 3.9×10−11=4s33.9 \times 10^{-11} = 4s^3 Solving for ss: s3=3.9×10−114=9.75×10−12s^3 = \frac{3.9 \times 10^{-11}}{4} = 9.75 \times 10^{-12} Therefore: s=9.75×10−123=2.1×10−4s = \sqrt[3]{9.75 \times 10^{-12}} = 2.1 \times 10^{-4} M, which is answer B. Answer A (1.3×10−111.3 \times 10^{-11} M) appears to confuse KspK_{sp} with solubility directly. Answer C (6.2×10−66.2 \times 10^{-6} M) likely results from incorrectly using Ksp=s2K_{sp} = s^2 instead of 4s34s^3. Answer D (1.2×10−51.2 \times 10^{-5} M) may come from calculation errors in the cube root step. Remember: always write the balanced dissolution equation first, then express each ion concentration in terms of molar solubility, accounting for stoichiometric coefficients.

Question 2

The KspK_{sp} for Mg(OH)2Mg(OH)_2 is 1.8×10−111.8 \times 10^{-11} at 25°C. What is the pH of a saturated solution of Mg(OH)2Mg(OH)_2 in pure water?

  1. 9.8
  2. 10.2
  3. 10.5 (correct answer)
  4. 11.1
  5. 12.4
Explanation: This question tests your ability to connect solubility equilibria with acid-base chemistry. When a sparingly soluble hydroxide dissolves, it releases OH⁻ ions that affect the solution's pH. Start by writing the dissolution equilibrium: Mg(OH)2(s)⇌Mg2+(aq)+2OH−(aq)Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq). The KspK_{sp} expression is Ksp=[Mg2+][OH−]2=1.8×10−11K_{sp} = [Mg^{2+}][OH^-]^2 = 1.8 \times 10^{-11}. Let the molar solubility of Mg(OH)2Mg(OH)_2 be ss. At equilibrium, [Mg2+]=s[Mg^{2+}] = s and [OH−]=2s[OH^-] = 2s (note the 2:1 stoichiometry). Substituting: Ksp=(s)(2s)2=4s3=1.8×10−11K_{sp} = (s)(2s)^2 = 4s^3 = 1.8 \times 10^{-11}. Solving for ss: s3=1.8×10−114=4.5×10−12s^3 = \frac{1.8 \times 10^{-11}}{4} = 4.5 \times 10^{-12}, so s=1.65×10−4s = 1.65 \times 10^{-4} M. Therefore, [OH−]=2s=3.3×10−4[OH^-] = 2s = 3.3 \times 10^{-4} M. Calculate pOH: pOH=−log⁡(3.3×10−4)=3.5pOH = -\log(3.3 \times 10^{-4}) = 3.5. Finally, pH=14−pOH=14−3.5=10.5pH = 14 - pOH = 14 - 3.5 = 10.5. Choice A (9.8) likely results from forgetting the factor of 2 in the hydroxide concentration. Choice B (10.2) might come from calculation errors in the cube root step. Choice D (11.1) could result from incorrectly assuming [OH−]=s[OH^-] = s instead of 2s2s, then making additional errors. Remember: always pay attention to stoichiometry when writing equilibrium expressions. The coefficient in the balanced equation becomes an exponent in KspK_{sp} and a multiplier for ion concentrations.

Question 3

Which of the following statements about solubility equilibria is correct?

  1. The KspK_{sp} expression includes the concentration of the solid compound
  2. A larger KspK_{sp} value always indicates a more soluble compound
  3. The KspK_{sp} expression includes only the concentrations of dissolved ions (correct answer)
  4. Temperature changes do not affect KspK_{sp} values significantly
  5. The KspK_{sp} can be calculated directly from molar mass differences
Explanation: Solubility equilibria questions test your understanding of how ionic compounds dissolve and reach equilibrium in solution. When a slightly soluble ionic compound dissolves, it establishes an equilibrium between the solid and its dissolved ions. The solubility product constant (KspK_{sp}) expression follows the same rules as other equilibrium expressions, but with one crucial difference: pure solids are never included in equilibrium expressions because their concentration remains constant. For example, when AgCl\text{AgCl} dissolves: AgCl(s)⇌Ag(aq)++Cl(aq)−\text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)}, the KspK_{sp} expression is Ksp=[Ag+][Cl−]K_{sp} = [\text{Ag}^+][\text{Cl}^-]. Only the concentrations of the dissolved ions appear in the expression, making option C correct. Option A is wrong because, as mentioned, pure solids never appear in equilibrium expressions since their "concentration" doesn't change during the reaction. Option B contains a common misconception: while larger KspK_{sp} values often indicate greater solubility, you can't directly compare KspK_{sp} values between compounds that produce different numbers of ions. For instance, Mg(OH)2\text{Mg(OH)}_2 produces three ions while AgCl\text{AgCl} produces two, so their KspK_{sp} values aren't directly comparable for solubility. Option D is incorrect because KspK_{sp} values are quite temperature-dependent, just like other equilibrium constants. Remember: KspK_{sp} expressions only include aqueous ions, never the solid reactant. When comparing solubilities, calculate actual molar solubilities rather than just comparing KspK_{sp} values.

Question 4

The molar solubility of Ag2CrO4Ag_2CrO_4 in pure water is 6.5×10−56.5 \times 10^{-5} M at 25°C. What is the KspK_{sp} for Ag2CrO4Ag_2CrO_4?

  1. 2.7×10−132.7 \times 10^{-13}
  2. 1.1×10−121.1 \times 10^{-12} (correct answer)
  3. 4.2×10−94.2 \times 10^{-9}
  4. 2.8×10−82.8 \times 10^{-8}
  5. 1.7×10−41.7 \times 10^{-4}
Explanation: When you encounter solubility equilibrium problems, you need to connect molar solubility (how much dissolves) to the solubility product constant (KspK_{sp}) through the dissolution equation and stoichiometry. For Ag2CrO4Ag_2CrO_4, the dissolution equilibrium is: Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq) The key insight is understanding the stoichiometric relationship. If the molar solubility is 6.5×10−56.5 \times 10^{-5} M, this means 6.5×10−56.5 \times 10^{-5} mol of Ag2CrO4Ag_2CrO_4 dissolves per liter. Since each formula unit produces 2 Ag+Ag^+ ions and 1 CrO42−CrO_4^{2-} ion:
  • [Ag+]=2×6.5×10−5=1.3×10−4[Ag^+] = 2 \times 6.5 \times 10^{-5} = 1.3 \times 10^{-4} M
  • [CrO42−]=6.5×10−5[CrO_4^{2-}] = 6.5 \times 10^{-5} M
The KspK_{sp} expression is: Ksp=[Ag+]2[CrO42−]K_{sp} = [Ag^+]^2[CrO_4^{2-}] Substituting: Ksp=(1.3×10−4)2×(6.5×10−5)=1.69×10−8×6.5×10−5=1.1×10−12K_{sp} = (1.3 \times 10^{-4})^2 \times (6.5 \times 10^{-5}) = 1.69 \times 10^{-8} \times 6.5 \times 10^{-5} = 1.1 \times 10^{-12} This matches answer B. Answer A (2.7×10−132.7 \times 10^{-13}) likely comes from incorrectly squaring the entire molar solubility. Answer C (4.2×10−94.2 \times 10^{-9}) suggests forgetting to account for the stoichiometric coefficient of 2 for Ag+Ag^+. Answer D (2.8×10−82.8 \times 10^{-8}) appears to involve calculation errors in the exponent arithmetic. Remember: always write the balanced dissolution equation first, then use stoichiometry to find individual ion concentrations before calculating KspK_{sp}.

Question 5

Which of the following ionic compounds would be expected to have the lowest molar solubility in pure water, based on the given KspK_{sp} values?

  1. AgBrAgBr (Ksp=5.4×10−13K_{sp} = 5.4 \times 10^{-13}) (correct answer)
  2. CaF2CaF_2 (Ksp=3.9×10−11K_{sp} = 3.9 \times 10^{-11})
  3. Ag2SO4Ag_2SO_4 (Ksp=1.2×10−5K_{sp} = 1.2 \times 10^{-5})
  4. PbI2PbI_2 (Ksp=7.1×10−9K_{sp} = 7.1 \times 10^{-9})
  5. Mg(OH)2Mg(OH)_2 (Ksp=1.8×10−11K_{sp} = 1.8 \times 10^{-11})
Explanation: When comparing ionic compounds' solubilities, you can't simply compare KspK_{sp} values directly—you must consider how each compound dissociates and calculate the actual molar solubility. For each compound, write the dissolution equation and express KspK_{sp} in terms of molar solubility (s): AgBr: AgBr⇌Ag++Br−AgBr \rightleftharpoons Ag^+ + Br^- Ksp=[Ag+][Br−]=s⋅s=s2K_{sp} = [Ag^+][Br^-] = s \cdot s = s^2 s=5.4×10−13=7.3×10−7 Ms = \sqrt{5.4 \times 10^{-13}} = 7.3 \times 10^{-7} \text{ M} CaF₂: CaF2⇌Ca2++2F−CaF_2 \rightleftharpoons Ca^{2+} + 2F^- Ksp=[Ca2+][F−]2=s⋅(2s)2=4s3K_{sp} = [Ca^{2+}][F^-]^2 = s \cdot (2s)^2 = 4s^3 s=3.9×10−1143=2.1×10−4 Ms = \sqrt[3]{\frac{3.9 \times 10^{-11}}{4}} = 2.1 \times 10^{-4} \text{ M} Ag₂SO₄: Ag2SO4⇌2Ag++SO42−Ag_2SO_4 \rightleftharpoons 2Ag^+ + SO_4^{2-} Ksp=[Ag+]2[SO42−]=(2s)2⋅s=4s3K_{sp} = [Ag^+]^2[SO_4^{2-}] = (2s)^2 \cdot s = 4s^3 s=1.2×10−543=1.4×10−2 Ms = \sqrt[3]{\frac{1.2 \times 10^{-5}}{4}} = 1.4 \times 10^{-2} \text{ M} PbI₂: PbI2⇌Pb2++2I−PbI_2 \rightleftharpoons Pb^{2+} + 2I^- Ksp=[Pb2+][I−]2=s⋅(2s)2=4s3K_{sp} = [Pb^{2+}][I^-]^2 = s \cdot (2s)^2 = 4s^3 s=7.1×10−943=1.2×10−3 Ms = \sqrt[3]{\frac{7.1 \times 10^{-9}}{4}} = 1.2 \times 10^{-3} \text{ M} Choice A (AgBr) has the lowest molar solubility at 7.3×10−77.3 \times 10^{-7} M. Choices B, C, and D all have significantly higher solubilities because their stoichiometry creates multiple ions, making them more soluble despite having larger KspK_{sp} values in some cases. Key strategy: Always convert KspK_{sp} to molar solubility when comparing compounds with different stoichiometries. A smaller KspK_{sp} doesn't automatically mean lower solubility if the dissolution produces different numbers of ions.

Question 6

A solution contains both 0.100.10 M Cl−Cl^- and 0.100.10 M Br−Br^-. If AgNO3AgNO_3 is slowly added, which silver halide will precipitate first? (KspK_{sp} for AgCl=1.8×10−10AgCl = 1.8 \times 10^{-10}; KspK_{sp} for AgBr=5.4×10−13AgBr = 5.4 \times 10^{-13})

  1. AgClAgCl because it has a larger KspK_{sp} value
  2. AgBrAgBr because it has a smaller KspK_{sp} value (correct answer)
  3. Both precipitate simultaneously because concentrations are equal
  4. AgClAgCl because chloride ion is smaller than bromide ion
  5. AgBrAgBr because bromide forms stronger ionic bonds with silver
Explanation: When you encounter a question about selective precipitation, you need to determine which compound will reach its solubility limit first as the precipitating agent is gradually added. To find which silver halide precipitates first, calculate the minimum Ag+Ag^+ concentration needed to start precipitation of each compound. Using the KspK_{sp} expression Ksp=[Ag+][X−]K_{sp} = [Ag^+][X^-], you can solve for [Ag+][Ag^+] when precipitation begins. For AgClAgCl: 1.8×10−10=[Ag+][0.10]1.8 \times 10^{-10} = [Ag^+][0.10], so [Ag+]=1.8×10−9[Ag^+] = 1.8 \times 10^{-9} M For AgBrAgBr: 5.4×10−13=[Ag+][0.10]5.4 \times 10^{-13} = [Ag^+][0.10], so [Ag+]=5.4×10−12[Ag^+] = 5.4 \times 10^{-12} M Since AgBrAgBr requires a much lower Ag+Ag^+ concentration to begin precipitating, it will form first when AgNO3AgNO_3 is slowly added. Choice A incorrectly assumes that a larger KspK_{sp} means precipitation occurs first, but actually the opposite is true—compounds with smaller KspK_{sp} values are less soluble and precipitate more readily. Choice C ignores the fundamental difference in solubilities despite equal concentrations. Choice D focuses on ion size, which doesn't directly determine precipitation order in this context. Study tip: Remember that in selective precipitation problems, the compound requiring the lowest concentration of the common ion (determined by KspK_{sp} calculations) will always precipitate first, regardless of the relative KspK_{sp} values themselves.

Question 7

For the dissolution reaction CaCO3(s)⇌Ca2+(aq)+CO32−(aq)CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq), which of the following correctly represents the relationship between the molar solubility (ss) and the equilibrium concentrations?

  1. [Ca2+]=s[Ca^{2+}] = s and [CO32−]=s[CO_3^{2-}] = s (correct answer)
  2. [Ca2+]=2s[Ca^{2+}] = 2s and [CO32−]=s[CO_3^{2-}] = s
  3. [Ca2+]=s[Ca^{2+}] = s and [CO32−]=2s[CO_3^{2-}] = 2s
  4. [Ca2+]=s2[Ca^{2+}] = s^2 and [CO32−]=s[CO_3^{2-}] = s
  5. [Ca2+]=[CO32−]=Ksp[Ca^{2+}] = [CO_3^{2-}] = K_{sp}
Explanation: When you encounter solubility equilibrium problems, the key is understanding that molar solubility represents the amount of solid compound that dissolves per liter of solution. You need to connect this dissolution amount to the concentrations of individual ions produced. For the dissolution CaCO3(s)⇌Ca2+(aq)+CO32−(aq)CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq), examine the stoichiometry carefully. When one mole of CaCO3CaCO_3 dissolves, it produces exactly one mole of Ca2+Ca^{2+} ions and one mole of CO32−CO_3^{2-} ions. If the molar solubility is ss (meaning ss moles of CaCO3CaCO_3 dissolve per liter), then ss moles of each ion are produced per liter. Therefore, [Ca2+]=s[Ca^{2+}] = s and [CO32−]=s[CO_3^{2-}] = s, making choice A correct. Choice B incorrectly suggests [Ca2+]=2s[Ca^{2+}] = 2s, which would only be true if each CaCO3CaCO_3 unit produced two calcium ions (like Ca2CO3Ca_2CO_3 would). Choice C makes the opposite error, suggesting [CO32−]=2s[CO_3^{2-}] = 2s, as if each dissolved unit produced two carbonate ions. Choice D shows [Ca2+]=s2[Ca^{2+}] = s^2, which confuses molar solubility with the solubility product constant (KspK_{sp}) calculation. Always trace the stoichiometry from the balanced equation: count how many of each ion comes from one formula unit of the dissolving compound. For 1:1 stoichiometry like this, ion concentrations equal molar solubility. This relationship changes only when compounds have different ion ratios, like Ca(OH)2Ca(OH)_2 producing two hydroxide ions per formula unit.

Question 8

A student measures the conductivity of a saturated AgClAgCl solution and finds it to be very low. Which of the following best explains this observation?

  1. AgClAgCl is a covalent compound that does not ionize in water
  2. The KspK_{sp} of AgClAgCl is very small, resulting in few ions in solution (correct answer)
  3. AgClAgCl forms ion pairs that do not conduct electricity effectively
  4. The solution is too concentrated, causing the conductivity to decrease
  5. AgClAgCl undergoes hydrolysis, removing ions from solution
Explanation: When you encounter questions about conductivity and ionic compounds, focus on the relationship between ion concentration and electrical conductivity. Conductivity depends on the number of ions present in solution - more ions mean better conductivity. Silver chloride (AgClAgCl) is an ionic compound that dissociates according to: AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq). However, AgClAgCl has an extremely small solubility product constant (Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10}), meaning very little dissolves. In a saturated solution, only tiny amounts of Ag+Ag^+ and Cl−Cl^- ions are present, resulting in very low conductivity. This explains why option B is correct. Option A is incorrect because AgClAgCl is definitely an ionic compound (formed between a metal and nonmetal with a large electronegativity difference), and it does ionize - just very sparingly. Option C misidentifies the issue; while ion pairing can occur in some solutions, the primary reason for low conductivity here is simply the extremely low concentration of ions due to poor solubility. Option D contradicts the scenario - you can't have a "too concentrated" saturated solution, and higher ion concentrations would actually increase conductivity, not decrease it. Remember this pattern: when you see low conductivity paired with an ionic compound, immediately think about solubility and KspK_{sp} values. Sparingly soluble salts like AgClAgCl, BaSO4BaSO_4, and PbI2PbI_2 will always show low conductivity because they produce few ions in solution.

Question 9

Which of the following experimental observations would provide the strongest evidence that a solution is saturated with respect to BaSO4BaSO_4?

  1. The solution has a very low conductivity compared to pure water
  2. Adding more solid BaSO4BaSO_4 does not increase the concentration of dissolved ions (correct answer)
  3. The solution appears cloudy due to suspended particles
  4. The pH of the solution is different from that of pure water
  5. The solution exhibits the Tyndall effect when a laser is shined through it
Explanation: When you encounter questions about solution saturation, focus on the defining characteristic: a saturated solution exists in dynamic equilibrium with its undissolved solute. At this point, the rates of dissolution and precipitation are equal, meaning no net change occurs in ion concentration. The strongest evidence for saturation is when adding more solid BaSO4BaSO_4 fails to increase the concentration of dissolved ions (B). This directly demonstrates that the solution has reached its maximum capacity to hold dissolved Ba2+Ba^{2+} and SO42−SO_4^{2-} ions. Any additional solid will simply remain undissolved, maintaining the equilibrium. Option A is misleading because BaSO4BaSO_4 has extremely low solubility regardless of saturation status - even an unsaturated solution would show minimal conductivity compared to pure water. Option C describes a supersaturated or unstable solution with excess solid, but cloudiness from suspended particles doesn't prove the dissolved portion is at saturation - you could have undissolved solid in an unsaturated solution due to slow dissolution kinetics. Option D is irrelevant since BaSO4BaSO_4 is a salt of a strong acid and strong base, so it won't significantly alter pH whether saturated or not. Remember this key principle: saturation is specifically about the dissolved species being at maximum concentration. Look for evidence that adding more solute fails to increase the amount that dissolves - this is the definitive test for a saturated solution and distinguishes it from simply having undissolved solid present.

Question 10

A chemistry student is investigating the effect of pH on the solubility of metal hydroxides. The student prepares several solutions with different pH values and attempts to dissolve various amounts of Mg(OH)₂ in each solution. The student observes that more Mg(OH)₂ dissolves in solutions with lower pH values.

Which of the following best explains the student's observation that Mg(OH)2Mg(OH)_2 is more soluble at lower pH?

  1. Lower pH increases the KspK_{sp} value for Mg(OH)2Mg(OH)_2
  2. H+H^+ ions react with OH−OH^- ions, reducing [OH−][OH^-] and allowing more Mg(OH)2Mg(OH)_2 to dissolve (correct answer)
  3. Lower pH provides more ions in solution, increasing the overall solubility
  4. Mg2+Mg^{2+} ions are more stable in acidic solutions than in basic solutions
  5. The formation of MgH2MgH_2 in acidic solution removes Mg2+Mg^{2+} from the equilibrium
Explanation: When you encounter questions about pH affecting the solubility of metal hydroxides, think about Le Châtelier's principle and how changing conditions shift equilibrium positions. The dissolution of Mg(OH)2Mg(OH)_2 establishes this equilibrium: Mg(OH)2(s)⇌Mg2+(aq)+2OH−(aq)Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq). When you lower the pH by adding acid, you're increasing the concentration of H+H^+ ions. These H+H^+ ions react with the OH−OH^- ions in solution: H++OH−→H2OH^+ + OH^- \rightarrow H_2O. This reaction removes OH−OH^- ions from the equilibrium, decreasing [OH−][OH^-]. According to Le Châtelier's principle, the system responds by shifting right to replace the consumed OH−OH^- ions, dissolving more Mg(OH)2Mg(OH)_2 in the process. Option A is incorrect because KspK_{sp} is a constant at a given temperature—it doesn't change with pH. The equilibrium expression Ksp=[Mg2+][OH−]2K_{sp} = [Mg^{2+}][OH^-]^2 remains the same, but the individual ion concentrations adjust. Option C misses the specific mechanism. While acidic solutions do contain more ions, this doesn't directly explain why Mg(OH)2Mg(OH)_2 becomes more soluble—it's specifically the consumption of OH−OH^- ions that matters. Option D suggests Mg2+Mg^{2+} stability changes with pH, but this isn't the driving force. The magnesium ion itself isn't significantly affected by pH changes in this context. Remember: When hydroxide salts encounter acidic conditions, look for the H++OH−→H2OH^+ + OH^- \rightarrow H_2O reaction. This "removes" one of the products from the dissolution equilibrium, driving more solid to dissolve.

Question 11

A solution is prepared by dissolving 0.0250.025 mol of KIKI in enough water to make 500500 mL of solution. If Pb(NO3)2Pb(NO_3)_2 is slowly added to this solution, at what [Pb2+][Pb^{2+}] will PbI2PbI_2 begin to precipitate? The KspK_{sp} for PbI2PbI_2 is 7.1×10−97.1 \times 10^{-9}.

  1. 7.1×10−67.1 \times 10^{-6} M
  2. 2.8×10−62.8 \times 10^{-6} M (correct answer)
  3. 1.4×10−71.4 \times 10^{-7} M
  4. 3.6×10−83.6 \times 10^{-8} M
  5. 1.8×10−71.8 \times 10^{-7} M
Explanation: When you encounter precipitation problems, you're working with solubility equilibria and the concept that precipitation begins when the ion product equals the solubility product constant (KspK_{sp}). First, find the iodide ion concentration. You have 0.025 mol of KI in 500 mL (0.500 L) of solution. Since KI dissociates completely: [I−]=0.025 mol0.500 L=0.050 M[I^-] = \frac{0.025 \text{ mol}}{0.500 \text{ L}} = 0.050 \text{ M} For PbI2PbI_2 precipitation, the equilibrium expression is: Ksp=[Pb2+][I−]2=7.1×10−9K_{sp} = [Pb^{2+}][I^-]^2 = 7.1 \times 10^{-9} Notice that the iodide concentration is squared because the formula shows two iodide ions per lead ion. Precipitation begins when this expression equals KspK_{sp}: [Pb2+]=Ksp[I−]2=7.1×10−9(0.050)2=7.1×10−92.5×10−3=2.8×10−6 M[Pb^{2+}] = \frac{K_{sp}}{[I^-]^2} = \frac{7.1 \times 10^{-9}}{(0.050)^2} = \frac{7.1 \times 10^{-9}}{2.5 \times 10^{-3}} = 2.8 \times 10^{-6} \text{ M} This confirms answer B. Looking at the distractors: A (7.1×10−67.1 \times 10^{-6} M) appears to result from not squaring the iodide concentration. C (1.4×10−71.4 \times 10^{-7} M) might come from incorrectly using the molarity as 0.1 M instead of 0.05 M. D (3.6×10−83.6 \times 10^{-8} M) could result from multiple calculation errors or using an incorrect volume conversion. Remember: always write the correct equilibrium expression with proper stoichiometry, and precipitation begins exactly when Q=KspQ = K_{sp}, not when it exceeds it.

Question 12

The KspK_{sp} values for several lead compounds are given below. Which compound has the highest molar solubility in pure water?

  1. PbCl2PbCl_2 (Ksp=1.6×10−5K_{sp} = 1.6 \times 10^{-5}) (correct answer)
  2. PbBr2PbBr_2 (Ksp=4.0×10−6K_{sp} = 4.0 \times 10^{-6})
  3. PbI2PbI_2 (Ksp=7.1×10−9K_{sp} = 7.1 \times 10^{-9})
  4. PbSPbS (Ksp=3.0×10−28K_{sp} = 3.0 \times 10^{-28})
  5. PbSO4PbSO_4 (Ksp=1.6×10−8K_{sp} = 1.6 \times 10^{-8})
Explanation: When you encounter questions comparing molar solubilities using KspK_{sp} values, you can't simply compare the KspK_{sp} numbers directly. The key is understanding that compounds with different stoichiometries require different mathematical relationships between KspK_{sp} and molar solubility. For compounds with the formula MX2MX_2 (like PbCl2PbCl_2, PbBr2PbBr_2, and PbI2PbI_2), the dissolution equation is MX2⇌M2++2X−MX_2 \rightleftharpoons M^{2+} + 2X^-. If the molar solubility is ss, then [M2+]=s[M^{2+}] = s and [X−]=2s[X^-] = 2s, giving Ksp=s×(2s)2=4s3K_{sp} = s \times (2s)^2 = 4s^3. Therefore, s=Ksp43s = \sqrt[3]{\frac{K_{sp}}{4}}. For PbSPbS with formula MXMX, the equation is MX⇌M2++X2−MX \rightleftharpoons M^{2+} + X^{2-}, so Ksp=s2K_{sp} = s^2 and s=Ksps = \sqrt{K_{sp}}. Calculating the molar solubilities:
  • A) PbCl2PbCl_2: s=1.6×10−543=1.6×10−2s = \sqrt[3]{\frac{1.6 \times 10^{-5}}{4}} = 1.6 \times 10^{-2} M
  • B) PbBr2PbBr_2: s=4.0×10−643=1.0×10−2s = \sqrt[3]{\frac{4.0 \times 10^{-6}}{4}} = 1.0 \times 10^{-2} M
  • C) PbI2PbI_2: s=7.1×10−943=1.2×10−3s = \sqrt[3]{\frac{7.1 \times 10^{-9}}{4}} = 1.2 \times 10^{-3} M
  • D) PbSPbS: s=3.0×10−28=5.5×10−15s = \sqrt{3.0 \times 10^{-28}} = 5.5 \times 10^{-15} M
PbCl2PbCl_2 has the highest molar solubility, making A correct. Study tip: Always write the dissolution equation first to determine the relationship between KspK_{sp} and molar solubility. Different stoichiometries require different formulas, so comparing KspK_{sp} values alone can be misleading.

Question 13

A solution contains 0.0100.010 M Ca2+Ca^{2+} and 0.0200.020 M SO42−SO_4^{2-}. The KspK_{sp} for CaSO4CaSO_4 is 2.4×10−52.4 \times 10^{-5}. What will happen when these solutions are mixed?

  1. No precipitation occurs because Q<KspQ < K_{sp}
  2. Precipitation occurs because Q=KspQ = K_{sp}
  3. Precipitation occurs because Q>KspQ > K_{sp} (correct answer)
  4. The solution becomes exactly saturated with no excess solid
  5. More information about temperature is needed to predict the outcome
Explanation: When you encounter a precipitation problem, you need to compare the reaction quotient (Q) to the solubility product constant (KspK_{sp}) to predict whether a precipitate will form. First, calculate the reaction quotient using the given concentrations. For the equilibrium CaSO4(s)⇌Ca2+(aq)+SO42−(aq)CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq), the expression is Q=[Ca2+][SO42−]Q = [Ca^{2+}][SO_4^{2-}]. Substituting the values: Q=(0.010)(0.020)=2.0×10−4Q = (0.010)(0.020) = 2.0 \times 10^{-4} Now compare Q to K_{sp}: Q=2.0×10−4Q = 2.0 \times 10^{-4} versus Ksp=2.4×10−5K_{sp} = 2.4 \times 10^{-5} Since 2.0×10−4>2.4×10−52.0 \times 10^{-4} > 2.4 \times 10^{-5}, we have Q>KspQ > K_{sp}, which means the solution is supersaturated and precipitation will occur. Looking at the wrong answers: Choice A incorrectly states that Q<KspQ < K_{sp}, which would mean no precipitation occurs, but our calculation shows Q is actually larger than K_{sp}. Choice B suggests Q=KspQ = K_{sp}, which would indicate equilibrium with no net precipitation, but Q and K_{sp} are clearly different values. Choice D implies the solution reaches exact saturation, which only occurs when Q=KspQ = K_{sp}, not when Q exceeds K_{sp}. Study tip: Always calculate Q first, then compare it to K_{sp}. Remember the rule: Q > K_{sp} means precipitation occurs, Q < K_{sp} means no precipitation, and Q = K_{sp} means the solution is at equilibrium (saturated).

Question 14

A solution contains 0.0150.015 M Pb2+Pb^{2+} ions. What minimum concentration of Cl−Cl^- ions is required to initiate precipitation of PbCl2PbCl_2? The KspK_{sp} for PbCl2PbCl_2 is 1.6×10−51.6 \times 10^{-5}.

  1. 1.1×10−61.1 \times 10^{-6} M
  2. 3.3×10−23.3 \times 10^{-2} M (correct answer)
  3. 1.1×10−31.1 \times 10^{-3} M
  4. 4.0×10−44.0 \times 10^{-4} M
  5. 6.5×10−36.5 \times 10^{-3} M
Explanation: When you encounter a precipitation problem, you're dealing with solubility equilibrium and the solubility product constant (KspK_{sp}). Precipitation begins when the ion product equals KspK_{sp}, meaning the solution becomes saturated. For PbCl2PbCl_2, the dissolution equilibrium is: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) The KspK_{sp} expression is: Ksp=[Pb2+][Cl−]2=1.6×10−5K_{sp} = [Pb^{2+}][Cl^-]^2 = 1.6 \times 10^{-5} Notice that the chloride concentration is squared because there are two Cl−Cl^- ions per formula unit of PbCl2PbCl_2. To find the minimum [Cl−][Cl^-] needed for precipitation, substitute the given [Pb2+]=0.015[Pb^{2+}] = 0.015 M into the KspK_{sp} expression: 1.6×10−5=(0.015)[Cl−]21.6 \times 10^{-5} = (0.015)[Cl^-]^2 Solving for [Cl−][Cl^-]: [Cl−]2=1.6×10−50.015=1.07×10−3[Cl^-]^2 = \frac{1.6 \times 10^{-5}}{0.015} = 1.07 \times 10^{-3} [Cl−]=1.07×10−3=3.3×10−2[Cl^-] = \sqrt{1.07 \times 10^{-3}} = 3.3 \times 10^{-2} M This matches answer choice B. Answer A (1.1×10−61.1 \times 10^{-6} M) likely comes from incorrectly dividing KspK_{sp} by [Pb2+][Pb^{2+}] without taking the square root. Answer C (1.1×10−31.1 \times 10^{-3} M) represents the intermediate step before taking the square root. Answer D (4.0×10−44.0 \times 10^{-4} M) might result from calculation errors or forgetting to account for the stoichiometry. Remember: always write the correct KspK_{sp} expression first, paying careful attention to stoichiometric coefficients that become exponents.

Question 15

The KspK_{sp} for BaSO4BaSO_4 is 1.1×10−101.1 \times 10^{-10} at 25°C. What is the solubility of BaSO4BaSO_4 in a 0.0500.050 M Na2SO4Na_2SO_4 solution?

  1. 2.2×10−92.2 \times 10^{-9} M (correct answer)
  2. 1.0×10−51.0 \times 10^{-5} M
  3. 5.0×10−65.0 \times 10^{-6} M
  4. 1.1×10−81.1 \times 10^{-8} M
  5. 2.2×10−62.2 \times 10^{-6} M
Explanation: This question tests the common ion effect, where the presence of a shared ion reduces the solubility of a sparingly soluble salt. When you see a solubility problem with an added salt that shares an ion, expect significantly reduced solubility compared to pure water. For BaSO4BaSO_4, the equilibrium is: BaSO4(s)⇌Ba2+(aq)+SO42−(aq)BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) with Ksp=[Ba2+][SO42−]=1.1×10−10K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10}. The Na2SO4Na_2SO_4 completely dissociates, providing 0.050 M SO42−SO_4^{2-} ions. When BaSO4BaSO_4 dissolves, it adds xx M of both Ba2+Ba^{2+} and SO42−SO_4^{2-}. Since xx will be very small compared to 0.050 M, the total sulfate concentration is approximately 0.050 M. Setting up the KspK_{sp} expression: 1.1×10−10=(x)(0.050)1.1 \times 10^{-10} = (x)(0.050) Solving for xx: x=1.1×10−100.050=2.2×10−9x = \frac{1.1 \times 10^{-10}}{0.050} = 2.2 \times 10^{-9} M This confirms answer A is correct. Answer B (1.0×10−51.0 \times 10^{-5} M) might result from incorrectly using the square root of KspK_{sp}, forgetting about the common ion effect. Answer C (5.0×10−65.0 \times 10^{-6} M) could come from calculation errors or wrong assumptions about ion concentrations. Answer D (1.1×10−81.1 \times 10^{-8} M) might arise from incorrectly manipulating the KspK_{sp} expression. Remember: when a common ion is present, always account for its initial concentration before adding the dissolving salt's contribution. The common ion effect dramatically reduces solubility.

Question 16

The KspK_{sp} for Cu(OH)2Cu(OH)_2 is 2.2×10−202.2 \times 10^{-20} at 25°C. In a solution where [OH−]=1.0×10−5[OH^-] = 1.0 \times 10^{-5} M, what is the maximum concentration of Cu2+Cu^{2+} that can exist without precipitation?

  1. 2.2×10−102.2 \times 10^{-10} M (correct answer)
  2. 2.2×10−152.2 \times 10^{-15} M
  3. 4.7×10−84.7 \times 10^{-8} M
  4. 1.5×10−71.5 \times 10^{-7} M
  5. 2.2×10−202.2 \times 10^{-20} M
Explanation: This question tests your understanding of solubility equilibrium and how to use KspK_{sp} expressions to find maximum ion concentrations before precipitation occurs. When Cu(OH)2Cu(OH)_2 dissolves, it establishes the equilibrium: Cu(OH)2(s)⇌Cu2+(aq)+2OH−(aq)Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^-(aq). The KspK_{sp} expression is: Ksp=[Cu2+][OH−]2K_{sp} = [Cu^{2+}][OH^-]^2. Notice that [OH−][OH^-] is squared because there are two hydroxide ions per formula unit. To find the maximum [Cu2+][Cu^{2+}] before precipitation, substitute the known values into the KspK_{sp} expression: 2.2×10−20=[Cu2+](1.0×10−5)22.2 \times 10^{-20} = [Cu^{2+}](1.0 \times 10^{-5})^2. Solving for [Cu2+][Cu^{2+}]: [Cu2+]=2.2×10−20(1.0×10−5)2=2.2×10−201.0×10−10=2.2×10−10[Cu^{2+}] = \frac{2.2 \times 10^{-20}}{(1.0 \times 10^{-5})^2} = \frac{2.2 \times 10^{-20}}{1.0 \times 10^{-10}} = 2.2 \times 10^{-10} M. Answer A (2.2×10−102.2 \times 10^{-10} M) is correct. Answer B (2.2×10−152.2 \times 10^{-15} M) results from incorrectly multiplying instead of dividing by [OH−]2[OH^-]^2. Answer C (4.7×10−84.7 \times 10^{-8} M) comes from forgetting to square the [OH−][OH^-] term, using (1.0×10−5)1(1.0 \times 10^{-5})^1 instead of (1.0×10−5)2(1.0 \times 10^{-5})^2. Answer D (1.5×10−71.5 \times 10^{-7} M) appears to involve calculation errors or incorrect substitution. Remember: always check the stoichiometric coefficients in the balanced equation to write the correct KspK_{sp} expression. The exponents in KspK_{sp} expressions must match the coefficients from the dissolution equation.

Question 17

A saturated solution of CaCO3CaCO_3 is in equilibrium with solid CaCO3CaCO_3. If HClHCl is added to this solution, which of the following will occur?

  1. More CaCO3CaCO_3 will precipitate due to the common ion effect
  2. The KspK_{sp} value will decrease due to the added acid
  3. More CaCO3CaCO_3 will dissolve due to consumption of CO32−CO_3^{2-} ions (correct answer)
  4. The solution will become supersaturated with CaCO3CaCO_3
  5. No change will occur since the system is already at equilibrium
Explanation: This question tests your understanding of how acid-base reactions affect solubility equilibria. When you encounter a problem involving a saturated solution and added acid, think about how the acid will react with the ions in solution and shift the equilibrium. The saturated CaCO3CaCO_3 solution exists in equilibrium: CaCO3(s)⇌Ca2+(aq)+CO32−(aq)CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq). When HClHCl is added, the H+H^+ ions react with carbonate ions to form bicarbonate and eventually carbonic acid: CO32−+H+→HCO3−+H+→H2CO3CO_3^{2-} + H^+ \rightarrow HCO_3^- + H^+ \rightarrow H_2CO_3. This removes CO32−CO_3^{2-} ions from solution, which shifts the equilibrium to the right according to Le Châtelier's principle, causing more solid CaCO3CaCO_3 to dissolve. This confirms answer C. Answer A incorrectly suggests precipitation due to a common ion effect, but HClHCl doesn't provide any ions common to the CaCO3CaCO_3 equilibrium. Answer B is wrong because KspK_{sp} is a constant that depends only on temperature, not on the presence of other substances. The acid doesn't change the fundamental solubility constant. Answer D suggests supersaturation, but removing carbonate ions actually decreases the ion product below KspK_{sp}, making the solution undersaturated. Remember: when acids are added to solutions containing basic anions like CO32−CO_3^{2-}, OH−OH^-, or S2−S^{2-}, they consume these anions through acid-base reactions, which increases the solubility of the corresponding salts by shifting equilibrium toward dissolution.

Question 18

A saturated solution of silver chloride (AgClAgCl) at 25°C has a concentration of Ag+Ag^+ ions equal to 1.3×10−51.3 \times 10^{-5} M. What is the solubility product constant (KspK_{sp}) for AgClAgCl at this temperature?

  1. 1.3×10−51.3 \times 10^{-5}
  2. 1.7×10−101.7 \times 10^{-10} (correct answer)
  3. 2.6×10−52.6 \times 10^{-5}
  4. 6.5×10−66.5 \times 10^{-6}
  5. 1.69×10−101.69 \times 10^{-10}
Explanation: When you encounter solubility product problems, you're dealing with equilibrium expressions for slightly soluble ionic compounds. The key is understanding how the dissolution equation relates to the KspK_{sp} expression. Silver chloride dissolves according to: AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) The solubility product expression is: Ksp=[Ag+][Cl−]K_{sp} = [Ag^+][Cl^-] Since the stoichiometry is 1:1, each mole of AgClAgCl that dissolves produces one mole of Ag+Ag^+ and one mole of Cl−Cl^-. Therefore, if [Ag+]=1.3×10−5[Ag^+] = 1.3 \times 10^{-5} M, then [Cl−]=1.3×10−5[Cl^-] = 1.3 \times 10^{-5} M as well. Substituting into the KspK_{sp} expression: Ksp=(1.3×10−5)(1.3×10−5)=1.69×10−10K_{sp} = (1.3 \times 10^{-5})(1.3 \times 10^{-5}) = 1.69 \times 10^{-10} This rounds to 1.7×10−101.7 \times 10^{-10}, which is answer B. Answer A (1.3×10−51.3 \times 10^{-5}) represents just the concentration of Ag+Ag^+, not the product. Answer C (2.6×10−52.6 \times 10^{-5}) is the sum of the ion concentrations, which isn't relevant to KspK_{sp}. Answer D (6.5×10−66.5 \times 10^{-6}) might result from incorrectly dividing instead of multiplying the concentrations. Remember: For KspK_{sp} calculations, always write the balanced dissolution equation first, then the equilibrium expression. Use stoichiometry to find all ion concentrations, then multiply them together according to the KspK_{sp} expression.

Question 19

The KspK_{sp} for PbCl2PbCl_2 is 1.6×10−51.6 \times 10^{-5} at 25°C. What is the molar solubility of PbCl2PbCl_2 in a 0.100.10 M NaClNaCl solution?

  1. 1.6×10−31.6 \times 10^{-3} M (correct answer)
  2. 4.0×10−44.0 \times 10^{-4} M
  3. 1.6×10−41.6 \times 10^{-4} M
  4. 1.6×10−51.6 \times 10^{-5} M
  5. 8.0×10−58.0 \times 10^{-5} M
Explanation: This question tests your understanding of the common ion effect in solubility equilibria. When a sparingly soluble salt dissolves in a solution containing one of its ions, the solubility decreases due to Le Châtelier's principle. For PbCl2PbCl_2, the equilibrium is: PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) The KspK_{sp} expression is: Ksp=[Pb2+][Cl−]2=1.6×10−5K_{sp} = [Pb^{2+}][Cl^-]^2 = 1.6 \times 10^{-5} In the 0.100.10 M NaClNaCl solution, the initial [Cl−]=0.10[Cl^-] = 0.10 M. Let ss be the molar solubility of PbCl2PbCl_2. At equilibrium:
  • [Pb2+]=s[Pb^{2+}] = s
  • [Cl−]=0.10+2s[Cl^-] = 0.10 + 2s
Since ss will be small compared to 0.100.10 M, we can approximate: [Cl−]≈0.10[Cl^-] \approx 0.10 M Substituting into the KspK_{sp} expression: 1.6×10−5=s×(0.10)21.6 \times 10^{-5} = s \times (0.10)^2 s=1.6×10−50.01=1.6×10−3s = \frac{1.6 \times 10^{-5}}{0.01} = 1.6 \times 10^{-3} M Answer A (1.6×10−31.6 \times 10^{-3} M) is correct. Answer B (4.0×10−44.0 \times 10^{-4} M) might result from calculation errors or incorrect stoichiometry. Answer C (1.6×10−41.6 \times 10^{-4} M) could come from using 0.100.10 instead of (0.10)2(0.10)^2 in the denominator. Answer D (1.6×10−51.6 \times 10^{-5} M) incorrectly assumes the molar solubility equals the KspK_{sp} value. Remember: when dealing with common ion effects, identify which ion is common, set up your ICE table carefully, and don't forget to square the chloride concentration in the KspK_{sp} expression.

Question 20

Consider the equilibrium: CaF2(s)⇌Ca2+(aq)+2F−(aq)CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq). Which of the following changes would increase the solubility of CaF2CaF_2?

  1. Adding solid NaFNaF to the solution
  2. Adding solid CaCl2CaCl_2 to the solution
  3. Adding HClHCl to the solution (correct answer)
  4. Increasing the temperature (assuming dissolution is exothermic)
  5. Adding more solid CaF2CaF_2 to the solution
Explanation: When you encounter equilibrium problems involving ionic solids, think about Le Châtelier's principle and how changes affect the position of equilibrium. The key is identifying which changes will shift the equilibrium toward more dissolution (right) versus precipitation (left). Adding HClHCl increases CaF2CaF_2 solubility because the added H+H^+ ions react with F−F^- ions to form HFHF, effectively removing F−F^- from solution. According to Le Châtelier's principle, when you remove a product (F−F^-), the equilibrium shifts right to replace it, dissolving more CaF2CaF_2. This is called the common ion effect in reverse—removing an ion rather than adding one. Choice A is wrong because adding NaFNaF increases the F−F^- concentration, creating a common ion effect that shifts equilibrium left, decreasing solubility. Choice B is incorrect because adding CaCl2CaCl_2 increases Ca2+Ca^{2+} concentration, another common ion effect that also shifts equilibrium left and decreases solubility. Choice D is wrong because if dissolution is exothermic (releases heat), increasing temperature shifts equilibrium left toward the solid, decreasing solubility. Remember this pattern: to increase solubility of an ionic compound, you need to remove one of its ions from solution (through chemical reaction) or add something that doesn't share common ions. Common ion effects always decrease solubility, while reactions that consume the dissolved ions increase it.