College Chemistry Quiz: Separations Of Solutions And Mixtures
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Separations Of Solutions And MixturesQuestion 1 of 15

A mixture contains 25.0 g of sodium chloride, 15.0 g of sand (silicon dioxide), and 10.0 g of iron filings. Which sequence of separation techniques would most efficiently isolate each component in pure form?

Dissolution in water, filtration, evaporation, magnetic separation
Magnetic separation, dissolution in water, filtration, evaporation
Filtration, magnetic separation, dissolution in water, evaporation
Evaporation, magnetic separation, dissolution in water, filtration
Dissolution in water, magnetic separation, filtration, evaporation
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College Chemistry Quiz

College Chemistry Quiz: Separations Of Solutions And Mixtures

Practice Separations Of Solutions And Mixtures 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 Separations Of Solutions And Mixtures, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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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

A mixture contains 25.0 g of sodium chloride, 15.0 g of sand (silicon dioxide), and 10.0 g of iron filings. Which sequence of separation techniques would most efficiently isolate each component in pure form?

  1. Dissolution in water, filtration, evaporation, magnetic separation
  2. Magnetic separation, dissolution in water, filtration, evaporation (correct answer)
  3. Filtration, magnetic separation, dissolution in water, evaporation
  4. Evaporation, magnetic separation, dissolution in water, filtration
  5. Dissolution in water, magnetic separation, filtration, evaporation
Explanation: When approaching mixture separation problems, you need to consider the unique physical properties of each component and design a sequence that efficiently exploits these differences without interfering with subsequent steps. The correct approach (B) starts with magnetic separation to remove the iron filings, since iron is ferromagnetic while sodium chloride and sand are not. This immediately isolates one pure component. Next, you dissolve the remaining mixture in water - sodium chloride is highly soluble while sand is insoluble. Filtration then separates the dissolved salt solution from the insoluble sand, giving you pure sand. Finally, evaporation removes the water from the filtrate, yielding pure sodium chloride crystals. Option A fails because it attempts magnetic separation after the other steps, but by then the iron filings would already be mixed with wet sand after filtration, making magnetic separation much less effective. Option C suggests starting with filtration, but there's nothing to filter initially since all components are dry solids. Option D begins with evaporation, which makes no sense since there's no liquid present in the original mixture to evaporate. The key insight is that magnetic separation works best on dry materials and should be done first when possible. Once you add water to dissolve the salt, subsequent magnetic separation becomes messy and inefficient. Study tip: Always start separation sequences by removing components with the most distinctive properties (like magnetism) first, then move to solubility-based separations. This prevents complications from wet materials interfering with physical separation methods.

Question 2

A student attempts to separate a mixture of acetone (boiling point 56°C) and water (boiling point 100°C) by simple distillation. The distillation is performed at standard atmospheric pressure. What is the most likely outcome of this separation attempt?

  1. Complete separation with pure acetone collected first, then pure water
  2. Partial separation with acetone-rich distillate followed by water-rich residue (correct answer)
  3. No separation because the compounds form an azeotropic mixture
  4. Decomposition of both compounds due to excessive heating
  5. Formation of a new compound from the reaction of acetone and water
Explanation: When you encounter distillation problems, focus on the relationship between boiling points and the nature of the liquid mixture. Simple distillation works best when components have significantly different boiling points (typically >25°C difference) and don't interact strongly with each other. Acetone and water have boiling points differing by 44°C, which seems favorable for separation. However, these compounds can form hydrogen bonds with each other, creating a non-ideal mixture that doesn't behave like two completely independent liquids. During distillation, the vapor will be enriched in the more volatile component (acetone) but won't be pure acetone. As distillation proceeds, you'll collect an acetone-rich distillate initially, then progressively more water-rich fractions, leaving a water-rich residue. This makes answer B correct. Answer A is wrong because the hydrogen bonding between acetone and water prevents complete separation by simple distillation - you won't get pure components. Answer C incorrectly suggests azeotrope formation. While acetone and water do interact, they don't form a true azeotropic mixture that would prevent any separation at standard conditions. Answer D is incorrect because neither compound decomposes at their respective boiling points under normal atmospheric pressure - both are thermally stable at these temperatures. Remember that "partial separation" is often the realistic outcome in distillation problems involving polar compounds that can hydrogen bond. Complete separation typically requires fractional distillation or other advanced techniques when dealing with interacting molecules.

Question 3

A laboratory mixture contains benzoic acid (C6H5COOHC_6H_5COOH, Ka=6.3×10−5K_a = 6.3 \times 10^{-5}), sodium chloride, and sand. The mixture is treated with dilute aqueous sodium hydroxide solution. Which statement best describes the result of this treatment?

  1. All three components dissolve completely in the basic solution
  2. Only benzoic acid dissolves, forming sodium benzoate in solution
  3. Only sodium chloride dissolves, leaving benzoic acid and sand as solids
  4. Both benzoic acid and sodium chloride dissolve, leaving only sand (correct answer)
  5. No components dissolve because sodium hydroxide is incompatible with organic acids
Explanation: When you encounter acid-base chemistry problems involving mixtures, think about how each component will behave individually in the given conditions. Here, you're adding a basic solution (NaOH) to a mixture, so consider what happens to each substance. Benzoic acid (C6H5COOHC_6H_5COOH) is a weak acid that will react with the strong base NaOH in an acid-base neutralization: C6H5COOH+NaOH→C6H5COONa+H2OC_6H_5COOH + NaOH \rightarrow C_6H_5COONa + H_2O. This forms sodium benzoate, which is highly soluble in water. Sodium chloride is an ionic compound that readily dissolves in water regardless of pH - the presence of NaOH doesn't prevent this. Sand (silicon dioxide) is insoluble in water and won't dissolve in basic conditions. Therefore, both benzoic acid (as sodium benzoate) and sodium chloride will be in solution, while sand remains undissolved. Looking at the wrong answers: Choice A incorrectly suggests sand dissolves - sand is essentially insoluble in aqueous solutions. Choice B misses that sodium chloride will also dissolve; this answer focuses only on the acid-base reaction while ignoring the other soluble component. Choice C incorrectly assumes benzoic acid won't dissolve and that only the salt dissolves, overlooking the acid-base neutralization that makes benzoic acid soluble as its sodium salt. Study tip: In separation problems, analyze each component's behavior independently. Remember that weak acids become soluble salts when neutralized by strong bases, and consider the inherent solubility of each substance in the given solvent conditions.

Question 4

A chromatography experiment is performed using silica gel as the stationary phase and a mixture of hexane and ethyl acetate (3:1 v/v) as the mobile phase. Three compounds are separated: compound A (Rf=0.85R_f = 0.85), compound B (Rf=0.45R_f = 0.45), and compound C (Rf=0.15R_f = 0.15). Which statement about the relative polarities of these compounds is correct?

  1. Compound A is most polar, compound C is least polar
  2. Compound C is most polar, compound A is least polar (correct answer)
  3. Compound B is most polar, compound A is least polar
  4. All compounds have similar polarity since they separated on the same system
  5. Polarity cannot be determined from RfR_f values alone
Explanation: When you encounter thin-layer chromatography (TLC) problems, the key relationship to remember is that RfR_f values reveal how compounds interact with the stationary phase. In normal-phase chromatography using silica gel (a polar stationary phase), polar compounds stick more strongly to the silica and travel shorter distances, resulting in lower RfR_f values. Looking at the RfR_f values: compound C (Rf=0.15R_f = 0.15) barely moved up the plate, indicating strong attraction to the polar silica gel, making it the most polar compound. Compound A (Rf=0.85R_f = 0.85) traveled nearly to the top, showing weak interaction with silica and strong affinity for the relatively nonpolar mobile phase, making it the least polar. Compound B (Rf=0.45R_f = 0.45) falls in between with intermediate polarity. Answer A incorrectly reverses the polarity relationship—it suggests the compound that traveled farthest (A) is most polar, when the opposite is true. Answer C correctly identifies A as least polar but wrongly places B as most polar, ignoring that C has the lowest RfR_f value. Answer D misses the fundamental point that different RfR_f values directly indicate different polarities—compounds separate precisely because they have different polarities. Answer B correctly states that compound C is most polar and compound A is least polar. Study tip: Remember the inverse relationship in normal-phase TLC: lower RfR_f = higher polarity. Polar compounds "stick" to polar stationary phases and don't travel far.

Question 5

An aqueous solution contains Ag+Ag^+, Cu2+Cu^{2+}, and Na+Na^+ ions, each at 0.10 M concentration. Concentrated hydrochloric acid is slowly added to this solution. Based on the solubility products (Ksp(AgCl)=1.8×10−10K_{sp}(AgCl) = 1.8 \times 10^{-10}, Ksp(CuCl2)=1.0×10−6K_{sp}(CuCl_2) = 1.0 \times 10^{-6}), what is the order of precipitation as [Cl−][Cl^-] increases?

  1. AgClAgCl precipitates first, followed by CuCl2CuCl_2, then NaClNaCl
  2. CuCl2CuCl_2 precipitates first, followed by AgClAgCl, then NaClNaCl
  3. AgClAgCl precipitates first, followed by CuCl2CuCl_2; NaClNaCl remains soluble (correct answer)
  4. All three compounds precipitate simultaneously when [Cl−]=0.10[Cl^-] = 0.10 M
  5. Only AgClAgCl precipitates; CuCl2CuCl_2 and NaClNaCl remain completely soluble
Explanation: When you encounter precipitation problems with multiple ions, you need to determine which compound will precipitate first by calculating the minimum chloride concentration required for each precipitation to begin. For precipitation to occur, the ion product must equal the KspK_{sp}. Using Ksp=[Mn+][Cl−]mK_{sp} = [M^{n+}][Cl^-]^m where m is the number of chloride ions: For AgClAgCl: Ksp=[Ag+][Cl−]=1.8×10−10K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10} With [Ag+]=0.10[Ag^+] = 0.10 M: [Cl−]=1.8×10−100.10=1.8×10−9[Cl^-] = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} M For CuCl2CuCl_2: Ksp=[Cu2+][Cl−]2=1.0×10−6K_{sp} = [Cu^{2+}][Cl^-]^2 = 1.0 \times 10^{-6} With [Cu2+]=0.10[Cu^{2+}] = 0.10 M: [Cl−]2=1.0×10−60.10=1.0×10−5[Cl^-]^2 = \frac{1.0 \times 10^{-6}}{0.10} = 1.0 \times 10^{-5} So [Cl−]=3.2×10−3[Cl^-] = 3.2 \times 10^{-3} M AgClAgCl precipitates first (at much lower [Cl−][Cl^-]), then CuCl2CuCl_2. NaClNaCl is highly soluble and won't precipitate under these conditions, making C correct. Answer A incorrectly suggests NaClNaCl will eventually precipitate, but sodium chloride is extremely soluble. Answer B reverses the precipitation order—this would happen if you incorrectly compared KspK_{sp} values directly without accounting for the different stoichiometries and ion concentrations. Answer D is wrong because the compounds have vastly different precipitation thresholds. Study tip: Always calculate the actual concentration needed for precipitation rather than just comparing KspK_{sp} values. Remember that compounds with different stoichiometries require different mathematical approaches, and highly soluble salts like NaClNaCl rarely precipitate from dilute solutions.

Question 6

A mixture of ethanol (C2H5OHC_2H_5OH) and diethyl ether (C2H5OC2H5C_2H_5OC_2H_5) needs to be separated. Both compounds have similar molecular weights (46 g/mol vs 74 g/mol) but different boiling points (78°C vs 35°C). A student considers using fractional distillation versus extraction with water. Which approach would be more effective and why?

  1. Fractional distillation, because the 43°C boiling point difference allows efficient thermal separation
  2. Water extraction, because ethanol's hydrogen bonding makes it preferentially water-soluble (correct answer)
  3. Both methods are equally effective since molecular weight differences are small
  4. Neither method works because both compounds are organic and have similar polarities
  5. Fractional distillation, because water extraction would cause both compounds to become immiscible
Explanation: When you encounter separation problems in organic chemistry, focus on the fundamental principle that "like dissolves like" and consider the intermolecular forces present in each compound. Water extraction is the superior method here because ethanol contains a hydroxyl group (-OH) that can form hydrogen bonds with water molecules, making it highly water-soluble. Diethyl ether, while it has oxygen atoms, cannot form hydrogen bonds as effectively due to its ether linkage, making it much less water-soluble. This dramatic difference in water solubility allows for clean separation: ethanol dissolves into the aqueous layer while ether remains in the organic layer. Let's examine why the other options fall short. Choice A incorrectly suggests fractional distillation is better. While the 43°C boiling point difference could work, it's actually less efficient than extraction because both compounds will vaporize to some degree, requiring multiple distillation cycles and careful temperature control. Choice C is wrong because molecular weight similarity doesn't determine separation effectiveness—intermolecular forces and solubility differences are what matter. Choice D incorrectly assumes that being organic compounds with similar polarities prevents separation, ignoring the crucial hydrogen bonding capability difference. The key study tip: when evaluating separation methods, always consider hydrogen bonding first. Compounds that can hydrogen bond with water (alcohols, carboxylic acids, amines) will generally be water-extractable, while those that cannot (alkanes, ethers, esters) will prefer organic solvents. This principle makes extraction often more efficient than distillation for polar/nonpolar mixtures.

Question 7

A student performs recrystallization of an impure solid using hot water as the solvent. After cooling the solution to room temperature, very few crystals form. Which modification to the procedure would most likely improve crystal formation?

  1. Use a larger volume of hot water to ensure complete dissolution
  2. Add a small amount of cold water to induce supersaturation
  3. Scratch the inside of the flask with a glass rod or add a seed crystal (correct answer)
  4. Heat the solution to a higher temperature before cooling
  5. Filter the hot solution before cooling to remove any remaining impurities
Explanation: Recrystallization relies on controlled crystal formation from a supersaturated solution. When you dissolve an impure solid in hot solvent and then cool it, the decreased solubility at lower temperature should cause pure crystals to precipitate out, leaving impurities in solution. However, sometimes crystals fail to form even when the solution is supersaturated. The key issue here is nucleation - crystals need a surface or starting point to begin forming. When very few crystals appear despite proper supersaturation, the solution lacks nucleation sites. Scratching the flask's interior with a glass rod creates tiny glass particles that serve as nucleation sites, while adding a seed crystal (a small amount of pure compound) provides a perfect template for crystal growth. Both techniques give dissolved molecules a surface to organize upon and begin crystallizing. Looking at the wrong answers: (A) Using more hot water would actually make the problem worse by creating a more dilute solution that's less likely to become supersaturated upon cooling. (B) Adding cold water would dilute the solution and reduce supersaturation, working against crystal formation. (D) Higher heating temperature doesn't address the nucleation problem and could potentially decompose your compound or increase impurity solubility. Study tip: Remember that successful recrystallization requires both supersaturation AND nucleation sites. If crystals won't form from a properly prepared solution, think nucleation first - scratching or seeding almost always solves the problem. This is a common lab issue you'll encounter in organic chemistry courses.

Question 8

A mixture contains naphthalene (C10H8C_{10}H_8, sublimes at 80°C) and sodium chloride (decomposes above 800°C). The mixture is heated to 100°C under reduced pressure. What separation technique is being employed, and what is the expected result?

  1. Distillation; both compounds vaporize and can be collected separately
  2. Sublimation; naphthalene vaporizes and resolidifies, leaving pure NaClNaCl behind (correct answer)
  3. Crystallization; both compounds recrystallize in different forms
  4. Thermal decomposition; naphthalene decomposes while NaClNaCl remains stable
  5. Fusion; both compounds melt and separate based on density differences
Explanation: When you encounter separation problems involving different physical properties, focus on the key temperatures and phase changes described. This question tests your understanding of sublimation as a purification technique. At 100°C under reduced pressure, naphthalene will sublime - it transitions directly from solid to gas without becoming liquid, since its sublimation point is 80°C. The vaporized naphthalene can then be collected on a cool surface where it resolidifies, leaving behind pure sodium chloride in the original container. This is a classic sublimation separation where one component has a much lower sublimation temperature than the decomposition temperature of the other. Answer A is incorrect because distillation involves liquid-to-vapor transitions, but naphthalene sublimates (solid-to-vapor) and NaClNaCl doesn't vaporize at 100°C - it would need over 800°C. Answer C is wrong because crystallization involves dissolving substances and allowing them to reform crystals, which isn't happening here with just heat application. Answer D incorrectly suggests naphthalene decomposes, but 100°C is far too low to break down this stable aromatic compound - it simply changes phase. Remember that sublimation works best when there's a large temperature gap between the sublimation point of one substance and the melting/decomposition point of another. Look for keywords like "reduced pressure" (which lowers sublimation temperatures) and check whether the heating temperature falls between the relevant transition points of the compounds involved.

Question 9

An aqueous solution contains Ba2+Ba^{2+}, Sr2+Sr^{2+}, and Ca2+Ca^{2+} ions. Dilute sulfuric acid is added dropwise until no more precipitate forms. Based on the KspK_{sp} values (BaSO4:1.1×10−10BaSO_4: 1.1 \times 10^{-10}, SrSO4:3.2×10−7SrSO_4: 3.2 \times 10^{-7}, CaSO4:9.1×10−6CaSO_4: 9.1 \times 10^{-6}), which ions remain in solution after precipitation is complete?

  1. All three ions precipitate completely; none remain in solution
  2. Only Ca2+Ca^{2+} remains in solution; Ba2+Ba^{2+} and Sr2+Sr^{2+} precipitate
  3. Only Ba2+Ba^{2+} remains in solution; Sr2+Sr^{2+} and Ca2+Ca^{2+} precipitate
  4. Ca2+Ca^{2+} and Sr2+Sr^{2+} remain in solution; only Ba2+Ba^{2+} precipitates
  5. The relative amounts depend on the initial concentrations of each ion (correct answer)
Explanation: When you encounter precipitation problems with multiple ions, the key is understanding that lower Ksp values indicate lower solubility and greater tendency to precipitate. All three ions form sulfate precipitates with the same stoichiometry (1:1), so you can directly compare their Ksp values to determine precipitation order. BaSO₄ has the smallest Ksp (1.1 × 10⁻¹⁰), making it least soluble and first to precipitate. SrSO₄ (Ksp = 3.2 × 10⁻⁷) precipitates next, while CaSO₄ (Ksp = 9.1 × 10⁻⁶) is most soluble. As sulfuric acid is added dropwise, BaSO₄ precipitates first and most completely. However, the question asks what remains after precipitation is "complete" - meaning when no more precipitate forms upon adding more acid. At this point, the solution is saturated with respect to all three sulfates, but their different solubilities mean different amounts remain dissolved. The higher the Ksp, the more ions remain in solution at equilibrium. Choice A incorrectly assumes complete precipitation means zero ions remain - this ignores equilibrium solubility. Choice B suggests only Ca²⁺ remains, missing that Sr²⁺ also has measurable solubility. Choice C incorrectly identifies Ba²⁺ as remaining when it's actually the least soluble. Choice D correctly identifies that both Ca²⁺ and Sr²⁺ remain in solution since they have higher Ksp values than Ba²⁺. Study tip: In precipitation problems, remember that "complete" precipitation doesn't mean zero concentration - it means equilibrium is established. Always compare Ksp values to predict which ions remain most abundant in solution.

Question 10

A student wants to separate a mixture of acetone (water-soluble) and toluene (water-insoluble) using a separatory funnel. Both solvents are less dense than water. After adding water and shaking, two distinct layers form. Which statement correctly describes the layer composition?

  1. Top layer contains toluene, bottom layer contains water with dissolved acetone (correct answer)
  2. Top layer contains water with dissolved acetone, bottom layer contains toluene
  3. Top layer contains both organic compounds, bottom layer contains pure water
  4. Both layers contain mixtures of all three components in different ratios
  5. The components cannot be separated this way because acetone acts as an emulsifier
Explanation: When you encounter separatory funnel problems, you need to consider two key principles: density differences and solubility patterns. The denser liquid always settles to the bottom, while the less dense liquid floats on top. In this mixture, water has the highest density (1.0 g/mL), so it will form the bottom layer. Acetone, being water-soluble, will dissolve into this aqueous layer. Toluene is water-insoluble and less dense than water, so it remains as a separate organic layer floating on top of the water-acetone solution. Choice A correctly identifies this arrangement: toluene forms the top layer, while the bottom layer contains water with dissolved acetone. This follows the fundamental principle that immiscible liquids separate by density, with miscible components dissolving into their preferred phase. Choice B incorrectly reverses the density relationship, suggesting water (with acetone) is less dense than toluene, which contradicts the given information that both organic solvents are less dense than water. Choice C wrongly assumes acetone won't dissolve in water and instead remains with toluene in the organic layer. This ignores acetone's high water solubility due to its polar carbonyl group. Choice D suggests incomplete separation with all components in both layers. While real separations aren't perfectly clean, the question describes "two distinct layers," indicating complete phase separation based on solubility and density differences. Remember: in liquid-liquid extractions, always identify the densest phase first (it's your bottom layer), then determine where each component will preferentially dissolve based on polarity and solubility rules.

Question 11

A mixture of ethyl acetate (b.p. 77°C) and ethanol (b.p. 78°C) forms an azeotropic mixture that boils at 72°C with a composition of 70% ethyl acetate and 30% ethanol. Simple distillation of this azeotropic mixture will produce:

  1. Pure ethyl acetate in the distillate and pure ethanol in the residue
  2. Pure ethanol in the distillate and pure ethyl acetate in the residue
  3. Azeotropic composition (70:30) in both distillate and residue
  4. Azeotropic composition (70:30) in the distillate with no residue remaining (correct answer)
  5. A mixture that cannot be distilled because azeotropes do not vaporize
Explanation: When you encounter azeotrope problems, remember that azeotropes behave as single compounds during distillation—they cannot be separated by simple distillation because both components have identical vapor pressures at the azeotropic composition. This ethyl acetate-ethanol azeotrope is a minimum-boiling azeotrope (boiling point 72°C is lower than either pure component). During simple distillation of the azeotropic mixture, the vapor produced will have the exact same composition as the liquid (70% ethyl acetate, 30% ethanol) because that's the fundamental property of azeotropes. Since you're distilling the azeotropic composition itself, the entire mixture will vaporize at 72°C with no change in composition, leaving no residue. Option A is wrong because azeotropes cannot be separated into pure components by simple distillation—this would require more advanced techniques like azeotropic distillation with a third component. Option B has the same fundamental flaw, incorrectly assuming separation is possible. Option C incorrectly suggests that some material remains behind as residue; however, when you distill an azeotropic mixture at its exact azeotropic composition, the entire mixture behaves as a single compound and will completely distill over. Study tip: Remember that azeotropes are "constant-boiling mixtures." If you start with the exact azeotropic composition, simple distillation will transfer the entire mixture unchanged to the distillate. Separation only becomes relevant when you have excess of one component beyond the azeotropic ratio.

Question 12

An aqueous solution contains Fe3+Fe^{3+}, Al3+Al^{3+}, and Mg2+Mg^{2+} ions. The solution is treated with aqueous ammonia (NH3NH_3) until the pH reaches 9.5. Based on the KspK_{sp} values for the hydroxides (Fe(OH)3:2.8×10−39Fe(OH)_3: 2.8 \times 10^{-39}, Al(OH)3:3.0×10−34Al(OH)_3: 3.0 \times 10^{-34}, Mg(OH)2:5.6×10−12Mg(OH)_2: 5.6 \times 10^{-12}), which ions will precipitate?

  1. Only Fe3+Fe^{3+} precipitates as Fe(OH)3Fe(OH)_3
  2. Both Fe3+Fe^{3+} and Al3+Al^{3+} precipitate as hydroxides (correct answer)
  3. All three ions precipitate as their respective hydroxides
  4. Only Mg2+Mg^{2+} precipitates because it has the highest KspK_{sp} value
  5. No precipitation occurs because ammonia is a weak base
Explanation: When you encounter precipitation problems involving multiple ions and pH changes, you need to determine which compounds will exceed their solubility limits under the given conditions. At pH 9.5, the hydroxide ion concentration is [OH−]=10−4.5=3.16×10−5[OH^-] = 10^{-4.5} = 3.16 \times 10^{-5} M. To find which hydroxides will precipitate, calculate the reaction quotient (Q) for each and compare it to the respective KspK_{sp}. For Fe(OH)3Fe(OH)_3: Q=[Fe3+][OH−]3Q = [Fe^{3+}][OH^-]^3. Assuming typical concentrations around 0.1 M for the metal ions, Q=(0.1)(3.16×10−5)3=3.16×10−15Q = (0.1)(3.16 \times 10^{-5})^3 = 3.16 \times 10^{-15}. Since this greatly exceeds Ksp=2.8×10−39K_{sp} = 2.8 \times 10^{-39}, Fe(OH)3Fe(OH)_3 precipitates. For Al(OH)3Al(OH)_3: Q=(0.1)(3.16×10−5)3=3.16×10−15Q = (0.1)(3.16 \times 10^{-5})^3 = 3.16 \times 10^{-15}, which exceeds Ksp=3.0×10−34K_{sp} = 3.0 \times 10^{-34}, so Al(OH)3Al(OH)_3 also precipitates. For Mg(OH)2Mg(OH)_2: Q=(0.1)(3.16×10−5)2=1.0×10−10Q = (0.1)(3.16 \times 10^{-5})^2 = 1.0 \times 10^{-10}, which is less than Ksp=5.6×10−12K_{sp} = 5.6 \times 10^{-12}, so Mg(OH)2Mg(OH)_2 remains soluble. Answer A is wrong because Al3+Al^{3+} also precipitates at this pH. Answer C is incorrect because Mg2+Mg^{2+} doesn't reach its precipitation threshold. Answer D shows a fundamental misunderstanding—higher KspK_{sp} values indicate greater solubility, not precipitation tendency. Remember: smaller KspK_{sp} values mean compounds precipitate more easily. Always calculate Q and compare to KspK_{sp} rather than just comparing the KspK_{sp} values alone.

Question 13

A precipitate of AgClAgCl is washed with distilled water to remove soluble impurities. However, some of the AgClAgCl dissolves during washing, reducing the yield. To minimize this loss while still removing impurities, what modification should be made to the washing procedure?

  1. Use hot distilled water to increase the washing efficiency
  2. Use a dilute HClHCl solution instead of pure water for washing (correct answer)
  3. Use a dilute AgNO3AgNO_3 solution instead of pure water for washing
  4. Use ice-cold distilled water to reduce AgClAgCl solubility
  5. Skip the washing step entirely to prevent any loss of product
Explanation: This question tests your understanding of the common ion effect and solubility equilibrium. When washing precipitates, you need to balance removing impurities while minimizing dissolution of your desired product. AgClAgCl establishes an equilibrium with its ions: AgCl(s)⇌Ag(aq)++Cl(aq)−AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}. In pure water, some AgClAgCl will dissolve according to its solubility product constant (KspK_{sp}). To minimize this dissolution, you can apply Le Chatelier's principle by adding one of the product ions to shift the equilibrium back toward the solid. Using dilute HClHCl solution (answer B) provides Cl−Cl^- ions, which creates a common ion effect. The excess Cl−Cl^- shifts the equilibrium to the left, suppressing the dissolution of AgClAgCl while still allowing water to wash away soluble impurities. Answer A is wrong because hot water increases solubility, making dissolution worse. Answer C would work in principle since Ag+Ag^+ is also a common ion, but AgNO3AgNO_3 is expensive and introduces new impurities (NO3−NO_3^-), making it impractical. Answer D incorrectly assumes temperature is the primary factor - while cold water slightly reduces solubility, the effect is minimal compared to the common ion effect, and impurity removal becomes less efficient at low temperatures. Remember: when washing ionic precipitates, use a dilute solution containing one of the precipitate's ions to suppress dissolution while maintaining washing efficiency. This is a classic application of the common ion effect in analytical chemistry.

Question 14

A mixture contains NH4ClNH_4Cl (sublimes at 340°C) and NaClNaCl (melts at 801°C). The mixture is heated to 400°C in a closed system with a cool condenser. What separation technique is being employed, and what is the expected outcome?

  1. Distillation; both salts vaporize and condense separately
  2. Sublimation; NH4ClNH_4Cl sublimes and condenses on the cool surface, leaving NaClNaCl behind (correct answer)
  3. Thermal decomposition; NH4ClNH_4Cl decomposes while NaClNaCl remains unchanged
  4. Fusion; both compounds melt and separate based on density
  5. Crystallization; both compounds recrystallize in different forms
Explanation: When you encounter separation technique problems, focus on the physical properties given and the conditions applied. Here, you have two compounds with very different behaviors at the heating temperature of 400°C. The key insight is understanding what happens at 400°C. NH4ClNH_4Cl sublimes at 340°C, meaning it transitions directly from solid to gas without melting. Since 400°C exceeds this temperature, NH4ClNH_4Cl will sublime completely. Meanwhile, NaClNaCl melts at 801°C, so at 400°C it remains solid since this temperature is well below its melting point. The cool condenser allows the gaseous NH4ClNH_4Cl to condense back to solid form on the cool surface, effectively separating it from the remaining solid NaClNaCl. This is sublimation separation. Choice A is incorrect because distillation involves boiling liquids, not subliming solids, and NaClNaCl doesn't vaporize at 400°C. Choice C is wrong because we're dealing with physical phase changes, not chemical decomposition—NH4ClNH_4Cl sublimes but doesn't break down into different compounds. Choice D fails because fusion refers to melting, and at 400°C, NaClNaCl doesn't melt (it needs 801°C), so no melting or density-based separation occurs. Study tip: Always compare the given temperature to the phase transition temperatures of each component. When one compound undergoes a phase change while another doesn't, you can usually separate them based on that difference.

Question 15

An aqueous solution contains both Ca2+Ca^{2+} and Mg2+Mg^{2+} ions. To selectively precipitate calcium while leaving magnesium in solution, which reagent should be added, and what principle governs this separation?

  1. Sodium carbonate; CaCO3CaCO_3 has lower solubility than MgCO3MgCO_3
  2. Sodium hydroxide; Ca(OH)2Ca(OH)_2 is less soluble than Mg(OH)2Mg(OH)_2
  3. Sodium sulfate; CaSO4CaSO_4 precipitates while MgSO4MgSO_4 remains soluble
  4. Sodium phosphate; calcium forms more stable phosphate complexes
  5. Ammonium oxalate; CaC2O4CaC_2O_4 is highly insoluble while MgC2O4MgC_2O_4 is soluble (correct answer)
Explanation: When you encounter selective precipitation problems, you're dealing with the principle that different compounds have different solubilities, and you can exploit these differences to separate ions. The key is finding conditions where one ion forms a precipitate while the other remains dissolved. For calcium and magnesium separation, sodium carbonate (Na2CO3Na_2CO_3) is the ideal reagent. When added to the solution, it provides carbonate ions that react with both metals, but here's the crucial difference: calcium carbonate (CaCO3CaCO_3) has a much lower solubility than magnesium carbonate (MgCO3MgCO_3). The KspK_{sp} of CaCO3CaCO_3 is approximately 3.4×10−93.4 \times 10^{-9}, while MgCO3MgCO_3 has a KspK_{sp} of about 6.8×10−66.8 \times 10^{-6} — nearly three orders of magnitude higher. This means Ca2+Ca^{2+} will precipitate first and completely, while Mg2+Mg^{2+} stays in solution. Option B is incorrect because both hydroxides have very low solubility, making selective precipitation difficult. Option C fails because both sulfates are actually quite soluble in water — neither would precipitate effectively. Option D is wrong because while calcium phosphates do form, the conditions required would likely precipitate magnesium as well, defeating the selectivity purpose. Remember this pattern: successful selective precipitation requires a significant difference in KspK_{sp} values between the target compounds. Always compare solubility constants when evaluating separation strategies — the larger the difference, the cleaner your separation will be.