Ions in Solutions (5A) - Chem & Physics

1. A neuron is bathed in extracellular fluid containing 145 mM NaCl and 4 mM KCl. A researcher adds 10 mM sucrose (non-electrolyte) to the extracellular fluid without changing ion amounts. Which statement best reflects the behavior of ions in the solution?
   Answer: Extracellular Na$^+$ and Cl$^-$ concentrations are essentially unchanged because sucrose does not dissociate into ions This question tests understanding of how non-electrolytes affect ion behavior in solutions. Sucrose is a non-electrolyte that dissolves in water but does not dissociate into ions. When sucrose is added to a solution containing NaCl and KCl, it increases the total solute concentration but does not change the amount of Na+, K+, or Cl- ions present. The correct answer B states that extracellular Na+ and Cl- concentrations are essentially unchanged because sucrose does not dissociate into ions. Answer A incorrectly suggests sucrose binds ions, while D incorrectly claims sucrose drives ion dissociation. When non-electrolytes are added to ionic solutions, they affect properties like osmolarity but not ion concentrations or ionic equilibria.
2. A lab prepares a saturated solution of PbI$_2$ for an analytical control: PbI$_2$(s) $\rightleftharpoons$ Pb$^{2+}$ + 2 I$^-$. In a second preparation, KI is added before equilibrating with PbI$_2$(s). Based on the information, which conclusion is most consistent with ion behavior?
   Answer: Adding KI decreases PbI$_2$ solubility because added I$^-$ shifts the equilibrium toward PbI$_2$(s). This question evaluates the common-ion effect in saturated solutions of lead salts. Excess iodide from KI reduces PbI₂ solubility by favoring the solid phase. The PbI₂ preparations differ by added KI in the second. Choice B logically states decreased solubility due to I⁻ shifting equilibrium. Choice D is wrong, claiming the common-ion effect increases solubility, a conceptual inversion. To approach similarly, incorporate added ion into Ksp calculations. Visualize equilibrium shifts with added reactants.
3. A biochemist prepares 1.0 L of a solution labeled 0.10 M MgCl$_2$ for an enzyme assay. Assume complete dissociation: $\text{MgCl}_2(aq)\rightarrow \text{Mg}^{2+}(aq)+2\text{Cl}^-(aq)$. Which statement best reflects the behavior of ions in the solution?
   Answer: The solution contains 0.10 M Mg$^{2+}$ and 0.20 M Cl$^-$ because each MgCl$_2$ produces one Mg$^{2+}$ and two Cl$^-$ ions. This question tests understanding of stoichiometric relationships in ionic dissociation and the resulting ion concentrations. When ionic compounds dissolve and dissociate completely, the concentration of each ion depends on the stoichiometry of the formula unit. MgCl2 dissociates according to: MgCl2(aq) → Mg2+(aq) + 2Cl-(aq), meaning each formula unit produces one Mg2+ ion and two Cl- ions. Starting with 0.10 M MgCl2, complete dissociation yields 0.10 M Mg2+ (1:1 ratio) and 0.20 M Cl- (1:2 ratio). Choice C correctly identifies both ion concentrations based on stoichiometry, while choice A incorrectly assumes only one Cl- per formula unit. To determine ion concentrations from ionic compounds, multiply the compound's molarity by the subscript of each ion in the formula - this gives the concentration of each ion type after complete dissociation.
4. A solubility test for a diagnostic reagent uses BaSO$_4$(s) $\rightleftharpoons$ Ba$^{2+}$ + SO$_4^{2-}$. Two samples are prepared: Sample 1 is pure water; Sample 2 contains added Na$_2$SO$_4$. After equilibration with excess BaSO$_4$(s), which outcome would be expected from the described ion interaction?
   Answer: Sample 2 has a lower dissolved Ba$^{2+}$ concentration due to the common-ion effect from SO$_4^{2-}$. This question explores the common-ion effect on salt solubility, focusing on how added anions suppress cation dissolution. Sparingly soluble salts have equilibria shifted by common ions, reducing solubility per Le Chatelier's. The BaSO₄ test compares pure water to Na₂SO₄-added samples. Choice A is correct as SO₄²⁻ decreases dissolved Ba²⁺ by favoring precipitation. Choice B incorrectly claims increased Ba²⁺ from Na⁺, ignoring common-ion dynamics. In similar cases, compute solubility using Ksp and common-ion concentration. Use equilibrium arrows to visualize shifts.
5. To model precipitation in renal tubules, a solution contains 1.0 mM Ca$^{2+}$ and 1.0 mM C$_2$O$_4^{2-}$ at 37°C. Calcium oxalate has $K_{sp}=2.0\times 10^{-9}$ for CaC$_2$O$_4$(s) $\rightleftharpoons$ Ca$^{2+}$(aq) + C$_2$O$_4^{2-}$(aq). Based on the information, which conclusion is most consistent with ion behavior?
   Answer: Precipitation is favored because the ionic product $[\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}]$ exceeds $K_{sp}$ This question tests understanding of solubility equilibria and precipitation in solutions. The solubility product constant (Ksp) determines whether precipitation occurs when the ionic product Q = [Ca2+][C2O42-] is compared to Ksp. With both ions at 1.0 mM (1.0 × 10-3 M), the ionic product Q = (1.0 × 10-3)(1.0 × 10-3) = 1.0 × 10-6, which exceeds Ksp = 2.0 × 10-9. The correct answer B states that precipitation is favored because the ionic product exceeds Ksp. Answer A incorrectly assumes millimolar concentrations are too low for precipitation, while C incorrectly suggests adding more Ca2+ increases solubility when it would actually decrease it. To solve precipitation problems, calculate Q and compare to Ksp: if Q > Ksp, precipitation occurs; if Q < Ksp, no precipitation.
6. A desalting step uses an anion-exchange resin in the OH$^-$ form. When a solution containing HCl passes through, exchange occurs: R–N$^+$(CH$_3$)$_3$ OH$^-$(s) + Cl$^-$(aq) $\rightleftharpoons$ R–N$^+$(CH$_3$)$_3$ Cl$^-$(s) + OH$^-$(aq). In the effluent, H$^+$(aq) + OH$^-$(aq) $\rightarrow$ H$_2$O(l). Which outcome would be expected from the described ion interaction?
   Answer: Effluent becomes less acidic because OH$^-$ is released and neutralizes H$^+$ This question tests understanding of anion exchange coupled with acid-base neutralization. The OH--form resin exchanges OH- for Cl- from the HCl solution, releasing OH- into the effluent. This OH- immediately neutralizes H+ from HCl to form water, effectively removing both H+ and Cl- from solution and decreasing acidity. The correct answer B states that effluent becomes less acidic because OH- is released and neutralizes H+. Answer A incorrectly suggests acidification, while D incorrectly claims ion exchange cannot alter pH. When using OH--form anion exchangers with acids, expect neutralization as OH- is released to react with H+.
7. To model calcium phosphate precipitation relevant to kidney stone formation, a researcher mixes equal volumes of two solutions at $25^\circ\text{C}$: Solution A contains $1.0\ \text{mM}$ CaCl$_2$; Solution B contains $1.0\ \text{mM}$ Na$_2$HPO$_4$. Assume complete dissociation of these soluble salts. The sparingly soluble solid is $\text{CaHPO}_4(s)$ with equilibrium: $\text{CaHPO}_4(s)\rightleftharpoons \text{Ca}^{2+}(aq)+\text{HPO}_4^{2-}(aq)$. Which outcome would be expected from the described ion interaction immediately after mixing?
   Answer: A precipitate of CaHPO$_4$ is favored because Ca$^{2+}$ and HPO$_4^{2-}$ are brought together, increasing the likelihood of exceeding the solubility equilibrium condition. This question tests understanding of precipitation equilibria and the common ion effect in biological systems. When solutions containing ions that can form a sparingly soluble compound are mixed, precipitation occurs if the ion product exceeds the solubility product (Ksp). Initially, Solution A contains 1.0 mM Ca2+ and Solution B contains 1.0 mM HPO42-; after mixing equal volumes, each ion concentration becomes 0.50 mM. The relevant equilibrium is CaHPO4(s) ⇌ Ca2+(aq) + HPO42-(aq), and bringing Ca2+ and HPO42- together increases the likelihood of exceeding Ksp for CaHPO4. Choice B correctly identifies that CaHPO4 precipitation is favored when these ions are brought together, while choice A incorrectly assumes dilution prevents all ion interactions. To predict precipitation, calculate the ion product Q = [Ca2+][HPO42-] and compare to Ksp - precipitation occurs when Q > Ksp, and the presence of both required ions in significant concentrations favors solid formation.
8. A protein purification protocol uses an anion-exchange column bearing fixed positive charges (Resin–N$^+$(CH$_3$)$_3$). A sample in low-salt buffer contains acetate (CH$_3$COO$^-$) and chloride (Cl$^-$). Which statement best reflects the behavior of ions in the solution during binding?
   Answer: Anions are attracted to the positively charged resin sites and can bind via electrostatic interactions. This question tests ion binding in chromatography, emphasizing electrostatic interactions between charged resins and solution ions. Anion-exchange resins with positive charges attract and bind negatively charged ions via Coulombic forces. In the protein purification, the positively charged resin interacts with anions like acetate and chloride. Choice A correctly describes anions binding to positive sites through attraction. Choice B errs by stating cations are attracted to positive sites, a charge polarity mistake. For verification, consider resin charge and ion signs. Reason about binding strength based on buffer conditions.
9. A lab compares two solutions used to rinse an ion-selective electrode: 10 mM NaCl and 10 mM AlCl$_3$, both fully dissociated. (AlCl$_3$(s) $\rightarrow$ Al$^{3+}$ + 3 Cl$^-$.) Which statement best reflects the behavior of ions in the solution relevant to conductivity?
   Answer: 10 mM AlCl$_3$ should generally have higher conductivity because it yields more ions per formula unit than 10 mM NaCl. This question tests the understanding of how dissociation and ion concentration influence the electrical conductivity of solutions containing electrolytes. Conductivity arises from the movement of ions in solution under an electric field, with greater numbers of ions generally increasing conductivity, assuming comparable ionic mobilities. In the given scenario, 10 mM NaCl dissociates into 10 mM Na⁺ and 10 mM Cl⁻, while 10 mM AlCl₃ dissociates into 10 mM Al³⁺ and 30 mM Cl⁻, resulting in a higher total ion concentration for AlCl₃. Therefore, the 10 mM AlCl₃ solution should have higher conductivity due to producing more ions per formula unit, as correctly indicated in choice B. Choice D fails by mistakenly equating the chloride concentrations in both solutions, ignoring that AlCl₃ provides three times as many Cl⁻ ions, which is a common error in overlooking dissociation stoichiometry. For similar questions, calculate the total ion concentration from the dissociation equation to compare conductivities. Additionally, remember that while ion mobility varies, the number of charge carriers often dominates in dilute solutions of strong electrolytes.
10. A protein purification step uses a cation-exchange resin bearing fixed negative sulfonate groups (Resin.). A sample in low-salt buffer contains Na+z, Cabz+z, and a positively charged peptide P+z. The resin is initially in the Na+z form. Ion exchange can be represented as: Resin.b7Na+z + P+z <c Resin.b7P+z + Na+z. Which outcome would be expected when the NaCl concentration in the buffer is increased substantially during elution?
    Answer: P+z is more likely to elute because Na+z competes with P+z for binding sites on the resin. This question tests understanding of ion exchange chromatography, where ions compete for binding sites based on concentration and charge. The principle is that cation exchange resins have fixed negative charges that attract and bind positively charged ions, with binding equilibria that can be shifted by changing ion concentrations. In this system, the peptide P+ is initially bound to the negatively charged resin, but increasing NaCl concentration floods the system with Na+ ions. The correct answer B follows because the mass action principle dictates that high Na+ concentration drives the equilibrium toward Na+ binding to the resin, displacing P+ into solution. Choice D incorrectly suggests that higher ionic strength increases binding, when actually it shields electrostatic interactions and promotes elution. A useful strategy is to apply Le Chatelier's principle: adding excess of one ion (Na+) shifts the equilibrium to consume that excess, displacing the originally bound ion.
11. A researcher prepares two 0.10 M solutions for a conductivity control: Solution 1 is NaCl; Solution 2 is NaCH$_3$COO. Both salts are fully dissociated: NaCl $\rightarrow$ Na$^+$ + Cl$^-$; NaCH$_3$COO $\rightarrow$ Na$^+$ + CH$_3$COO$^-$. Which statement best reflects the behavior of ions in the solution?
    Answer: Both solutions should conduct because each contains mobile ions in comparable concentrations. This question compares conductivity of different salts with similar ion concentrations, based on dissociation. Fully dissociated salts provide mobile ions for conduction regardless of anion identity. The researcher's NaCl and NaCH₃COO solutions are both 0.10 M strong electrolytes. Choice B follows as both yield comparable Na⁺ and anion ions for conduction. Choice A fails by calling acetate neutral, confusing conjugate bases with non-ions. To verify, confirm salt dissociation in water. Compare anion mobilities if needed for fine differences.
12. A formulation contains Ca$^{2+}$ for cell signaling studies. The solubility equilibrium for calcium fluoride is: CaF$_2$(s) $\rightleftharpoons$ Ca$^{2+}$ + 2 F$^-$. If NaF is added to the solution, which statement best reflects the behavior of ions in the solution?
    Answer: CaF$_2$ becomes less soluble because added F$^-$ shifts the equilibrium toward CaF$_2$(s). This question probes the common-ion effect in solubility, focusing on how additional anions reduce the solubility of calcium salts in solution. Solubility equilibria respond to added common ions by shifting toward the solid phase to maintain Ksp. The Ca²⁺-containing formulation with added NaF introduces excess F⁻ to the CaF₂ equilibrium. Choice B correctly states reduced solubility as F⁻ shifts the equilibrium toward CaF₂(s). Choice A fails by confusing ionic strength with the common-ion effect, an overgeneralization error. For analogous questions, compare solubility in pure water versus with added salts. Use Le Chatelier's principle to reason directional shifts.
13. A neuron-like cell is placed in an external solution where Na$^+$ is replaced with an equimolar amount of glucose (osmolarity maintained). Measured conductivity of the external solution decreases sharply. Which statement best reflects the behavior of ions in the solution?
    Answer: Conductivity decreases because glucose does not dissociate into ions, reducing charge carriers. This question evaluates how replacing ionic solutes with nonelectrolytes affects ion concentration and conductivity in physiological solutions. Nonelectrolytes like glucose do not dissociate, thus not contributing to ion-based conductivity, unlike salts. In the neuron-like cell setup, substituting Na⁺ with glucose maintains osmolarity but reduces ions. Choice A is correct because glucose's lack of dissociation decreases charge carriers, lowering conductivity. Choice B fails by suggesting glucose binds Na⁺, an invented interaction error. For similar scenarios, calculate total ion molarity before and after changes. Consider osmolarity separately from conductivity contributions.
14. A patch-clamp experiment uses an extracellular solution containing 150 mM NaCl and 2 mM CaCl. A chelator is added that selectively binds free Cababa, decreasing its free concentration without changing total chloride. Which outcome is most consistent with ion behavior in solution and charge balance?
    Answer: The solution can remain overall electroneutral because Cababa is replaced by an equivalent amount of bound (nonfree) Cababa within the chelator complex. This question tests understanding of electroneutrality and the distinction between free ions and complexed ions in solution. The principle of electroneutrality states that the sum of positive charges must equal the sum of negative charges in any solution. When the chelator binds Ca2+, it forms a complex where Ca2+ is still present but no longer free - the calcium is now part of a larger complex that maintains the same overall charge. The correct answer B follows because the Ca2+ isn't removed from solution entirely, just converted from free Ca2+ to chelator-bound Ca2+, maintaining the same total positive charge to balance the negative charges from Cl- and other anions. Choice A incorrectly suggests Cl- concentration must change even though no chloride-binding occurs, violating mass conservation. A key concept is that electroneutrality considers all charges present, whether in free ions or complexes, and chelation changes the form but not the total charge balance.
15. An ion-exchange resin in the H+ form is used to acidify a protein sample by exchanging cations. The resin contains fixed sulfonate groups (R–SO3−) with mobile counterions. When a solution containing 10 mM NaCl is passed through, the exchange is represented as: $\mathrm{R{-}SO_3^-H^+ + Na^+ \rightleftharpoons R{-}SO_3^-Na^+ + H^+}$. Which outcome would be expected from this ion interaction in the effluent (the liquid leaving the column)?
    Answer: Effluent pH decreases because released $\mathrm{H^+}$ replaces $\mathrm{Na^+}$ on the resin. This question tests understanding of ion exchange processes and their effects on solution properties. Ion exchange resins contain fixed charged groups with mobile counterions that can be exchanged with ions in solution. When Na+ from the NaCl solution encounters the H+-form resin, the Na+ ions bind to the sulfonate groups, releasing H+ ions into the effluent according to the given equilibrium. This exchange increases the H+ concentration in the effluent, thereby decreasing its pH (making it more acidic). Choice B incorrectly suggests H+ would bind Cl- to form neutral HCl, but HCl is a strong acid that remains fully dissociated in aqueous solution. The key principle is that ion exchange maintains charge balance while swapping one mobile ion for another. To approach similar problems, track which ions are being exchanged and consider their effect on solution pH.
16. A biochemistry lab prepares a solution containing 0.050 M MgCl$_2$ and measures strong conductivity. The dissolution is MgCl$_2$(s) $\rightarrow$ Mg$^{2+}$ + 2 Cl$^-$. Which statement best reflects the behavior of ions in the solution?
    Answer: The solution conducts because MgCl$_2$ produces mobile ions, including two moles of Cl$^-$ per mole of MgCl$_2$. This question tests electrolyte dissociation and its role in solution conductivity, emphasizing complete ion formation in strong electrolytes. Strong electrolytes like MgCl₂ fully dissociate into mobile ions that enable charge transport in aqueous solutions. The biochemistry lab's 0.050 M MgCl₂ solution shows strong conductivity due to this dissociation. Choice A is correct as it highlights the production of Mg²⁺ and two Cl⁻ ions, facilitating conduction. Choice B is incorrect, mistakenly attributing conduction to undissociated molecules, a nonelectrolyte confusion. To solve similar items, classify the solute as strong, weak, or nonelectrolyte. Relate ion count to expected conductivity levels.
17. A neuron is placed in an isotonic solution containing 140 mM NaCl and 5 mM KCl. A toxin selectively increases membrane permeability to K+z without changing Na+z permeability. Considering only ion diffusion tendencies (ignoring active transport and electrical potential), which outcome is expected for K+z movement immediately after permeability increases?
    Answer: K+z will move out of the neuron because its intracellular concentration is typically higher than extracellular concentration. This question tests understanding of ion concentration gradients and passive diffusion across membranes. The principle is that ions move down their concentration gradient when membrane permeability allows, from high to low concentration. In neurons, intracellular K+ concentration (~140 mM) is typically much higher than extracellular K+ concentration (5 mM in this case), creating a strong outward gradient. The correct answer C follows because increasing K+ permeability allows K+ to follow its concentration gradient out of the cell, regardless of other ions present. Choice A incorrectly suggests that changing permeability reverses the concentration gradient itself, which is impossible - permeability only allows existing gradients to drive ion movement. A key strategy is to identify the direction of concentration gradients first, then determine how changes in permeability allow ions to follow those gradients.
18. A clinical chemistry lab compares electrolyte behavior by dissolving $0.010\ \text{mol}$ of each solute in separate beakers containing $1.0\ \text{L}$ deionized water at $25^\circ\text{C}$. The measured conductivities are: Solution 1 (NaCl) $\kappa=1.2\ \text{mS/cm}$, Solution 2 (CH$_3$COOH) $\kappa=0.10\ \text{mS/cm}$, Solution 3 (glucose) $\kappa\approx 0$. Relevant equations: $\text{NaCl}(s)\rightarrow \text{Na}^+(aq)+\text{Cl}^-(aq)$; $\text{CH}_3\text{COOH}(aq)\rightleftharpoons \text{H}^+(aq)+\text{CH}_3\text{COO}^-(aq)$. Based on the information, which conclusion is most consistent with ion behavior in these solutions?
    Answer: CH$_3$COOH produces fewer mobile ions than NaCl at the same formal concentration, consistent with partial dissociation. This question tests understanding of ion dissociation and its relationship to electrical conductivity in aqueous solutions. Electrical conductivity depends on the concentration of mobile ions in solution - complete dissociation produces more ions than partial dissociation. NaCl completely dissociates into Na+ and Cl- ions (2 ions per formula unit), CH3COOH partially dissociates into H+ and CH3COO- ions (weak acid with Ka << 1), and glucose does not dissociate at all (molecular compound). The conductivity pattern (NaCl >> CH3COOH > glucose ≈ 0) directly reflects the number of ions produced: NaCl produces ~0.020 M total ions, CH3COOH produces far fewer due to partial dissociation, and glucose produces none. Choice B correctly identifies that CH3COOH produces fewer mobile ions than NaCl due to partial dissociation, while choice A incorrectly assumes all solutes produce similar ion concentrations. To verify ion behavior in solutions, compare the dissociation patterns: strong electrolytes (ionic compounds, strong acids/bases) dissociate completely, weak electrolytes partially dissociate, and nonelectrolytes remain molecular.
19. To remove excess K$^+$ from a dialysate preparation, a clinician tests a cation-exchange resin initially in the H$^+$ form (Resin–SO$_3^-$·H$^+$). When dialysate containing KCl is passed through, the exchange is represented as: Resin–SO$_3^-$·H$^+$ + K$^+$ $\rightleftharpoons$ Resin–SO$_3^-$·K$^+$ + H$^+$. Which outcome would be expected from the described ion interaction?
    Answer: K$^+$ concentration in the effluent decreases because K$^+$ binds to the resin in exchange for H$^+$. This question evaluates ion exchange processes in solutions, emphasizing how charged resins interact with ions to alter concentrations. Cation-exchange resins bind cations, releasing others like H⁺ in exchange, based on electrostatic attraction and equilibrium. In the dialysate, the resin targets excess K⁺ from KCl, exchanging it for H⁺. Choice B is logical because K⁺ binds to the resin, decreasing its effluent concentration while releasing H⁺. Choice D fails by incorrectly suggesting increased K⁺ from dissociation, ignoring the exchange mechanism. To approach similar questions, write the exchange reaction and track ion concentrations before and after. Consider charge balance to predict pH or concentration changes.
20. A researcher prepares 0.10 M solutions of Na$_2$SO$_4$ and NaCl at the same temperature. Assume complete dissociation and similar ion mobilities. Which statement best reflects the behavior of ions in the solution regarding conductivity?
    Answer: Na$_2$SO$_4$ should conduct better because it produces three ions per formula unit This question tests understanding of how ion production affects solution conductivity. Na2SO4 dissociates into 2Na+ + SO42- (3 ions total per formula unit), while NaCl dissociates into Na+ + Cl- (2 ions total). At equal molar concentrations, Na2SO4 produces 50% more ions than NaCl, resulting in higher conductivity. The correct answer B states that Na2SO4 should conduct better because it produces three ions per formula unit. Answer A incorrectly generalizes about ion mobility without considering total ion count, while C incorrectly focuses only on the common Na+ ion. To compare conductivities of strong electrolytes, count total ions produced per formula unit at the same molarity.

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