Ions in Solutions (5A)
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MCAT Chemical and Physical Foundations of Biological Systems › Ions in Solutions (5A)
A protein purification step uses a cation-exchange resin bearing fixed negative sulfonate groups (Resine). A sample in low-salt buffer contains Naba, Cababa, and a positively charged peptide Pba. The resin is initially in the Naba form. Ion exchange can be represented as: Resineb7Naba + Pba cc Resineb7Pba + Naba. Which outcome would be expected when the NaCl concentration in the buffer is increased substantially during elution?
Pba is more likely to remain bound because added NaCl decreases the number of ions in solution.
Pba is more likely to elute because added Cle binds to the negative resin and neutralizes it.
Pba is more likely to elute because Naba competes with Pba for binding sites on the resin.
Pba is more likely to remain bound because higher ionic strength increases electrostatic attraction to the resin.
Explanation
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.
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?
The solution becomes net negative because removing free Cababa removes positive charge from the solution entirely.
Free Cle concentration must decrease to maintain electroneutrality, even though no chloride-binding agent is added.
The solution can remain overall electroneutral because Cababa is replaced by an equivalent amount of bound (nonfree) Cababa within the chelator complex.
The solution becomes net positive because binding Cababa releases additional Cle into solution.
Explanation
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.
A lab compares ion behavior in three 0.10 M aqueous solutions at 25C: NaCl, CHCOOH, and glucose. Conductivity measurements (same cell constant) show: NaCl = high, CHCOOH = low, glucose e 0. Relevant equilibria: NaCl(s) d2 Naba(aq) + Cle(aq); CHCOOH(aq) cc CHCOOe(aq) + Hba(aq). Which conclusion is most consistent with ion behavior in these solutions?
Glucose produces negligible conductivity because it does not form ions in water.
NaCl has high conductivity because it remains mostly undissociated but is highly polar.
CHCOOH has low conductivity because it produces fewer ions in solution than NaCl at the same molarity.
CHCOOH has low conductivity because it fully dissociates into ions but ions move slowly.
Explanation
This question tests understanding of ion behavior in solutions, specifically how ion concentration affects electrical conductivity. The principle is that electrical conductivity in aqueous solutions depends on the concentration of mobile ions present - more ions mean higher conductivity. In this system, NaCl fully dissociates into Na+ and Cl- ions, CH3COOH is a weak acid that only partially dissociates into CH3COO- and H+ ions, and glucose is a molecular compound that doesn't form ions at all. The correct answer D follows logically because at the same molarity, NaCl produces many more ions (complete dissociation) than CH3COOH (partial dissociation), explaining the conductivity difference. Choice B incorrectly claims CH3COOH fully dissociates, which contradicts its nature as a weak acid. A key strategy for similar questions is to identify whether compounds are strong electrolytes (complete dissociation), weak electrolytes (partial dissociation), or non-electrolytes (no dissociation), then relate ion concentration to the measured property.
A researcher studies precipitation risk when mixing electrolyte solutions at 25C. The solubility equilibrium for silver chloride is: AgCl(s) cc Agba(aq) + Cle(aq), with $K_{sp} = 1.8 \times 10^{-10}$. Two solutions are mixed to equal final volumes: 1.0d710d4 M AgNO and 1.0d710d4 M NaCl. Which statement is most consistent with ion behavior immediately after mixing?
AgCl dissolves more as mixing increases total ion concentration, shifting equilibrium to ions.
Precipitation is prevented because Naba and NOe form an insoluble salt first.
No precipitate forms because both salts are strong electrolytes and remain fully dissolved.
Precipitation is favored because the product $[\text{Ag}^+][\text{Cl}^-]$ exceeds $K_{sp}$ after dilution.
Explanation
This question tests understanding of precipitation equilibria and the solubility product constant (Ksp). The principle is that precipitation occurs when the ion product [Ag+][Cl-] exceeds Ksp for AgCl. In this system, mixing equal volumes of 1.0×10^-4 M AgNO3 and 1.0×10^-4 M NaCl dilutes each concentration by half, giving [Ag+] = [Cl-] = 5.0×10^-5 M after mixing. The correct answer B follows because the ion product (5.0×10^-5)×(5.0×10^-5) = 2.5×10^-9, which exceeds Ksp = 1.8×10^-10, causing precipitation. Choice A incorrectly assumes strong electrolytes cannot precipitate, ignoring that even fully dissociated salts can exceed their solubility limits. A key check is to always calculate the ion product Q and compare to Ksp: if Q > Ksp, precipitation occurs; if Q < Ksp, the solution remains unsaturated.
A chromatography resin with fixed negative charges (Re) is equilibrated with Cababa (Reb7Cababa). A sample containing Naba is applied. A simplified exchange is: Reb7Cababa + 2Naba cc 2(Reb7Naba) + Cababa. Which outcome is expected when a high concentration of NaCl is flowed through the column?
No change occurs because ion exchange requires a change in pH, not salt concentration.
Cle binds to Re and displaces Cababa because opposite charges repel on the resin surface.
More Cababa remains bound because high [Naba] drives the equilibrium left by the common-ion effect.
Cababa is displaced into solution because increased [Naba] drives the exchange equilibrium to the right.
Explanation
This question tests understanding of ion exchange equilibria in chromatography systems. The principle is that ion exchange follows mass action laws, where high concentrations of competing ions can displace bound ions from the resin. In this system, Ca2+ is initially bound to the negatively charged resin sites, but flowing high concentration NaCl through provides excess Na+ ions. The correct answer B follows because the equilibrium R-·Ca2+ + 2Na+ ⇌ 2(R-·Na+) + Ca2+ is driven to the right by the high Na+ concentration, displacing Ca2+ into solution for elution. Choice A incorrectly invokes the common-ion effect, which applies to solubility equilibria not ion exchange, and gets the direction wrong - high [Na+] drives the equilibrium right, not left. A key strategy is to apply Le Chatelier's principle to ion exchange: excess of the competing ion (Na+) shifts equilibrium to consume that excess, displacing the originally bound ion.
A neuron is placed in an isotonic solution containing 140 mM NaCl and 5 mM KCl. A toxin selectively increases membrane permeability to Kba without changing Naba permeability. Considering only ion diffusion tendencies (ignoring active transport and electrical potential), which outcome is expected for Kba movement immediately after permeability increases?
Kba will move out of the neuron because its intracellular concentration is typically higher than extracellular concentration.
Kba will show no net movement because the solution is isotonic.
Kba will move into the neuron because increasing permeability reverses the concentration gradient.
Kba will move out of the neuron because Cle cannot follow, so diffusion must stop.
Explanation
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.
In a microbiology assay, BaSO precipitation is used to remove sulfate from solution. The equilibrium is: BaSO(s) cc Bababa(aq) + SOb2e(aq), with $K_{sp} = 1.1 \times 10^{-10}$. A solution already contains 1.0d710d3 M SOb2e from NaSO. Which statement is most consistent with ion behavior if small amounts of BaCl are added?
BaSO becomes more soluble due to the common-ion effect from SOb2e already present.
BaSO is less likely to precipitate because added Cle complexes Bababa and increases its solubility.
BaSO will not precipitate unless the solution is acidic, because sulfate exists only at low pH.
BaSO is more likely to precipitate because added Bababa increases the ion product $[\text{Ba}^{2+}][\text{SO}_4^{2-}]$.
Explanation
This question tests understanding of precipitation equilibria and the common-ion effect. The principle is that precipitation occurs when the ion product [Ba2+][SO42-] exceeds Ksp, and adding either ion to a solution already containing the other increases the likelihood of precipitation. In this system, the solution already contains 1.0×10^-3 M SO42-, and adding BaCl2 introduces Ba2+ ions. The correct answer B follows because as Ba2+ is added, the ion product [Ba2+][SO42-] increases, and once it exceeds Ksp = 1.1×10^-10, BaSO4 precipitates. Choice D incorrectly states the common-ion effect increases solubility, when it actually decreases solubility - the presence of SO42- makes BaSO4 less soluble, not more. A useful calculation check is that precipitation begins when [Ba2+] > Ksp/[SO42-] = 1.1×10^-10/1.0×10^-3 = 1.1×10^-7 M, which is easily exceeded with small BaCl2 additions.
A conductivity probe is placed in two solutions at 25C, each prepared to the same formal concentration (0.10 M): Solution 1 contains NaSO; Solution 2 contains NaCl. Both salts fully dissociate: NaSO d2 2Naba + SOb2e; NaCl d2 Naba + Cle. Ignoring differences in ionic mobility, which outcome is expected for conductivity?
Solution 1 has higher conductivity because it produces a greater total concentration of ions per formula unit dissolved.
Solution 2 has higher conductivity because monovalent ions always carry charge more efficiently than divalent ions.
Both solutions have identical conductivity because formal concentration fixes the number of charge carriers.
Solution 1 has lower conductivity because sulfate ions are larger and therefore eliminate current flow.
Explanation
This question tests understanding of how dissociation stoichiometry affects ion concentration and conductivity. The principle is that conductivity depends on the total concentration of ions, which varies with how many ions each formula unit produces upon dissociation. In this system, 0.10 M Na2SO4 dissociates to give 2(0.10) = 0.20 M Na+ and 0.10 M SO42- for 0.30 M total ions, while 0.10 M NaCl gives 0.10 M Na+ and 0.10 M Cl- for 0.20 M total ions. The correct answer B follows because Na2SO4 produces 3 ions per formula unit (2 Na+ + 1 SO42-) while NaCl produces only 2 ions per formula unit, giving Na2SO4 solution 50% more total ion concentration and thus higher conductivity. Choice A incorrectly generalizes about monovalent versus divalent ions without considering the actual ion count. A reliable strategy is to calculate total ion concentration by multiplying molarity by the number of ions produced per formula unit.
A solubility study examines CaF in the presence of added NaF. The equilibrium is: CaF(s) cc Cababa(aq) + 2Fe(aq). A researcher adds NaF to a saturated CaF solution while keeping temperature constant. Which outcome would be expected from the described ion interaction?
CaF solubility increases because adding Fe shifts the equilibrium right to produce more Cababa.
CaF solubility decreases because added Naba forms insoluble NaF(s), removing Fe from solution.
CaF solubility is unchanged because NaF is a strong electrolyte and does not affect equilibria.
CaF solubility decreases because added Fe drives the equilibrium left, favoring precipitation of CaF(s).
Explanation
This question tests understanding of the common-ion effect on solubility equilibria. The principle is that adding a common ion (one already present in the equilibrium) shifts the equilibrium according to Le Chatelier's principle, typically decreasing solubility of sparingly soluble salts. In this system, CaF2 is in equilibrium with Ca2+ and F- ions, and adding NaF increases the F- concentration. The correct answer B follows because the added F- from NaF shifts the equilibrium CaF2(s) ⇌ Ca2+ + 2F- to the left, favoring the solid form and decreasing CaF2 solubility - this is the classic common-ion effect. Choice A incorrectly predicts the equilibrium shift direction, suggesting added F- would somehow produce more Ca2+, which violates Le Chatelier's principle. A key strategy is recognizing that adding a product of an equilibrium (F-) always shifts the equilibrium toward reactants (CaF2(s)), reducing solubility.
To compare electrolyte strength, equal volumes of 0.050 M solutions are prepared at 25C: HCl, NH, and MgCl. Relevant equilibria: HCl(aq) d2 Hba + Cle; NH(aq) + Hbae cc NHba + OHe; MgCl(s) d2 Mgbaba + 2Cle. Which statement best reflects ion behavior affecting conductivity at equal molarity?
HCl will conduct poorly because Hba is covalently bound to water and not a charge carrier.
NH will conduct best because it produces both NHba and OHe, doubling charge carriers.
MgCl will conduct poorly because multivalent ions reduce mobility enough to dominate conductivity.
HCl and MgCl will both conduct strongly because they produce substantial ion concentrations in solution.
Explanation
This question tests understanding of how ion concentration from different electrolytes affects solution conductivity. The principle is that conductivity depends on the total concentration of all ions in solution, with strong electrolytes producing more ions than weak electrolytes. In this system, HCl is a strong acid producing 0.050 M H+ and 0.050 M Cl-, NH3 is a weak base producing very few ions, and MgCl2 dissociates completely to give 0.050 M Mg2+ and 0.100 M Cl-. The correct answer B follows because both HCl and MgCl2 are strong electrolytes that fully dissociate, producing substantial ion concentrations (0.100 M total for HCl, 0.150 M total for MgCl2). Choice C incorrectly assumes multivalent ions have such reduced mobility that it dominates over their contribution to conductivity, when in reality the higher total ion concentration from MgCl2 more than compensates. A useful approach is to calculate total ion concentration for each solution, recognizing that conductivity generally increases with total ion concentration.