MCAT Chemical and Physical Foundations of Biological Systems Quiz: 5a Ions In Solutions

What this quiz covers

This quiz focuses on 5a Ions In Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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.

Question 1

1. Effluent becomes more acidic because OH$^-$ is consumed by the resin
2. Effluent becomes less acidic because OH$^-$ is released and neutralizes H$^+$ (correct answer)
3. Chloride concentration increases because the resin releases Cl$^-$ into solution
4. No change occurs because ion exchange cannot alter pH

Explanation: 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+.

Question 2

1. Pba is more likely to remain bound because added NaCl decreases the number of ions in solution.
2. Pba is more likely to elute because Naba competes with Pba for binding sites on the resin. (correct answer)
3. Pba is more likely to elute because added Cle binds to the negative resin and neutralizes it.
4. 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.

Question 3

1. Free Cle concentration must decrease to maintain electroneutrality, even though no chloride-binding agent is added.
2. The solution can remain overall electroneutral because Cababa is replaced by an equivalent amount of bound (nonfree) Cababa within the chelator complex. (correct answer)
3. The solution becomes net negative because removing free Cababa removes positive charge from the solution entirely.
4. 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.

Question 4

1. More Cababa remains bound because high [Naba] drives the equilibrium left by the common-ion effect.
2. Cababa is displaced into solution because increased [Naba] drives the exchange equilibrium to the right. (correct answer)
3. Cle binds to Re and displaces Cababa because opposite charges repel on the resin surface.
4. No change occurs because ion exchange requires a change in pH, not salt concentration.

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.

Question 5

1. Solution 2 has higher conductivity because monovalent ions always carry charge more efficiently than divalent ions.
2. Solution 1 has higher conductivity because it produces a greater total concentration of ions per formula unit dissolved. (correct answer)
3. Both solutions have identical conductivity because formal concentration fixes the number of charge carriers.
4. 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.

Question 6

1. CaF solubility increases because adding Fe shifts the equilibrium right to produce more Cababa.
2. CaF solubility decreases because added Fe drives the equilibrium left, favoring precipitation of CaF(s). (correct answer)
3. CaF solubility is unchanged because NaF is a strong electrolyte and does not affect equilibria.
4. CaF solubility decreases because added Naba forms insoluble NaF(s), removing Fe from solution.

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.

Question 7

1. NH will conduct best because it produces both NHba and OHe, doubling charge carriers.
2. HCl and MgCl will both conduct strongly because they produce substantial ion concentrations in solution. (correct answer)
3. MgCl will conduct poorly because multivalent ions reduce mobility enough to dominate conductivity.
4. HCl will conduct poorly because Hba is covalently bound to water and not a charge carrier.

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.

Question 8

1. Side 1 retains extra $\mathrm{Na^+}$ relative to Side 2 to balance impermeant $\mathrm{P^{3-}}$, with corresponding redistribution of $\mathrm{Cl^-}$. (correct answer)
2. All ions equalize to identical concentrations on both sides because diffusion eliminates any gradient regardless of impermeant ions.
3. Side 1 becomes net negative because impermeant $\mathrm{P^{3-}}$ cannot be balanced by mobile ions across a membrane.
4. Side 2 accumulates $\mathrm{Na^+}$ because cations always diffuse toward the side with fewer total solutes.

Explanation: This question tests understanding of the Donnan equilibrium effect with impermeant ions. When a membrane separates solutions containing both permeant ions (Na⁺, Cl⁻) and an impermeant ion (P³⁻), the impermeant ion creates an unequal distribution of permeant ions at equilibrium. Side 1 contains impermeant P³⁻ which carries negative charge that must be balanced by mobile cations. This causes Side 1 to retain extra Na⁺ relative to Side 2, and Cl⁻ redistributes accordingly to maintain electroneutrality on both sides and satisfy the Donnan equilibrium condition. Choice B incorrectly ignores the constraint imposed by the impermeant ion, while choice C violates electroneutrality. The key principle is that impermeant ions create asymmetric distributions of permeant ions while maintaining electroneutrality. To solve Donnan problems, remember that the product [cation]×[anion] must be equal on both sides for permeant ions.

Question 9

1. $\mathrm{OH^-}$ consumes $\mathrm{H^+}$, shifting $\mathrm{BH^+ \rightleftharpoons B + H^+}$ toward neutral B and decreasing the fraction of charged species. (correct answer)
2. $\mathrm{OH^-}$ increases the fraction of $\mathrm{BH^+}$ because bases protonate amines, increasing ionization.
3. NaOH decreases solubility by precipitating $\mathrm{NaCl}$, removing counterions required for dissolution.
4. NaOH has no effect because $\mathrm{Cl^-}$ is a spectator ion and spectators fix the $\mathrm{BH^+}$/$\mathrm{B}$ ratio.

Explanation: This question tests understanding of acid-base equilibria and ion speciation in drug formulations. The hydrochloride salt BH⁺Cl⁻ dissociates completely to give BH⁺ (protonated base) and Cl⁻. When NaOH is added, the OH⁻ ions react with H⁺ from the equilibrium BH⁺ ⇌ B + H⁺, removing H⁺ and shifting the equilibrium toward the neutral base B. This decreases the fraction of charged BH⁺ species in solution. Choice B incorrectly suggests OH⁻ would protonate the amine, but bases deprotonate acids, not protonate bases. The principle is that adding base shifts weak acid equilibria toward their conjugate bases. For similar problems involving drug ionization, track how pH changes affect the protonation state using Henderson-Hasselbalch concepts.

Question 10

1. Free $\mathrm{Ca^{2+}}$ decreases because complex formation sequesters calcium into a soluble anionic complex. (correct answer)
2. Free $\mathrm{Ca^{2+}}$ increases because binding to EDTA releases additional $\mathrm{Ca^{2+}}$ from the complex by mass action.
3. EDTA decreases conductivity to zero by removing all ions, including $\mathrm{Na^+}$ and $\mathrm{Cl^-}$, from plasma.
4. EDTA causes calcium to precipitate as $\mathrm{Ca(OH)_2(s)}$ because chelation always reduces solubility of metal ions.

Explanation: This question tests understanding of chelation and complex ion formation in biological systems. EDTA is a hexadentate ligand that forms stable complexes with metal ions like Ca²⁺. When EDTA binds Ca²⁺, it forms the soluble complex [CaEDTA]²⁻, effectively sequestering the calcium and reducing the concentration of free Ca²⁺ ions in solution. This prevents calcium from participating in clotting reactions that require free Ca²⁺. Choice B incorrectly suggests complex formation would increase free Ca²⁺, which contradicts the equilibrium direction. Choice D wrongly claims chelation causes precipitation, but EDTA complexes are highly soluble. The key concept is that chelation removes free metal ions by incorporating them into soluble complexes. To approach chelation problems, recognize that complex formation reduces free ion concentration without precipitation.

Question 11

1. NH4Cl is expected to be more acidic than ammonium acetate because $\mathrm{Cl^-}$ does not consume $\mathrm{H^+}$, whereas acetate can act as a base. (correct answer)
2. Ammonium acetate is expected to be more acidic because it contains two ions, doubling $\mathrm{H^+}$ production compared with NH4Cl.
3. Both solutions must be neutral because they are salts and salts do not affect pH in water.
4. NH4Cl is expected to be basic because $\mathrm{NH_4^+}$ is the conjugate acid of a weak base and therefore consumes $\mathrm{H^+}$.

Explanation: This question tests understanding of salt hydrolysis and its effect on solution pH. NH₄Cl contains NH₄⁺ (weak acid, from weak base NH₃) and Cl⁻ (conjugate base of strong acid HCl, essentially neutral). The NH₄⁺ releases H⁺ making the solution acidic, while Cl⁻ has no effect on pH. In contrast, ammonium acetate contains both NH₄⁺ (weak acid) and CH₃COO⁻ (weak base from weak acid CH₃COOH). The acetate ion can accept H⁺ from water or from NH₄⁺, partially neutralizing the acidity. Therefore, NH₄Cl is more acidic than ammonium acetate. Choice D incorrectly states NH₄⁺ consumes H⁺, but as a weak acid it releases H⁺. The strategy for predicting salt solution pH is to identify whether the cation and anion act as acids or bases. When comparing salts, those with neutral anions (from strong acids) will be more acidic if the cation is acidic.

Question 12

1. AgCl becomes more soluble because added Cl$^-$ shifts dissolution to the right.
2. More AgCl precipitates because added Cl$^-$ shifts the equilibrium toward solid formation. (correct answer)
3. Less AgCl precipitates because Na$^+$ complexes Ag$^+$ and prevents precipitation.
4. No change occurs because NaCl is a strong electrolyte and does not affect equilibria.

Explanation: This question examines the common-ion effect on solubility equilibria, highlighting how added ions suppress dissociation of sparingly soluble salts. In solubility equilibria, excess common ions shift the balance toward precipitation per Le Chatelier's principle. The lactated Ringer's solution with added NaCl introduces extra Cl⁻ before AgNO₃ addition, affecting AgCl formation. Choice B is accurate because the common Cl⁻ ion promotes more AgCl precipitation by shifting equilibrium leftward. Choice A is a distractor, wrongly applying Le Chatelier's to suggest increased solubility, a reversal error. In similar problems, calculate the ion product versus Ksp to predict precipitation. Identify common ions and their impact on equilibrium position.

Question 13

1. The solution conducts because MgCl$_2$ produces mobile ions, including two moles of Cl$^-$ per mole of MgCl$_2$. (correct answer)
2. The solution conducts primarily because undissociated MgCl$_2$ molecules carry charge through the solvent.
3. The solution does not conduct because Mg$^{2+}$ and Cl$^-$ neutralize each other immediately upon dissolving.
4. The solution conducts weakly because MgCl$_2$ is a nonelectrolyte at 0.050 M.

Explanation: 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.

Question 14

1. Conductivity decreases because glucose does not dissociate into ions, reducing charge carriers. (correct answer)
2. Conductivity decreases because glucose binds Na$^+$ and removes it from solution.
3. Conductivity decreases because glucose increases the number of ions by hydrolyzing into H$^+$ and OH$^-$.
4. Conductivity decreases because replacing Na$^+$ with glucose increases total ion concentration.

Explanation: 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.

Question 15

1. The salt is more soluble in 0.20 M NaCl because added Cl$^-$ pulls BH$^+$ into solution.
2. The salt is less soluble in 0.20 M NaCl because added Cl$^-$ shifts the dissolution equilibrium toward the solid. (correct answer)
3. The salt solubility is unchanged because NaCl is a strong electrolyte and does not affect equilibria.
4. The salt is less soluble because Na$^+$ reacts with BH$^+$ to form a covalent complex.

Explanation: This question investigates the common-ion effect on drug salt solubility in saline solutions. Added common ions decrease solubility by shifting dissolution equilibria toward the solid. The BH⁺Cl⁻ drug in 0.20 M NaCl experiences excess Cl⁻. Choice B correctly indicates reduced solubility due to Cl⁻ shifting equilibrium leftward. Choice A errs by suggesting increased solubility from Cl⁻, reversing the common-ion effect. For analogous problems, set up Ksp expressions with added ion concentrations. Predict qualitative changes using Le Chatelier's principle.

Question 16

1. Solution X provides more total dissolved ions because NaCl produces three ions per formula unit.
2. Solution Y provides more total dissolved ions per mole of solute because CaCl$_2$ dissociates into three ions. (correct answer)
3. Solution X and Y provide the same number of ions because both contain chloride.
4. Solution Y provides fewer charge carriers because Ca$^{2+}$ cannot move in water due to its higher charge.

Explanation: This question compares total ion concentrations from different salts at equal molarities, affecting charge carrier availability. Salts with higher ion yields per formula unit provide more charge carriers in solution. The patch-clamp solutions X (NaCl) and Y (CaCl₂) are at 150 mM. Choice B is correct as CaCl₂ dissociates into three ions, yielding more per mole than NaCl's two. Choice A errs by stating NaCl produces three ions, a dissociation miscount. For verification, compute total ion molarity using van't Hoff factors. Consider implications for osmotic or electrical properties.

Question 17

1. Na$^+$ and K$^+$ will diffuse until their concentrations are equal on both sides, but charge separation will oppose extensive net cation movement. (correct answer)
2. Cl$^-$ will diffuse rapidly to equalize charge because it is the counterion for both salts.
3. Na$^+$ will move entirely to Side 2 because K$^+$ has a higher charge.
4. No ions can move because solutions of strong electrolytes do not permit diffusion.

Explanation: This question tests ion diffusion across semipermeable membranes, considering charge separation effects. Cations can diffuse if permeable, but impermeable anions create electrostatic opposition to net movement. The dialysis setup allows Na⁺ and K⁺ but not Cl⁻ diffusion. Choice A accurately predicts equalization tempered by charge imbalance. Choice C errs by suggesting complete Na⁺ movement due to K⁺ charge, a charge misunderstanding. In similar problems, consider Donnan equilibrium principles. Predict limiting factors like potential differences.

Question 18

1. More Ca$_3$(PO$_4$)$_2$ precipitates because added Ca$^{2+}$ shifts the equilibrium toward the solid. (correct answer)
2. Less Ca$_3$(PO$_4$)$_2$ precipitates because added Ca$^{2+}$ pulls PO$_4^{3-}$ into solution.
3. No change occurs because Cl$^-$ is not in the equilibrium expression.
4. Precipitation is prevented because CaCl$_2$ is a strong electrolyte and therefore increases solubility of all salts.

Explanation: This question explores common-ion effects on phosphate salt precipitation in media. Added Ca²⁺ shifts the equilibrium toward Ca₃(PO₄)₂(s), increasing precipitation. The cell culture medium with extra CaCl₂ affects the phosphate equilibrium. Choice A is correct as common Ca²⁺ promotes solid formation. Choice B is incorrect, claiming less precipitation from pulling PO₄³⁻, reversing the effect. For analogous questions, apply Le Chatelier's to ion additions. Calculate if Q exceeds Ksp with added concentrations.

Question 19

1. NH$_4$OH has lower conductivity because it produces fewer ions due to incomplete ionization compared with NaOH. (correct answer)
2. NH$_4$OH has lower conductivity because OH$^-$ cannot conduct electricity in water.
3. NaOH has higher conductivity because Na$^+$ is more acidic than NH$_4^+$.
4. Both should have the same conductivity because both contain the same functional group, OH.

Explanation: This question contrasts strong and weak base ionization and resulting conductivity in solutions. Weak bases like NH₄OH ionize incompletely, producing fewer ions than strong bases like NaOH. The student's measurements show NaOH's higher conductivity. Choice A is correct as NH₄OH's partial ionization yields fewer charge carriers. Choice D fails by equating them based on OH group, ignoring dissociation differences. For related items, reference base strength via Kb. Estimate ion fractions from equilibrium constants.

Question 20

1. Beaker 2 has higher [Ag$^+$] because added Cl$^-$ drives dissolution forward
2. Beaker 2 has lower [Ag$^+$] due to the common-ion effect from Cl$^-$ (correct answer)
3. Both beakers have the same [Ag$^+$] because $K_{sp}$ depends only on temperature
4. AgCl dissolves completely in Beaker 2 because NaCl is highly soluble

Explanation: This question tests understanding of the common-ion effect on solubility equilibria. When a common ion (Cl- from NaCl) is added to a saturated solution of AgCl, it shifts the equilibrium AgCl(s) ⇌ Ag+(aq) + Cl-(aq) to the left according to Le Chatelier's principle. This reduces the concentration of Ag+ in solution to maintain the constant Ksp value. The correct answer B states that Beaker 2 has lower [Ag+] due to the common-ion effect from Cl-. Answer A incorrectly suggests added Cl- increases dissolution, while C incorrectly claims both have the same [Ag+]. To solve common-ion problems, recognize that adding a common ion always decreases the solubility of the sparingly soluble salt.