MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5c Extraction Distillation

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This deck focuses on 5c Extraction Distillation, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: In a separatory funnel, which layer is on top when organic solvent is less dense than water?

Answer: The organic layer is on top. Density differences cause the less dense organic solvent to float above the denser aqueous phase in the funnel.

Flashcard 2: What is the defining feature of fractional distillation that improves separation?

Answer: A fractionating column enables repeated vaporization–condensation (more theoretical plates). The column provides multiple vapor-liquid equilibria stages, enriching the vapor in the more volatile component for better separation.

Flashcard 3: What is the defining feature of simple distillation that determines when it is appropriate?

Answer: It is best when boiling points differ greatly (roughly $\ge 25\,^{\circ}\text{C}$). Significant boiling point differences allow the more volatile component to vaporize and condense separately without extensive overlap.

Flashcard 4: Which observation indicates sufficient drying agent has been added to an organic solution?

Answer: Drying agent remains free-flowing and no longer clumps. Free-flowing drying agent indicates excess capacity remains after all trace water has been absorbed, preventing clumping from hydration.

Flashcard 5: What is the main purpose of using a drying agent (for example, $\text{MgSO}_4$) after extraction?

Answer: To remove trace water from the organic layer. Anhydrous salts like MgSO4 absorb residual water through hydration, ensuring the organic layer is dry for further processing or analysis.

Flashcard 6: Identify the condition for a base to be mostly protonated in water in terms of pH and $\text{p}K_a$.

Answer: Mostly protonated when $\text{pH} < \text{p}K_a$ (of the conjugate acid). Below the conjugate acid's pKa, the base is more likely to accept a proton, favoring the charged, water-soluble form.

Flashcard 7: Identify the condition for an acid to be mostly deprotonated in water in terms of pH and $\text{p}K_a$.

Answer: Mostly deprotonated when $\text{pH} > \text{p}K_a$. When pH exceeds pKa, the equilibrium shifts toward the deprotonated conjugate base form according to Henderson-Hasselbalch.

Flashcard 8: What is the relationship between pH and ionization of an acid using Henderson–Hasselbalch?

Answer: $\text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right)$. This equation relates pH to the ratio of conjugate base to acid, determining the extent of ionization at equilibrium.

Flashcard 9: Which form of an amine preferentially partitions into the aqueous layer: $\text{RNH}_2$ or $\text{RNH}_3^+$?

Answer: $\text{RNH}_3^+$ partitions into the aqueous layer. Protonation creates a charged ammonium ion that is highly soluble in water due to its polarity and ability to hydrogen bond.

Flashcard 10: Which form of a carboxylic acid preferentially partitions into the aqueous layer: $\text{RCOOH}$ or $\text{RCOO}^-$?

Answer: $\text{RCOO}^-$ partitions into the aqueous layer. The charged carboxylate ion forms strong ion-dipole interactions with water, enhancing its solubility in the aqueous phase.

Flashcard 11: What is the chemical principle behind acid–base extraction of amines or carboxylic acids?

Answer: Convert neutral compounds to ionic salts to change solubility between phases. pH adjustment protonates or deprotonates the compound, altering its polarity and solubility to favor one phase over the other.

Flashcard 12: What does “salting out” accomplish in an aqueous extraction mixture?

Answer: It decreases solubility of organics in water, pushing them into the organic phase. High salt concentration hydrates water molecules, reducing their availability to solvate organics and forcing them into the nonpolar phase.

Flashcard 13: Which change most directly helps break an emulsion during extraction?

Answer: Add brine (salt) to increase ionic strength and promote phase separation. Brine increases the aqueous phase's ionic strength, reducing solubility of organic droplets and promoting their coalescence into a separate layer.

Flashcard 14: What is an emulsion in liquid–liquid extraction?

Answer: A stable dispersion of tiny droplets that prevents clean phase separation. Vigorous mixing can create fine, stable droplets of one phase in the other, impeding coalescence and clear layer formation.

Flashcard 15: What is the purpose of venting a separatory funnel during extraction?

Answer: To release built-up pressure from volatile solvents or gas-forming reactions. Venting prevents excessive pressure buildup that could dislodge the stopper or rupture the funnel during shaking.

Flashcard 16: In a separatory funnel, which layer is on the bottom when organic solvent is denser than water?

Answer: The organic layer is on the bottom. Higher density of the organic solvent positions it below the less dense aqueous phase due to gravitational settling.

Flashcard 17: What is the formula for fraction of solute remaining after $n$ identical extractions?

Answer: $q^n = \left(\frac{V_{\text{aq}}}{V_{\text{aq}} + K_D V_{\text{org}}}\right)^n$. For independent successive extractions, the fraction remaining multiplies with each step, resulting in an exponential decrease in solute left behind.

Flashcard 18: Identify the formula for fraction of solute remaining in the aqueous phase after one extraction.

Answer: $q = \frac{V_{\text{aq}}}{V_{\text{aq}} + K_D V_{\text{org}}}$. Derived from the distribution coefficient, this formula calculates the proportion of solute that remains unextracted in the aqueous phase after equilibration.

Flashcard 19: What is the key requirement for a solvent to be effective in liquid–liquid extraction?

Answer: It must preferentially dissolve the target solute and be immiscible with the other phase. This ensures selective partitioning of the solute into the extracting phase while maintaining two distinct layers for easy separation.

Flashcard 20: What distribution coefficient expression defines partitioning in extraction?

Answer: $K_D = \frac{[S]_{\text{org}}}{[S]_{\text{aq}}}$ at equilibrium. This ratio quantifies the equilibrium concentration of the solute in the organic phase relative to the aqueous phase, indicating partitioning efficiency.

Flashcard 21: Which extraction strategy removes more solute: one large extraction or several small extractions?

Answer: Several small extractions remove more solute (greater overall recovery). Multiple extractions successively remove portions of the remaining solute, leading to higher total recovery than a single extraction with the same total volume.

Flashcard 22: What is steam distillation primarily used to isolate?

Answer: High-boiling, water-immiscible, heat-sensitive compounds (often aromatic oils). Steam lowers the effective boiling point of immiscible mixtures, allowing isolation of sensitive compounds without thermal degradation.

Flashcard 23: In steam distillation of immiscible liquids, what relationship gives the total vapor pressure?

Answer: $P_{\text{total}} = P_1^* + P_2^*$ (sum of pure-component vapor pressures). For immiscible liquids, Dalton's law states the total vapor pressure is the sum of each component's vapor pressure, enabling lower boiling temperatures.

Flashcard 24: How does lowering external pressure affect a liquid’s boiling point in vacuum distillation?

Answer: Lower pressure lowers the boiling point. Boiling occurs when vapor pressure equals external pressure, so reduced pressure allows boiling at lower temperatures to avoid decomposition.

Flashcard 25: What is the physical basis of distillation for separating liquid mixtures?

Answer: Separation by differences in volatility (vapor pressure) and boiling points. Components with higher vapor pressure or lower boiling points evaporate preferentially, enabling separation upon controlled condensation.