Biochemistry Lecture: Thermodynamics, Ionization, and Acid-Base Chemistry

Solute-Water Interactions and Entropic Costs

• Polar and Ionic Solute Behavior:

  • Positively charged ions, such as sodium ($\text{Na}^+$), - interact favorably with water molecules in aqueous solution.

  • Water molecules reorient themselves around the solute to facilitate favorable electrostatic interactions, aligning their partial negatively charged oxygen atoms ($\text{O}^{\delta-}$) toward the positively charged sodium ion ($\text{Na}^+$).

• Hydrophobic Solute Behavior:

  • Non-polar solutes, such as benzene, exhibit no favorable interactions with water molecules.

  • Because there are no favorable solute-water interactions, water molecules line up around the surface of the non-polar solute to form a highly ordered structure known as a water cage.

• Thermodynamics of Water Cage Formation:

  • Water molecules in bulk liquid exist in a less ordered, higher entropy state.

  • Forcing water molecules into an ordered cage around a hydrophobic solute transition them from a less ordered state to a highly ordered state, resulting in a negative change in entropy ($\Delta S < 0$).

  • Forming a water cage carries a massive entropic cost (entropic penalty), rendering the solvation of hydrophobic molecules thermodynamically unfavorable unless compensated elsewhere.

• Amphipathic Molecules:

  • Molecules containing both hydrophilic (polar or charged) and hydrophobic (non-polar) regions within the same chemical structure are defined as amphipathic.

  • The hydrophilic portion can favorably interact with aqueous solutions, whereas the hydrophobic portion cannot.

  • Example — Phenylalanine: Contains polar/charged amino and carboxyl functional groups (hydrophilic portion) attached to an aromatic phenyl ring (hydrophobic portion). Consequently, phenylalanine is an amphipathic molecule.

  • Amino Acid Properties in Proteins: Memorization of the complete structures of all $20$ standard amino acids is required for this course (students may begin early or wait until Chapter 5). Within a folded protein, main-chain charged groups are participating in peptide bonds; thus, the overall chemical properties of each amino acid within a protein are dominated by its side chain.

Amphipathic Molecules and Micelle Dynamics

• Mechanism of Clustering:

  • When amphipathic molecules are dissolved in water, two distinct solvation phenomena occur simultaneously.

  • To minimize energy and reduce the entropic penalty associated with water cage formation, amphipathic molecules spontaneously cluster together in solution.

  • The hydrophobic portions aggregate away from water into the interior of the cluster, while the polar/charged hydrophilic portions face outward toward the surrounding aqueous environment to interact favorably with water molecules.

• Structure and Function of Micelles:

  • The resulting organized spherical structure is called a micelle.

  • Practical Application — Dishwasher Detergent: Detergent molecules are amphipathic and spontaneously form micelles when dissolved in water.

  • Grease is completely non-polar and hydrophobic, making it impossible to wash away using water alone because water molecules cannot interact with it favorably.

  • In a detergent solution, non-polar grease molecules become trapped within the hydrophobic interior of the detergent micelle, allowing the grease to be sequestered and washed away in aqueous solution.

Proton Hopping and Water Autoionization

• Proton Jumps (Proton Hopping):

  • Protons ($\text{H}^+$) in an aqueous medium are delocalized due to the continuous hydrogen-bonding network of water.

  • When a chemical reaction requires a proton, it does not depend on a specific $\text{H}^+$ ion diffusing across long distances.

  • Instead, a proton is transferred extremely rapidly along a chain of hydrogen-bonded water molecules by grabbing a proton from an adjacent water molecule and passing it forward. This dynamic transfer mechanism is termed a proton jump.

• Autoionization Constant of Water ($K_w$):

  • The molar concentration of pure water is $55.5\,\text{M}$.

  • Water undergoes autoionization (meaning it reacts with itself to form ions, which is seen when water molecules transfer a proton to each other from a hydronium ion H3O+ and a hydroxide ion OH-) according to the equilibrium:

$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

*   The ion product constant of water ($K_w$) at standard state ($25^{\circ}\text{C}$) is:

$K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$

*   Under standard conditions, knowing the concentration of {H}^+ allows direct determination of the corresponding {OH}^- concentration.

• Definitions of Acids and Bases:

  • Acid: Any molecule capable of releasing a hydrogen atom / proton ($\text{H}^+$).

  • Base: Any molecule capable of accepting or combining with a proton ($\text{H}^+$) when needed. ‘

The pH Scale and Quantitative Calculations

• Mathematical Definition of pH:

  • To eliminate the tedious manipulation of very small exponential numbers (e.g., $10^{-7}\,\text{M}$, $10^{-6}\,\text{M}$, $10^{-13}\,\text{M}$), hydrogen ion concentration is expressed on a logarithmic scale defined as:

$pH = -\log_{10}[\text{H}^+]$

• pH Thresholds at Standard Conditions:

  • Neutral water: $pH = 7$

  • Acidic solution: $pH < 7$ (smaller pH values denote higher $\text{H}^+$ concentrations and greater acidity).

  • Basic solution: $pH > 7$

• Sample Calculations:

  • Calculation 1: If $[\text{H}^+] = 0.1\,\text{M} = 10^{-1}\,\text{M}, then:\n\npH = -\log_{10}(10^{-1}) = 1\n\n * *Calculation 2*: If $[\text{H}^+] = 1\,\mu\text{M} = 10^{-6}\,\text{M}, then:

$pH = -\log_{10}(10^{-6}) = 6$

*   *Calculation 3*: If $[\text{OH}^-] = 10^{-1}\,\text{M}$, solving for $[\text{H}^+]$ gives:

$[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-1}} = 10^{-13}\,\text{M}$

$pH = -\log_{10}(10^{-13}) = 13$

Temperature Dependence of Water Dissociation and Neutrality

• Thermodynamics of Water Autoionization:

  • The dissociation of water into $\text{H}^+$ and $\text{OH}^-$ is an endothermic process.

• Le Chatelier's Principle & Temperature Shifts:

  • Increasing Temperature: Adding heat shifts the equilibrium toward the right (products), increasing the concentration of $\text{H}^+$ and $\text{OH}^-$. Consequently, the $pH$ of neutral water decreases as temperature rises.

    • At $40^{\circ}\text{C}$, the $pH$ of neutral water decreases to $6.7$

  • Decreasing Temperature: Removing heat shifts the equilibrium toward the left (reactants), decreasing the concentration of OH- and$\text{H}^+$.

    • At $0^{\circ}\text{C}$, the $pH$ of neutral water increases to $7.47$

• Preservation of Neutrality:

  • Despite changes in $pH$ at non-standard temperatures, the water remains strictly neutral because $\text{H}^+$ and $\text{OH}^-$ concentrations increase or decrease in equal molar proportions ($[\text{H}^+] = [\text{OH}^-]$ at all times in pure water).

Strong vs. Weak Acids and ICE Table Calculations

• Strong Acid Dissociation:

  • Example Calculation: Calculate the $pH$ of a $1\,\text{L}$ aqueous solution after adding $10\,\mu\text{L}$ of $1\,\text{M}$ hydrochloric acid ($\text{HCl}$).

  • Convert volume to liters: $10\,\mu\text{L} = 10 \times 10^{-6}\,\text{L} = 10^{-5}\,\text{L}$.

  • Calculate moles of $\text{HCl}$ added: $10^{-5}\,\text{L} \times 1\,\text{M} = 10^{-5}\,\text{mol}$.

  • Calculate final $\text{H}^+$ concentration in $1\,\text{L}$: $[\text{H}^+] = 10^{-5}\,\text{M}$.

  • Determine $pH$: $pH = -\log_{10}(10^{-5}) = 5$

• Weak Acid Dissociation & Acid Constant ($K_a$):

  • Weak acids do not completely dissociate in water and establish an equilibrium between molecular acid ($\text{HA}$) and its conjugate base ($\text{A}^-$):

$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

*   The acid dissociation constant ($K_a$) is defined by the ratio of equilibrium concentrations:

$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

*   For acetic acid ($\text{CH}_3\text{COOH}$ or $\text{HAc}$), the acid dissociation constant is $K_a = 1.76 \times 10^{-5}$.

• Setting Up an ICE (Initial, Change, Equilibrium) Table:

  • To determine $\text{H}^+$ and conjugate base concentrations for a $0.1\,\text{M}$ solution of weak acid (e.g., acetic acid):

    • Initial: $[\text{HA}] = 0.1\,\text{M}$, $[\text{H}^+] = 0\,\text{M}$, $[\text{A}^-] = 0\,\text{M}$

    • Change: $[\text{HA}] = -x$, $[\text{H}^+] = +x$, $[\text{A}^-] = +x$

    • Equilibrium: $[\text{HA}] = 0.1 - x$, $[\text{H}^+] = x$, $[\text{A}^-] = x$

• Equilibrium Expression and Approximation:

  • Substitute equilibrium terms into the $K_a$ definition:

$K_a = \frac{x \cdot x}{0.1 - x} = 1.76 \times 10^{-5}$

*   *Simplifying Approximation*: Because acetic acid is a very weak acid, the amount of ionization $x$ is extremely small relative to $0.1\,\text{M}$. Therefore, $0.1 - x \approx 0.1\,\text{M}$.
*   The expression simplifies to:

$x^2 \approx 0.1 \times K_a = 0.1 \times (1.76 \times 10^{-5}) = 1.76 \times 10^{-6}$

$x = \sqrt{1.76 \times 10^{-6}} = 1.326 \times 10^{-3}\,\text{M}$

The Henderson-Hasselbalch Equation and Buffer Theory

• Derivation of the Henderson-Hasselbalch Equation:

  • Starting from the acid dissociation equilibrium expression:

$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

*   Rearrange to isolate $\text{H}^+$ concentration on the left side:

$[\text{H}^+] = K_a \cdot \frac{[\text{HA}]}{[\text{A}^-]}$

*   Take the negative logarithm ($-\log_{10}$) of both sides:

$-\log_{10}[\text{H}^+] = -\log_{10} K_a - \log_{10}\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)$

*   Applying the definitions $pH = -\log_{10}[\text{H}^+]$ and $pK_a = -\log_{10} K_a$, and inverting the log term quotient yields the **Henderson-Hasselbalch equation**:

$pH = pK_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$

[H+]=10^-ph

-lower PKA value means a stronger acid

• Buffer Rules and Properties:

  • Weak Acid Constraint: The Henderson-Hasselbalch equation and buffer chemistry apply exclusively to weak acids and their conjugate bases. Strong acids cannot be used to create buffer systems.

  • $50\%$ Ionization Condition: When the concentration of conjugate base ($[\text{A}^-]$) equals the concentration of molecular acid ($[\text{HA}]$), exactly $50\%$ of the initial acid has ionized.

  • Under the condition $[\text{A}^-] = [\text{HA}]$:

$\frac{[\text{A}^-]}{[\text{HA}]} = 1 \implies \log_{10}(1) = 0 \implies pH = pK_a$

• Sample Buffer Calculation Problem:

  • Problem Setup: Determine the final $pH$ of a $2\,\text{L}$ aqueous solution containing $10\,\text{mL}$ of weak acid and conjugate base components.

  • Calculation Strategy: Identify which chemical species contribute directly to the concentration of molecular weak acid ($[\text{HA}]$) and which species contribute to the conjugate base concentration ($[\text{A}^-]$), then solve using the Henderson-Hasselbalch equation.

    
    

Course Logistics, Recitations, and Midterm Exam Preparation

• Midterm Examination Warning:

  • Buffer problems are notoriously challenging for students and represent a frequent spot for lost points.

  • Dedicated practice and extra effort are mandatory.

  • Buffer calculation questions are guaranteed to appear on the first midterm examination.

• Recitation Schedule:

  • Buffer calculation problems will be covered extensively during recitation sessions in the upcoming week.

• Audience Interactions & Asides:

  • Questions regarding class schedule timings on Thursdays and lunch plans were addressed during lecture.

  • Seating arrangements and student proximity during class meetings were noted.