Biochem exam 1

Physical and Chemical Properties of Water

• Water possesses abnormally high melting points, boiling points, and heats of vaporization compared to other common organic and inorganic solvents of similar molecular size.

• Heat of vaporization definition: The heat energy required to convert $1.0\,\text{g}$ of a liquid at its boiling point and at atmospheric pressure into its gaseous state at the same temperature. It provides a direct quantitative measure of the energy required to overcome attractive intermolecular forces in the liquid phase.t{J\,g}^{-1}$</p></li></ul><h3 id="fe886c3f-13e0-43ad-aaaa-f285fc98ce8e" data-toc-id="fe886c3f-13e0-43ad-aaaa-f285fc98ce8e" collapsed="false" seolevelmigrated="true">Molecular Structure, Electronegativity, and Hydrogen Bonding</h3><ul><li><p>Molecular Geometry of Water:</p><ul><li><p>Bond angle between$\text{H-O-H}$:$104.5^{\circ}$</p></li><li><p>Covalent bond length ($\text{O-H}$):$0.0965\,\text{nm}$</p></li><li><p>Hydrogen bond length ($\text{O}\cdots\text{H}$between adjacent molecules):$0.177\,\text{nm}$</p></li><li><p>Charge distribution: Oxygen carries a partial negative charge ($\delta^{-}$), while each hydrogen atom carries a partial positive charge ($\delta^{+}$).</p></li><li><p>Because oxygen possesses two unshared electron pairs and two bonded hydrogens, each water molecule functions as a perfect hydrogen bond donor (2 hydrogen sites) and hydrogen bond acceptor (2 lone pair sites).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/1.jpg" data-width="50$4$hydrogen bonds with neighboring molecules.</p></li><li><p><strong>Liquid Phase:</strong> Water molecules form an average of$3.4$hydrogen bonds per$\text{H}_2\text{O}$molecule at any given instant. These hydrogen bonds are highly dynamic, transient, and continuously breaking and reforming ("flickering clusters").</p></li><li><p><strong>Density Implications:</strong> Because the open, tetrahedral crystal structure of ice holds water molecules further apart to satisfy all 4 hydrogen bonds, solid ice is less dense than liquid water ($\text{density of liquid water} > \text{density of ice}$).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/3.jpg" data-width="50$\text{O-H}$or$\text{N-H}$).</p></li><li><p><strong>Hydrogen acceptor:</strong> An electronegative atom containing an unshared pair of non-bonding electrons (e.g.,$\text{C=O}$,$\text{-O-}$, or$\text{-N=}$).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/5.jpg" data-width="50$\text{R-O-H} \cdots \text{OH}_2$).</p></li><li><p>Between the carbonyl group of a ketone and water ($\text{R}^1\text{R}^2\text{C=O} \cdots \text{H-O-H}$).</p></li><li><p>Between peptide groups in polypeptide chains ($\text{C=O} \cdots \text{H-N}$).</p></li><li><p>Between complementary nitrogenous base pairs in DNA (such as the double hydrogen bonds formed between Thymine and Adenine).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/6.jpg" data-width="50$\text{O-H} \cdots \text{O}$($180^{\circ}$) geometry.</p></li><li><p><strong>Weaker Hydrogen Bond:</strong> Occurs when the bonded atoms are oriented at an angle (non-linear configuration).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/7.jpg" data-width="50$\text{NaCl}$):</p></li><li><p><strong>Hydrated Chloride Ion ($\text{Cl}^-$):</strong> Partially positive hydrogen atoms ($\delta^{+}$) of surrounding water molecules align towards the negatively charged$\text{Cl}^-$ion.</p></li><li><p><strong>Hydrated Sodium Ion ($\text{Na}^+$):</strong> Partially negative oxygen atoms ($\delta^{-}$) of surrounding water molecules align towards the positively charged$\text{Na}^+$ion.</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/8.jpg" data-width="50$\Delta S < 0$), representing an energetically unfavorable condition ($\Delta G > 0$).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/9.jpg" data-width="50$\Delta S > 0$).</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/10.jpg" data-width="50$\text{-C=O} \cdots \text{H-O-}$)</p></li><li><p>Peptide group interactions (e.g.,$\text{-C=O} \cdots \text{H-N-}$)</p></li><li><p><strong>Ionic Interactions:</strong></p></li><li><p>Electrostatic attraction between opposite charges (e.g.,$\text{-NH}_3^+ \cdots ^-\text{OOC-}$)</p></li><li><p>Electrostatic repulsion between like charges (e.g.,$\text{-NH}_3^+ \cdots \text{}^+\text{H}_3N-$)</p></li><li><p><strong>Hydrophobic Interactions:</strong></p></li><li><p>Association of non-polar functional groups (e.g., leucine side chains or benzene ring residues) driven by water entropy gain.</p></li><li><p><strong>van der Waals Interactions:</strong></p></li><li><p>Transient weak attractive forces operating between any two uncharged atoms in close proximity.</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/12.jpg" data-width="50$\text{ATP} + \text{H}_2\text{O} \rightleftharpoons \text{ADP} + \text{P}_i$</p></li><li><p>Specific Cleavage Reaction:$\text{R-O-PO}_2^-\text{-O-PO}_3^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{R-O-PO}_3^{2-} + \text{HO-PO}_3^{2-}$</p></li><li><p>Standard Free Energy Change:$\Delta G^{\circ\prime} = -30\,\text{kJ\,mol}^{-1}$</p></li><li><p>Energetic mechanism: Breaking covalent bonds intrinsically requires energy input. Hydrolysis releases energy overall because the product molecules (Adenosine Diphosphate and inorganic phosphate) form new bonds that exhibit significantly higher thermodynamic stability than the initial reactants. Stabilization is achieved via resonance delocalization of charge, reduced electrostatic repulsion among phosphate groups, and enhanced solvation of the reaction products by water.</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/13.jpg" data-width="50$\text{H}^+$) and a hydroxide ion ($\text{OH}^-$):$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$</p></li><li><p>Equilibrium constant expression:$K_{\text{eq}} = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]} = 1.8 \times 10^{-16}\,\text{M}$</p></li><li><p>Concentration of pure water:$[\text{H}_2\text{O}] = \frac{1000\,\text{g/L}}{18.015\,\text{g/mol}} = 55.5\,\text{M}$</p></li><li><p>Ion Product Constant of Water ($K_w$):$K_w = K_{\text{eq}} \times [\text{H}_2\text{O}] = (1.8 \times 10^{-16}\,\text{M}) \times (55.5\,\text{M}) = 1.0 \times 10^{-14}\,\text{M}^2 \quad (\text{at } 25\,^{\circ}\text{C})$</p></li><li><p>In pure neutral water:$[\text{H}^+] = [\text{OH}^-] = \sqrt{K_w} = 1.0 \times 10^{-7}\,\text{M}$</p></li></ul></li><li><p>Quantitative Relationship for Strong Acids and Bases:</p><ul><li><p>Since$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\,\text{M}^2$, calculating the concentration of one ion directly reveals the concentration of the other.</p></li><li><p>Example: For a$0.1\,\text{M}$solution of fully ionized$\text{HCl}$($[\text{H}^+] = 10^{-1}\,\text{M}$):$[\text{OH}^-] = \frac{1.0 \times 10^{-14}}{10^{-1}} = 10^{-13}\,\text{M}$</p></li></ul></li><li><p>Definition of the pH Scale:</p><ul><li><p>Biological hydrogen ion concentrations vary across several orders of magnitude ($1.5 \times 10^{-3}\,\text{M}$to$\sim 1 \times 10^{-8}\,\text{M}$).</p></li><li><p>Mathematical definition of pH:$\text{pH} = -\log_{10}[\text{H}^+]$</p></li><li><p>The symbol "p" represents the negative common logarithm ($-\log_{10}$) of a given value.</p></li><li><p>Neutral water at$25\,^{\circ}\text{C}$:$\text{pH} = -\log_{10}(1.0 \times 10^{-7}) = 7.0$</p></li></ul></li></ul><h3 id="8a51cfe9-452a-4963-ab74-31448b379f36" data-toc-id="8a51cfe9-452a-4963-ab74-31448b379f36" collapsed="false" seolevelmigrated="true">Weak Acid-Base Equilibria and the Henderson-Hasselbalch Equation</h3><ul><li><p>Definitions of Acids and Bases:</p><ul><li><p><strong>Acid:</strong> Proton donor (Brønsted-Lowry) or electron pair acceptor (Lewis).</p></li><li><p><strong>Base:</strong> Proton acceptor (Brønsted-Lowry) or electron pair donor (Lewis).</p></li></ul></li><li><p>Behavior of Weak Acids vs. Strong Acids:</p><ul><li><p>Strong acids ionize completely upon dissolution in water.</p></li><li><p>Weak acids ionize only partially, establishing an equilibrium between the weak acid ($\text{HA}$) and its conjugate base ($\text{A}^-$):$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$</p></li><li><p>$\text{HA}$and$\text{A}^-$form a conjugate acid-base pair. Stronger acids possess larger dissociation constants ($K_a$) and greater proton-releasing tendencies.</p></li></ul></li><li><p>Acid Dissociation Constant ($K_a$) and$\text{p}K_a$:</p><ul><li><p>Acid dissociation equilibrium constant:$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$</p></li><li><p>Logarithmic expression:$\text{p}K_a = -\log_{10}(K_a)$</p></li></ul></li><li><p>Complete Derivation of the Henderson-Hasselbalch Equation:</p><ol><li><p>$\displaystyle K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$</p></li><li><p>$\displaystyle [\text{H}^+] = K_a \cdot \frac{[\text{HA}]}{[\text{A}^+]}$</p></li><li><p>$\displaystyle -\log_{10}[\text{H}^+] = -\log_{10}(K_a) - \log_{10}\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)$</p></li><li><p>$\displaystyle \text{pH} = \text{p}K_a - \log_{10}\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)$</p></li><li><p>Inverting the log term yields the standard equation:$\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$</p></li></ol></li><li><p>Dissociation Constants ($K_a$) and$\text{p}K_a$Values of Representative Acids (at$25\,^{\circ}\text{C}$):</p><ul><li><p><strong>Acetic Acid</strong> ($\text{CH}_3\text{COOH}$):</p></li><li><p>$K_a = 1.74 \times 10^{-5}\,\text{M}$</p></li><li><p>$\text{p}K_a = 4.76$</p></li><li><p><strong>Phosphoric Acid</strong> ($\text{H}_3\text{PO}_4$):</p></li><li><p>$K_a = 7.25 \times 10^{-3}\,\text{M}$</p></li><li><p>$\text{p}K_a = 2.14$</p></li><li><p><strong>Dihydrogen Phosphate</strong> ($\text{H}_2\text{PO}_4^-$):</p></li><li><p>$K_a = 1.38 \times 10^{-7}\,\text{M}$</p></li><li><p>$\text{p}K_a = 6.86$</p></li><li><p><strong>Monohydrogen Phosphate</strong> ($\text{HPO}_4^{2-}$):</p></li><li><p>$K_a = 3.98 \times 10^{-13}\,\text{M}$</p></li><li><p>$\text{p}K_a = 12.4$</p></li><li><p><strong>Carbonic Acid</strong> ($\text{H}_2\text{CO}_3$):</p></li><li><p>$K_a = 1.7 \times 10^{-4}\,\text{M}$</p></li><li><p>$\text{p}K_a = 3.77$</p></li><li><p><strong>Bicarbonate</strong> ($\text{HCO}_3^-$):</p></li><li><p>$K_a = 6.31 \times 10^{-11}\,\text{M}$</p></li><li><p>$\text{p}K_a = 10.2$</p></li><li><p><strong>Ammonium</strong> ($\text{NH}_4^+$):</p></li><li><p>$K_a = 5.62 \times 10^{-10}\,\text{M}$</p></li><li><p>$\text{p}K_a = 9.25$</p></li></ul></li></ul><h3 id="c727db62-1af7-4358-a59a-4a9fbb86dc11" data-toc-id="c727db62-1af7-4358-a59a-4a9fbb86dc11" collapsed="false" seolevelmigrated="true">Titration Curves and Buffering Principles</h3><ul><li><p>Titration Curve Analysis (Acetic Acid Titration):</p><ul><li><p>Starting Point ($0$equivalents$\text{OH}^-$): All solute exists as$\text{CH}_3\text{COOH}$.</p></li><li><p>Midpoint ($0.5$equivalents$\text{OH}^-$added /$50\%$titrated):</p></li><li><p>$[\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]$</p></li><li><p>$\text{pH} = \text{p}K_a + \log_{10}(1) = \text{p}K_a = 4.76$</p></li><li><p>Buffering Region: Defined as the plateau centered around the midpoint extending$\pm 1.0$pH unit relative to the$\text{p}K_a$. For acetic acid, the effective buffering region spans from$\text{pH } 3.76$to$\text{pH } 5.76$</p></li><li><p>Endpoint ($1.0$equivalent$\text{OH}^-$added /$100\%$titrated):$\text{CH}_3\text{COOH}$is fully deprotonated to$\text{CH}_3\text{COO}^-$.</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/15.jpg" data-width="50$0.1\,\text{M}$acetic acid ($\text{HA}$) and$0.1\,\text{M}$sodium acetate ($\text{A}^-$).$\text{pH} = 4.76 + \log_{10}\left(\frac{0.1}{0.1}\right) = 4.76 + 0 = 4.76$</p></li><li><p>Addition of Strong Acid: An equal volume of$0.05\,\text{M}\,\text{HCl}$is added.</p></li><li><p>Because total volume doubles, initial concentrations halve before chemical reaction:$[\text{HA}] = 0.05\,\text{M}$,$[\text{A}^-] = 0.05\,\text{M}$, and added$[\text{H}^+] = 0.025\,\text{M}$.</p></li><li><p>Added$\text{H}^+$reacts quantitatively with$\text{A}^-$to form$\text{HA}$.</p></li><li><p>New Acid Concentration:$[\text{HA}] = 0.05 + 0.025 = 0.075\,\text{M}$</p></li><li><p>New Base Concentration:$[\text{A}^-] = 0.05 - 0.025 = 0.025\,\text{M}$</p></li><li><p>Recalculated pH:$\text{pH} = 4.76 + \log_{10}\left(\frac{0.025}{0.075}\right) = 4.76 + \log_{10}\left(\frac{1}{3}\right) = 4.76 - 0.48 = 4.28$</p></li><li><p>Conclusion: Addition of strong acid produces only a minor drop in pH (from$4.76$to$4.28$), demonstrating effective buffering capacity.</p></li></ul></li></ul><h3 id="015df987-dba0-4293-85ea-5b63768ab0d5" data-toc-id="015df987-dba0-4293-85ea-5b63768ab0d5" collapsed="false" seolevelmigrated="true">Biological Buffer Systems</h3><ul><li><p>Biological Necessity of Buffering:</p><ul><li><p>Intracellular and extracellular environments must strictly regulate pH because enzyme structure and metabolic activity depend on protonation states.</p></li><li><p>Human blood plasma pH is held constant at$\sim 7.4$</p></li></ul></li><li><p>Phosphate Buffer System:</p><ul><li><p>Operates via the dihydrogen phosphate / monohydrogen phosphate equilibrium:$\text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-}$</p></li><li><p>$\text{p}K_a = 6.86$</p></li><li><p>Effective buffering range:$\text{pH } 5.86$to$\text{pH } 7.86$($\text{p}K_a \pm 1$</p></li><li><p>Serves as a major physiological buffer system within intracellular cytoplasm.</p></li></ul></li><li><p>Bicarbonate Buffer System:</p><ul><li><p>Primary buffer system in blood plasma, involving an open equilibrium between capillary blood aqueous phase and alveolar gas phase.</p></li><li><p>Linked Reversibility Steps:</p></li><li><p><strong>Reaction 1 (Aqueous Dissociation):</strong>$\text{H}^+ + \text{HCO}_3^- \rightleftharpoons \text{H}_2\text{CO}_3 \quad (\text{p}K_a = 3.77)$</p></li><li><p><strong>Reaction 2 (Aqueous Hydration/Dehydration):</strong>$\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}_2\text{O} + \text{CO}_2(d)$</p></li><li><p><strong>Reaction 3 (Gas-Liquid Phase Exchange):</strong>$\text{CO}_2(d) \rightleftharpoons \text{CO}_2(g)$</p></li><li><p>Complete System Equilibrium:$\text{CO}_2(g) \rightleftharpoons \text{CO}_2(d) + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$</p></li></ul></li></ul><img src="https://assets.knowt.com/pdf-flow-prod/6d508e21-25df-4bfe-b7cf-94ce010ae75b-figures/16.jpg" data-width="50$\text{density of liquid water} > \text{density of ice}$) due to liquid water forming an average of$3.4$dynamic hydrogen bonds per molecule compared to the open$4.0$hydrogen-bonded crystalline lattice of ice. When ice melts, the volume decreases, causing the water level to drop below the rim.</p></li></ul></li><li><p><strong>Question 2:</strong> What is the pH of a$0.1\,\text{M}$($10^{-1}\,\text{M}$)$\text{HCl}$solution?</p><ul><li><p>Options:     A. 0.1     B. 1     C. 10</p></li><li><p><strong>Answer:</strong> B. 1</p></li><li><p><strong>Explanation:</strong> Hydrochloric acid ($\text{HCl}$) is a strong acid that ionizes completely in water, producing$[\text{H}^+] = 0.1\,\text{M} = 10^{-1}\,\text{M}$. Applying the pH formula:$\text{pH} = -\log_{10}(10^{-1}) = 1$.</p></li></ul></li><li><p><strong>Question 3:</strong> Breaking chemical bonds requires an input of energy. Why does ATP hydrolysis release energy ($\Delta G^{\circ\prime} = -30\,\text{kJ\,mol}^{-1}$$)?

  • Answer & Explanation: While breaking the phosphoanhydride bond requires energy input, the overall reaction is exergonic because the newly formed bonds in the products (ADP and inorganic phosphate) are significantly more stable than the reactants. Product stabilization is achieved through charge resonance delocalization, decreased electrostatic repulsion among negative oxygen charges, and higher hydration energy of the resulting products.