Bioc220 2 Water PDF

2. The Cell Environment Lehininger Chapter 2: Water, the solvent of life Weak Interactions in Aqueous (water-based) Systems Cellular Equilibria – Environmental Control Ionization of Water, Weak Acids, and Weak Bases  Weak Interactions in Aqueous (water-based) Systems (1) Why Study Water: -...

2. The Cell Environment Lehininger Chapter 2: Water, the solvent of life Weak Interactions in Aqueous (water-based) Systems Cellular Equilibria – Environmental Control Ionization of Water, Weak Acids, and Weak Bases  Weak Interactions in Aqueous (water-based) Systems (1) Why Study Water: - Life exists in an aqueous environment - Intra- and extra-cellular fluids very similar to sea water and the primordial soup (prebiotic soup) The primordial soup is a generic term that describes the aqueous solution of organic compounds that accumulated in primitive water bodies of the early Earth as a result of endogenous abiotic syntheses and the extraterrestrial delivery by cometary and meteoritic collisions, and from which some have assumed that the first living systems evolved. (2) Why Life is based on Water, not other liquid? abundant highly polarized, high dielectric constant  good solvent for polar mol or salts, zwitterion amino acids, the exterior of globular proteins  H-bonding liquid over a large range of T  optimal for Carbon chemistry nonpolar mol not soluble  Hydrophobic effects  biomembranes, protein folding excellent thermal properties (high heat capacity, high heat of evaporation, etc) Electronegativity: the tendency for an atom to attract shared electrons when forming a covalent bond.  Polar covalent bond between atoms with different electronegativities. (3) Characteristics of Water: Why water is all that special?  The water molecule is bent.  O (3.5) is more electronegative than H (2.2). H2O is highly polar with a net dipole.   O O H H  H  H  104.5oC Figure 2-1a  H2O interacts through hydrogen bonds. b.p. (oC) m.p. Water 100 0 MeOH 65 ‒98 BuOH 117 Butane –0.5 1 Å (Angstrom) = 10‒8 cm = 0.1 nm (Figure 2-1b) Hydrogen bond: A type of dipole force. An electrostatic attractive force between molecules. Found when a H atom is bonded to N, O or F. Weaker than covalent bonds (23 kJ/mol in water). C—C: 350 kJ/mol, O—H: 470 kJ/mol. Can a H atom bonded to C form a H-bond? Yes, very weak. Not considered in most Intro Biochem. C (2.5) is only slightly more electronegative than H (2.2).  Hydrogen bonding in ice. Each H2O forms 4 H-bonds. In liquid water, each H2O forms, on average, 3.4 H-bonds. This is why less dense ice floats in liquid water, and why water expands when frozen. · Directionality of the H-bond: The bond strength depends not only on the distance between the H or X (N or O), but also the angle of the atoms in space. Thus, Hydrogen bonds can vary in strength from very weak (1-2 kJ mol−1) to strong (20~30 kJ mol−1). - Solubility in water is enhanced for molecules that can form H-bonds. i.e. hydroxy, keto, carboxy, amino groups, and these are termed hydrophilic (water loving) groups  Common H-bonds in biological systems: (Figure 2-3) - Solubility is reduced for molecules that can not form H- bonds (i.e. nonpolar hydrocarbons such as alkanes, alkenes), and these are termed hydrophobic (water fearing) compounds. - Note: C is only slightly more electronegative than H Polar Nonpolar Amphipathic phospholipids (4) Aqueous (Water-based) Solutions Water is a polar solvent. · Solvation of crystalline salt: Water breaks up the salt crystal lattice by H-bonding with the individual ions. hydration High dielectric constant () of water makes ionic interactions weaker. (F= Q1Q2/ r2) · Entropy-driven: G < 0  H ≥ 0, S ≫ 0 (of ions) (Figure 2-6) · Nonpolar gases (O2, CO2) and nonpolar compounds (benzene, hexane) are not soluble in water. H > 0 breaking up the water H-bonds. S < 0 cage-like, highly ordered H2O molecules around the hydrophobic compounds. Overall, G > 0 (non-spontaneous; not favorable; requires E input). This is why we need a water soluble O2 carrier, hemoglobin.  · Amphipathic (Amphiphilic) compounds contain both polar (hydrophilic) & non-polar (hydrophobic) regions (ex. fatty acids) In aqueous solution: water hydrates the hydrophilic portion but excludes the hydrophobic region to give micelles and lipid bilayers This process reduces entropy (S) or increases the order of the lipid molecules, but in doing so increases the entropy of surrounding water with a net entropic increase (the basis of cell membranes) Amphipathic compounds in aqueous solution form structures that increase entropy (Figure 2-7)  Not favorable G > 0 G = H  TS Favorable G < 0 Hydrophobic effects (5) Hydrophobic Effects (Read the addition pdf on URCourses) The tendency of hydrophobic (lipophilic) groups to form intermolecular aggregates in an aqueous medium, and analogous intramolecular interactions. Often called hydrophobic interactions, which is misleading. No repulsion between water and hydrophobic compounds, per se, but effects of the hydrophobic groups on the water- water interactions are the driving force for the hydrophobic interactions. Hydrophobic effect is driven by S ≫ 0.  “Clathrate” structure (Ordered water structure surrounding a compound.) First hydration shell of MeOH/H2O solution. either the negative end nor the positive end of the dipole ants to be pointed to the non-polar molecule, hence, he water molecules in the hydration shell tend to be oriented ith their dipole moments oriented tangentially to the non-po eutral molecule, resulting in a cage-like, or 'clathrate' structure ound the non-polar group. http://www.youtube.com/watch?v=ETMmH2trTpM Hydrophobic effects are Important in protein folding, biomembrane organization, and enzyme-substrate binding. effects  (6) Noncovalent interactions are: -individually weak, but their cumulative effects are critical in biological functions. effect (7) Water can be part of a protein’s structure Tightly bound water molecules (red spheres) in hemoglobin.  Water can play an important role in protein function. Water chain in cytochrome f. Protons move across the membrane, probably through “proton hopping.” Water molecule found at Active site of Stilbene Synthase the active site of stilbene synthase. The trapped water molecule is involved in the hydrolysis reaction catalyzed by the enzyme. Other water molecules forms H-bonds network that hold the amino acid residues in correct positions in space. (8) Water as a Reactant Hydrolysis (exergonic) are catalyzed by hydrolases. ( Condensation)  Cellular Equilibria – Environmental Control Cells require a strongly controlled environment: Sites of Control: 1. Membranous organelles (compartments with barriers to limit movement to and from) - Water moves fairly freely across the cell membrane (aquaporins) - Not all molecules can cross the membrane, which is semi-permeable - The cytoplasm contains: proteins, organic polyphosphates etc. that can not permeate the plasma membrane - The cytoplasm also contains ions (Na+, K+, Cl-, etc.) to which the plasma membrane is semi-permeable (ion channels) - Protein transporters transport molecules 2. Aqueous Properties (Osmotic pressure, pH etc.)  Ionization of Water, Weak Acids, and Weak Bases (1) Ionization of Water: Slight tendency of H2O to undergo reversible ionization to hydrogen ion (proton) and hydroxide ion: H2O ⇄ H+ + OH‒ BUT, free protons do not exist in aqueous solution: protons are hydrated to hydronium ions: H2O + H+ ⇄ H3O+ (very fast due to proton hopping) The ionization is expressed by an equilibrium constant (Keq) H2O ⇄ H+ + OH‒ [H+][OH‒] Keq = [H2O] At 25 °C, [H2O] = 55.5 M.  (55.5 M)(Keq) = [H+][OH‒] = Kw (ion product of water) Keq is 1.8 x 10‒16 M as determined by electro-conductivity exp. (Question) Calculate the pH of a neutral soln. Kw = [H+][OH‒] = [H2O]Keq = (55.5 M)(1.8 x 10‒16 M) = 1.0 x 10‒14 M2 At neutral pH, [H+] =[OH‒] and [H+][OH‒] = [H+]2  [H+] = 1.0 x 10‒7 M in pure water at 25°C. Neutral pH =  log(10‒7) = 7 Since Kw is constant, when [H+] > 10 ‒7 M (add vinegar) ⇒ [OH‒] < 10 ‒7 M (acidic) when [H+] < 10 ‒7 M (add NaOH) ⇒ [OH‒] > 10 ‒7 M (basic, alkaline) (2) The pH Scale 1 pH = log [H+] = ‒ log [H+] pH 7 is neutral, [ H+] = [OH‒] pH < 7 is acidic, [ H+] > [OH‒] pH > 7 is basic, [OH‒] > [ H+] pH + pOH = 14  The pH must be controlled in an organism; where the breakdown in pH regulation can lead to serious metabolic disturbances: Strong electrolytes (acids/bases) are 100% dissociated in solution. Some common strong acids (often a proton with inorganic halide): HCl hydrochloric acid HNO3 nitric acid H2SO4 sulfuric acid HBr, HI, HClO4, HF Some common strong bases (often hydroxide with an inorganic metal): NaOH sodium hydroxide LiOH lithium hydroxide KOH potassium hydroxide These strong acids/bases can easily make solutions of 1.0 M and are 100% dissociated at that concentration. Weak electrolytes will dissociate in solution, but they do so less than 100%. Weak acids/bases in bioc will dissociate only 1 to 5%. Common weak acids (often protons with organic weak bases): H3PO4 phosphoric acid CH3COOH acetic acid (vinegar) HCO3‒ carbonic acid pyruvic acid Common weak bases (organic molecules): Salts of polyatomic ions of weak acids … NH ammonia and all nitrogen relatives i.e. CH NH  What is the pH of 0.1 M NaOH solution ? NaOH is a strong base and completely ionized in dilute aq. solution. \[OH‒] = 0.1 M. From Kw = [H+][OH‒] = 1.0  10‒14 M2, [H+] = 10‒14 M2/0.1 M = 10‒13 M.  pH = ‒log10‒13 = 13. Or, from [OH‒] = 0.1 M, pOH = ‒log 10‒1 = 1. Since pH + pOH = 14, pH of the solution is 13. What is [H+] in a solution with [OH‒] of 1.3 x 10‒4 M? [H+] = Kw/[OH‒] = 10‒14 M2/1.3  10‒4 M = 0.769....  10‒10 M = 7.7  10‒11 M (Notice the number of significant figures.) pH meter: History and How it works https://www.acs.org/content/acs/en/education/whatischemistry/ landmarks/beckman.html (3) Weak Acids and Weak Bases · Acids – proton (H+) donors; Bases – proton acceptors When acid is added to water, it dissociates to the hydronium ion (H3O+) and a conjugate base (A‒): Keq HA + H2O ⇄ H3O+ + A‒ (A‒ : conjugate base) The strength of an acid is determined by the degree of dissociation; large – strong; small – weak; as represented: Keq = [H3O+][A‒] [HA][H2O] In dilute solutions, [H2O] is essentially constant (55.5 M), and [H+] [A‒] · Ka (acid ionization constant) = [HA] Ka values are small, and cumbersome. Often transformed to pKa values (similar to pH): pKa = ‒ log Ka Fig 2-15 (4) The Henderson-Hasselbalch (H-H) Equation Q) If I mix an acid and a base at certain amt (1:1?), what pH will the resulting soln be at? Q) If I add a weak acid (pKa = 5) to a soln of pH = 7, dissociation%? How much in its acid form, and how much in its base form? The pH of a solution, and the concentration of an acid and its conjugate base are related by the H-H equation: [A‒] pH = pKa + log [HA] When the molar concentration of an acid (HA) and its conjugate base (A‒) are equal ([A‒] = [HA]), [A‒]/[HA] = 1; and log[A‒]/[HA] = log1 = 0 So the pH of the solution simply equals the pKa of the acid. When [A‒] > [HA], pH > pKa. When [A‒] < [HA], pH < pKa.  Derivation of the Henderson-Hasselbalch (H-H) Equation Consider the disassociation of a weak acid in water Ka = [H+][A‒]; rearrange: Ka = [A‒] [HA] [H+] [HA] Take the log of both sides log Ka = log [A‒] [H+] [HA] log Ka – log [H+] = log [A‒] – log [HA]; rearrange – log [H+] = – log Ka + log [A‒] ; recall, p scale is the negative log! [HA] pH = pKa + log [A‒] [HA] [proton acceptor] pH = pKa + log [proton donor] Calculate the pH of a 2 L solution containing 10 mL of 5 M acetic acid (CH3COOH) and 10 mL of 1 M sodium acetate (CH3COONa). pKa of CH3COOH = 4.76. 2. Calculate [HA] = cHA x VHA / VT = (5 M)(0.01 L)/2 L = 0.025 M 3. Calculate [A‒] = cA- x VA- / VT = (1 M)(0.01 L)/2 L = 0.005 M 4. Look up pKa for acetic acid; pKa = 4.76 1. Which equation relates the information given to that asked? pH = pKa + log [A‒] = 4.76 + log (0.005 M)/(0.025 M) [HA] = 4.76 – 0.70 = 4.06 = 4.1 (round all pH values to 1 decimal Therefore the pH of the solution is 4.1  (5) Titration Curve Reveals the pKa of Weak Acids - How do pH values of an acetic acid solution vary with added [OH‒]? ® a titration curve - Constructed by: a) experiment b) H-H equation midpoint pH of titration = pKa of corresponding acid: b/c pH = pKa when [HA] = [A‒] * Slope lower near midpoint When [HA]  [A‒], pH is relatively insensitive to addition of strong acid or base i.e. buffered solution Buffering capacity is maximal when pH = pKa. The useful range of a buffer is within one pH unit of its pKa. Above or below this, the pH will change rapidly. (6) Buffers Buffer: A system whose pH changes only slightly when small amounts of acid or base is added. A buffer ordinarily consists of a weak acid and its conjugate base, present in roughly equal amounts (at pH = pKa of the acid) Used to control the pH within a system Take an example of pure water: The addition of a 0.01 mL drop of 1 M HCl to 1 L of pure water will change the pH from 7 to 5! If we add a small concentration (i.e. 10 mM) of buffer to the water, the addition of acid will cause virtually no change in the pH, even with much larger amounts of acid added … “Buffers are weak acids or bases that react with acid or base to create their conjugate base or acid, respectively, such that even relatively large additions of an acid or base do not lead to large changes in pH.” Biological systems use the same strategy to control pH!  The pH must be controlled in an organism; where the breakdown in pH regulation can lead to serious metabolic disturbances: The pH of blood is normally kept within 7.35~7.45. Outside the narrow range, the organism can not function. The pH of the cytosol of most cells is ~ 7.4, however, in the lysosomal organelles the pH is ~ 5.0. This is the pH at which the degradative enzymes (proteases) of the lysosome function best, and they are actually inactive at cytosolic pH! Blood acidosis (metabolic, respiratory) leads to blood acidemia, which can lead to mental confusion, fatique/lethargy, breathing difficulty, shock or death. Blood alkalosis (disease-related or hyperventilation) which can cause muscular weakness, myalgia, and muscle cramps (disturbed function of the skeletal muscles), and muscle spasms (disturbed function of smooth muscles). How buffer works? (Fig 2-18) Buffer works because the added H+ or OH‒ ions are consumed and do not directly affect the pH. HAc + OH‒ ⇄ H2O + Ac‒ Ac‒ + H+ ⇄ HAc [buffer] >> added [H+] or [OH‒] (7) Polyprotic Acids: Substances that have more than one acid/base group. H3PO4 ⇄ H2PO4‒ + H+ ⇄ HPO42‒ + H+ ⇄ PO43‒ + H+ pKa1 = 2.14 pKa2 = 6.86 pKa3 = 12.4 Example: 1.00 mole of phosphoric acid (H3PO4) and 1.75 moles of NaOH are added to 1 L of water. Calculate the pH. Step 1: 1 mol of H3PO4 + 1.75 mol OH- 1 mol H2PO4- + 1 mol H2O + 0.75 mol OH- Step 2: 1 mol of H2PO4- + 0.75 mol OH- 0.25 mol H2PO4- + 0.75 mol HPO42- + 0.75 mol H2O Step 3: In the end, we have 0.25 moles of H2PO4- and 0.75 moles of HPO42-, we can calculate the pH using the H-H equation: Step 4: Look up the pKa of the reaction: pKa for H2PO4- ⇄ HPO42- + H+, is 6.86 Step 5: Calculate [HA] = [H2PO4-]: 0.25 mol/1 L = 0.25 M Step 6: calculate [A-] = [HPO42-] : 0.75 mol/1 L = 0.75 M Therefore: pH = pKa+ log[A-]/[HA] = 6.86 + log((0.75 M)/(0.25 M)) = 7.34 Practice Questions 1. At 37oC, ion product for water (Kw) has a value of 2.4 x 10-14. What is the pH of pure water at 37oC? (6.81) 2. Hydroxide ions were released during an enzyme reaction performed at pH 6.8. Circle the buffer you would select for the enzyme assay. CH3COOH/CH3COO- pKa = 4.76 H2PO4-/HPO42- pKa = 7.21 HCO3-/CO32- pKa = 10.0 Example: The addition of a 0.01 mL drop of 1 M HCl to 1 L of water will change the pH from 7 to 5. A small concentration of buffer can alter this so that there is virtually no change in the pH, even with much larger amounts of acid added! Question: A buffer contains 0.010 mol of lactic acid (pKa=3.86) and 0.050 mol of sodium lactate per liter. (a) Calculate the pH of the buffer. (b) Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L of the buffer. (c) What pH change would you expect if you added the same quantity of HCl to 1 L of pure water (pH=7). Work on the textbook problems at the end of Chapter 2. (8) Biological buffers: HCO3‒ : H2CO3 pKa = 6.35 HPO42‒ : H2PO4‒ pKa = 6.86 pKa of amino acid, histidine = 6.0 Common buffers used in lab: Buffer pKa (25oC) Effective pH Range succinate (pK1) 4.21 3.2-5.2 acetate 4.76 3.6-5.6 citrate (pK2) 4.76 3.0-6.2 malate (pK2) 5.13 4.0-6.0 succinate (pK2) 5.64 5.5-6.5 MES 6.10 5.5-6.7 carbonate (pK1) 6.35 6.0-8.0 citrate (pK3) 6.40 5.5-7.2 imidazole 6.95 6.2-7.8 MOPS 7.14 6.5-7.9 phosphate (pK2) 7.20 5.8-8.0 HEPES 7.48 6.8-8.2 Trizma (Tris) 8.06 7.5-9.0 glycine (pK2) 9.78 8.8-10.6 Bicarbonate Buffer System in Blood and Lung Note: Report pH values to no more than one decimal place. 1. At 37C ion product for water (Kw) has a value of 2.4 x 10-14 M2. What is the pH of a neutral solution of pure water at 37C? 2. Hydroxide ions were released during an enzyme reaction performed at pH 6.8. Circle the buffer you would select for the enzyme assay. CH3COOH/CH3COO- pKa = 4.8 H2PO4-/HPO42- pKa = 6.9 HCO3-/CO332- pKa = 10.0  In this lecture series we learned about: 1. The cellular environment, including properties of water, hydrogen bonding, how hydrophobic environments (like the cell membrane) are created. 2. How electronegativity relates to molecular dipoles, hydrogen bonding and acidity. 3. How cellular environmental controls relate to the law of diffusion, osmolarity, pH and buffer systems. 4. How to solve biochemical problems.  Negative pH ? YES http://en.wikipedia.org/wiki/PH Most substances have a pH in the range 0 to 14, although extremely acidic or extremely basic substances may have pH less than 0 or greater than 14. An example is acid mine runoff, with a pH = –3.6. * Slope lower near midpoint Why? When [HA] = [A‒], pH is relatively insensitive to addition of strong acid or base i.e. buffered solution Buffering power is maximal when pH = pKa. The useful range of a buffer is within one pH unit of its pKa. Above this, the pH will change rapidly. Why does the titration curve of a week acid look the way it looks? sing the H-H equation, let’s follow the change of pH as we increase the ratio of [acid]/[base]. [CH3COO] H = pKa + log [CH3COOH] CH3COO]/[CH3COOH] log([CH3COO]/[CH3COOH]) pH 1/10 log(1/10) = log10 = 1 4.76  1 = 3.76 1/4 log(1/4) = log4 = 0.602 4.76  0.602 = 4.16 1/2 log(1/2) = log2 = 0.301 4.76  0.301 = 4.46 3/4 log(3/4) = 0.125 4.76  0.125 = 4.64 1.0 log1 = 0 4.76 4/3 log(4/3) = 0.125 4.76 + 0.125 = 4.89 2/1 log2 = 0.301 4.76 + 0.301 = 5.06 4/1 log4 = 0.602 4.76 + 0.602 = 5.36 10/1 log10 = 1.0 4.76 + 1 = 5.76 nd, so on.... om the H-H eq. we know that the pH is related to the ratio of [acid] and [conjugate base]. nce the relationship is logarithmic, ) pH decreases as the ratio approaches to 1.0 (titration midpoint), and ) the titration curve is symmetrical on both sides of the midpoint. onclusion: A solution made of a weak acid and its conjugate base has its highest buffer pacity at pH equal to its pKa.