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CHAPTER 2 Water and Aqueous Solutions Learning goals: • • • • • • • What kind of interactions occur between molecules Why water is a good medium for life Why nonpolar moieties aggregate in water How dissolved molecules alter properties of water How weak acids and bases behave in water How buffers work and why we need them How water participates in biochemical reactions Structure of the Water Molecule • Octet rule dictates that there are four electron pairs around an oxygen atom in water • These electrons are in four sp3 orbitals • Two of these pairs covalently link two hydrogen atoms to a central oxygen atom • The two remaining pairs remain nonbonding (lone pairs) • Water geometry is a distorted tetrahedron • The electronegativity of the oxygen atom induces a net dipole moment • Because of the dipole moment, water can serve as both a hydrogen bond donor and acceptor Hydrogen Bonds • Strong dipole-dipole interaction that arises between an acid (proton donor) and a base (proton acceptor) • Typically 4–6 kJ/mol for bonds with neutral atoms, and 6–10 kJ/mol for bonds with one charged atom • Can only form between H and N, O, F • Hydrogen bonds are strongest when the bonded molecules are oriented to maximize electrostatic interaction • Ideally the three atoms involved are in a line Hydrogen Bonding in Water • Water can serve as both – an H donor – an H acceptor • Up to four H-bonds per water molecule gives water its – anomalously high boiling point – anomalously high melting point – unusually large surface tension • Hydrogen bonding in water is cooperative • Hydrogen bonds between neighboring molecules are weak (20 kJ/mol) relative to the H–O covalent bonds (420 kJ/mol) Water as a Solvent • Water is a good solvent for charged and polar substances – amino acids and peptides – small alcohols – carbohydrates • Water is a poor solvent for nonpolar substances – nonpolar gases – aromatic moieties – aliphatic chains Why is this? The interaction energies and entropies are too low Water dissolves many salts • High dielectric constant reduces attraction between oppositely charged ions in salt crystal; almost no attraction at large (> 40 nm) distances • Essentially, the water “shorts out” the interaction b/w two ions • Strong electrostatic interactions between the solvated ions and water molecules lower the energy of the system • Entropy increases as ordered crystal lattice is dissolved Backup: Intermolecular Forces Four Basic Types of Non Covalent Interactions 1.Ion-Dipole (z1m2/r3) 2.Dipole-Dipole (m1m2/r3) 3.Dipole-Induced dipole (m1a2/r6) 4.London Forces (a1a2/r6) The strength of the interactions in each force is inversely proportional to the distance between the interacting species (That’s the 1/r term next to each name) Coulomb’s Law and Interacting Species The force between two interacting ions is equal to the magnitude of the charges divided by the square of the distance between them. kq1q2 F= 2 Dr k = proportionality constant = 8.99x109 JmC-2 D is the dielectric constant The dielectric constant is a measure of the ability of the solvent or medium containing the charges to keep them apart (D for a vacuum is 1, D for water is ~80, D for a brick wall is ∞) Interion and Intermolecular Forces Physics of Noncovalent Interactions Noncovalent interactions do not involve sharing a pair of electrons. Based on their physical origin, one can distinguish between: • Ionic (Coulombic) Interactions (Ionic and Ion/Dipole) – Electrostatic interactions between permanently charged species, or between the ion and a permanent dipole • Dipole Interactions (Dipole/Dipole and Dipole/Ind. Dipole) – Electrostatic interactions between uncharged, but polar molecules • van der Waals Interactions (London Forces) – Weak interactions between all atoms, regardless of polarity – Attractive (dispersion) and repulsive (steric) component • Hydrophobic Effect (Rationale behind a lot of Biochemistry) – Complex phenomenon associated with the ordering of water molecules around nonpolar substances Examples of Noncovalent Interactions H-bonds are a special type of dipole/dipole interaction Between amino acids = Salt Bridges “Hydrophobic Exclusions” Hydrophobic, Dipole/Induced Dipole and London Force Interactions Combined make van der Waals Importance of Hydrogen Bonds • • • • • • • Source of unique properties of water Structure and function of proteins Structure and function of DNA Structure and function of polysaccharides Binding of substrates to enzymes Binding of hormones to receptors Matching of mRNA and tRNA “I believe that as the methods of structural chemistry are further applied to physiological problems, it will be found that the significance of the hydrogen bond for physiology is greater than that of any other single structural feature.” –Linus Pauling, The Nature of the Chemical Bond, 1939 Hydrogen Bonds: Examples Biological Relevance of Hydrogen Bonds van der Waals Interactions • van der Waals interactions have two components: – Attractive force (London dispersion) depends on the polarizability – Repulsive force (Steric repulsion) depends on the size of atoms • Attraction dominates at longer distances (typically 0.4–0.7 nm) • Repulsion dominates at very short distances • There is a minimum energy distance (van der Waals contact distance) Origin of the London Dispersion Force • Quantum mechanical origin • Instantaneous polarization by fluctuating charge distributions • Universal and always attractive • Stronger in polarizable molecules • Important only at a short range Biochemical Significance of van der Waals Interactions • Weak individually – easily broken, reversible • Universal – occur between any two atoms that are near each other • Importance – determines steric complementarity – stabilizes biological macromolecules (stacking in DNA) – facilitates binding of polarizable ligands The Hydrophobic Effect • Refers to the association or folding of nonpolar molecules in the aqueous solution • Is one of the main factors behind: – protein folding – protein-protein association – formation of lipid micelles – binding of steroid hormones to their receptors • Does not arise because of some attractive direct force between two nonpolar molecules Solubility of Polar and Nonpolar Solutes Why are nonpolar molecules poorly soluble in water? Low solubility of hydrophobic solutes can be explained by entropy • Bulk water has little order: – high entropy • Water near a hydrophobic solute is highly ordered: – low entropy Low entropy is thermodynamically unfavorable, thus hydrophobic solutes have low solubility. Remember to consider the Entropy of the Solute AND the Entropy of the Solvent since both are part of the System •Seemingly unfavorable or unlikely Solute behaviour can often be explained when considering the changes in the Solvent behaviour. Water surrounding nonpolar solutes has lower entropy Clathrate Cages Origin of the Hydrophobic Effect (1) • Consider amphipathic lipids in water • Lipid molecules disperse in the solution; nonpolar tail of each lipid molecule is surrounded by ordered water molecules • Entropy of the system decreases • System is now in an unfavorable state Origin of the Hydrophobic Effect (2) • Nonpolar portions of the amphipathic molecule aggregate so that fewer water molecules are ordered • The released water molecules will be more random and the entropy increases • All nonpolar groups are sequestered from water, and the released water molecules increase the entropy further • Only polar “head groups” are exposed and make energetically favorable H-bonds Hydrophobic effect favors ligand binding • Binding sites in enzymes and receptors are often hydrophobic • Such sites can bind hydrophobic substrates and ligands such as steroid hormones • Many drugs are designed to take advantage of the hydrophobic effect Ionization of Water H+ + OHH2O • O-H bonds are polar and can dissociate heterolytically • Products are a proton (H+) and a hydroxide ion (OH–) • Dissociation of water is a rapid reversible process • Most water molecules remain un-ionized, thus pure water has very low electrical conductivity (resistance: 18 M•cm) • The equilibrium is strongly to the left • Extent of dissociation depends on the temperature Proton Hydration • Protons do not exist free in solution. • They are immediately hydrated to form hydronium (oxonium) ions. • A hydronium ion is a water molecule with a proton associated with one of the non-bonding electron pairs. • Hydronium ions are solvated by nearby water molecules. • The covalent and hydrogen bonds are interchangeable. This allows for an extremely fast mobility of protons in water via “proton hopping.” Proton Hopping Water bound to proteins is essential for their function Water bound to proteins is essential for their function Water bound to proteins is essential for their function Ionization of Water: Quantitative Treatment Concentrations of participating species in an equilibrium process are not independent but are related via the equilibrium constant: +]•[OH-] [H H+ + OHH2O Keq = ———— [H2O] Keq can be determined experimentally, it is 1.8•10–16 M at 25C. [H2O] can be determined from water density, it is 55.5 M. • Ionic product of water: K w K eq [H 2 O] [H ][ OH - ] 1 10 14 M 2 •In pure water [H+] = [OH–] = 10–7 M What is pH? pH = -log[H+] • pH is defined as the negative logarithm of the hydrogen ion concentration • Simplifies equations K w [H ][OH- ] 11014 M 2 • The pH and pOH must always add to 14 log[ H ] log[ OH- ] 14 • In neutral solution, [H+] = [OH–] and the pH is 7 pH pOH 14 • pH can be negative ([H+] = 6 M) pH scale is logarithmic: 1 unit = 10-fold Dissociation of Weak Electrolytes: Principle O H3C + H2O O Keq H3C OH + O- H3O+ K a Keq [H 2O] [H ][CH 3COO - ] Ka 1.74 10 5 M [CH 3COOH] [H ] Ka [CH 3COOH ] [CH 3COO ] • Weak electrolytes dissociate only partially in water. • Extent of dissociation is determined by the acid dissociation constant Ka. • We can calculate the pH if the Ka is known. But some algebra is needed! Dissociation of Weak Electrolytes: Example What is the final pH of a solution when 0.1 moles of acetic acid is added to water to a final volume of 1L? The Ka of acetic acid is 1.74x10-5 O O Ka • We assume that the + H3C H H3C + only source of H+ is OH Othe weak acid 0.1 – x x x Ka [x][ x] 1.74 10 5 M [0.1 - x] x 2 1.74 10 6 1.74 10 5 x x 2 1.74 10 5 x 1.74 10 6 0 x = 0.001310, pH = 2.883 • To find the [H+], a quadratic equation must be solved Dissociation of Weak Electrolytes: Simplification O H3C Ka O + H+ H3C O- OH 0.1 – x 0.1 Ka x x x x [x][ x] 1.74 10 5 M [0.1] x 2 1.74 10 6 • The equation can be simplified if the amount of dissociated species is much less than the amount of undissociated acid • Approximation works for sufficiently weak acids and bases • Check that x < [total acid] x = 0.00132, pH = 2.880 Nearly identical to the answer determined by the quadratic formula! pKa measures acidity • pKa = –log Ka (strong acid large Ka small pKa) Buffers are mixtures of weak acids and their anions (conjugate base) • Buffers resist change in pH • At pH = pKa, there is a 50:50 mixture of acid and anion forms of the compound • Buffering capacity of acid/anion system is greatest at pH = pKa • Buffering capacity is lost when the pH differs from pKa by more than 1 pH unit Acetic Acid-Acetate as a Buffer System Weak acids have different pKas Henderson–Hasselbalch Equation: How to Find the pH of a weak acid during its titration For a weak acid: HA H+ + A- and, therefore [HA] [H ] K a [A - ] + [H ][ A - ] Ka [HA] We want to know the pH, so solve for [H+] [HA] - log[H ] -logK a log [A-] - [A ] pH pK a log [HA] How to solve any weak acid/base problem 1) Identify the Acid and the Base and the Ionization Reaction 2) If it is a titration: (Eg: Find the pH at each of the following points in the titration of 25 mL of 0.3 M Acetic acid with 0.3 M NaOH. The Ka value is 1.74x10-5: The initial pH, After adding 10 mL of 0.3 M NaOH, After adding 12.50 mL of 0.3 M NaOH, After adding 25 mL of 0.3 M NaOH, After adding 26 mL of 0.3 M NaOH) • Calculate how much of your starting material was reacted and subtract that from the starting number of moles of acid or base • Determine the new total volume and calculate the molarity of the acid or base • Use the Ka equation or the Henderson-Hasselbach equation to determine the molarity of the remaining species in the beaker or the pH 3) If it is not a titration (Eg: What is the concentration of acetate in 500 mL of a 0.05M Acetate buffer at pH 5.5?) • Determine how many moles of acid or base you have • Use the Ka equation or the Henderson-Hasselbach equation to determine the molarity of the remaining species in the beaker or the pH Biological Buffer Systems • Maintenance of intracellular pH is vital to all cells – Enzyme-catalyzed reactions have optimal pH – Solubility of polar molecules depends on H-bond donors and acceptors – Equilibrium between CO2 gas and dissolved HCO3– depends on pH • Buffer systems in vivo are mainly based on – phosphate, concentration in millimolar range – bicarbonate, important for blood plasma – histidine, efficient buffer at neutral pH • Buffer systems in vitro are often based on sulfonic acids of cyclic amines – HEPES – PIPES – CHES HO N N SO3Na Chapter 2: Summary In this chapter, we learned about: • • • • The nature of intermolecular forces The properties and structure of liquid water The behavior of weak acids and bases in water The way water can participate in biochemical reactions