Water: The Solvent for Biochemical Reactions PDF

BIOCHEMISTRY Water: The Solvent for Biochemical Reactions Topic Outline ❖ Water and Polarity ❖ Intermolecular Forces of Attraction ❖ Structure of Water ❖ Hydrogen Bonding in Water ❖ Water as Solvent ❖ Acids, Bases, and pH ❖ Buf...

BIOCHEMISTRY Water: The Solvent for Biochemical Reactions Topic Outline ❖ Water and Polarity ❖ Intermolecular Forces of Attraction ❖ Structure of Water ❖ Hydrogen Bonding in Water ❖ Water as Solvent ❖ Acids, Bases, and pH ❖ Buffers Water and Polarity Water: The Medium for Life Organisms typically contain 70-90% water. Chemical reactions occur in aqueous medium. Water is a critical determinant of the structure and function of proteins, nucleic acids, and membranes. Water and Polarity There are two kinds of forces, or attractions, that operate in a molecule—intramolecular and intermolecular. Intramolecular Forces of Attraction: hold atoms together in a molecule; also known as chemical bond. Intermolecular Forces of Attraction: attractive forces between molecules or Ions. Water and Polarity Polarity: Separation of electric charge leading to a molecule having an electric dipole moment. Electronegativity – is a measure of the tendency of an atom to attract a bonding pair of electrons. Water and Polarity Polar: uneven distribution of electrons. Examples are ammonia and water. Nitrogen atom is more electronegative than Hydrogen atom. Most of the electrons were attracted to N atom causing an uneven distribution of electron density. Water and Polarity Non-polar: equal sharing of electrons. Examples are O2, CO2, CH4. O2 is composed of the same atom which have the same Carbon dioxide has polar methane is a nonpolar molecule electronegativity covalent C=O bond because O is because its regular tetrahedron more electronegative than C shape leads to a symmetrical atom. However, both sides of distribution of electron density the molecule have C=O bond causing the cancellation of dipole moment. Water and Polarity Polar or Non-polar? Ethane Water Lauric acid Water and Polarity Ethane is non-polar Water is a polar molecule because O atom is more electronegative than H atom causing an uneven distribution of electron density Polar Lauric acid is composed of polar and Non-polar non-polar parts. Amphipathic molecule: a molecule that has both polar and non-polar parts. Intermolecular Forces of Attraction Ion Interactions: attractive forces that involve ions Ion-ion interaction: attraction that exist between two ions of opposite charges. Ex: Na+(aq) and CH3COO-(aq) Ion-dipole Interaction: the attraction between an ion and a polar molecule to each other. Ex: Na+ and H2O; Cl- and H2O Intermolecular Forces of Attraction Van der Waals Forces: attraction without ions, applicable to molecules. Dipole-dipole Interaction: attractive forces between polar molecules. Ex: H2S Hydrogen bonding: electrostatic attraction Dipole-dipole interaction between a molecule with a highly electronegative element (O, F, N) to the hydrogen atom of another neighboring molecule, special type of dipole-dipole interaction. Hydrogen bonding Ex: water, HF, ammonia Structure of Water ▪ Four electron pairs on four sp3 orbitals (distorted tetrahedron) ▪ Two pairs covalently link hydrogen atoms to a central oxygen atom. ▪ Two remaining pairs remain nonbonding (lone pairs) ▪ The electronegativity of the oxygen atom induces a net dipole moment Hydrogen Bonding in Water Water can serve as both a hydrogen bond donor and acceptor. 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 a substrates to enzymes Binding of hormones to receptors Matching of mRNA and tRNA Water as 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 Hydrophobic Effect ▪Refers to the association or folding of nonpolar molecules in the aqueous solution ▪Is one of the main factors behind Protein folding Acids, Bases, and pH The biochemical behaviour of many important compounds depends on their acid–base properties. Acid: Proton donor Base: Proton acceptor Acids, Bases, and pH Acid Strength: tendency of an acid to dissociate into a proton. Acids, Bases, and pH Acid Dissociation Constant (Ka): a quantitative measure of the strength of an acid in solution. Ka Acid Strength Acids, Bases, and pH @ Campbell, M. K., & Farrell, S. O. (2012). In Biochemistry. Australia: Brooks/Cole Cengage Learning. Acids, Bases, and pH pH: a measure of the hydrogen ion concentration of a solution. pH = -log [H+] ❖ pH is equal to the negative logarithm of the hydrogen ion concentration. ❖ The pH and pOH must always add to 14 @ 25 °C ❖ In neutral solution, [H+] =[OH-] and the pH is 7 Acids, Bases, and pH Problems: Calculating the pH of Solution 1. What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 × 10–3 M? HCl → H+ + Cl- pH = -log [H+] = -log [1.2 x 10-3 M] pH = 2.92 Acids, Bases, and pH Problems: Calculating the pH of Solution 2. Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline). pH = -log [H+] 7.3 = -log [H+] [H+] = 10-7.3 M [H+] = 5.01 x 10-8 M [H+] = [H3O+] [H3O+] = 5.01 x 10-8 M Acids, Bases, and pH Problems: Calculating the pH of Solution 3. What are the pOH and the pH of a 0.0095-M solution of barium hydroxide, Ba(OH)2 @ 25oC. Ba(OH)2 → Ba2+ + 2OH- 0.0095 M 0.019 M pOH = -log [OH-] pH + pOH = 14 pOH = -log [0.019 M] pH = 14 - pOH pOH = 1.72 pH = 14 – 1.72 pH = 12.28 Acids, Bases, and pH pH and pKa Relationship: The Henderson-Hasselbalch Equation H+ [A− ] Ka= [HA] take the common logarithm of both sides of the equation: H+ [A− ] log Ka = log [HA] [A− ] log Ka = log H+ + log [HA] [A− ] -log H + = -log Ka + log [HA] Since, pH = -log H + and pka = -logka [𝑨− ] pH = pKa + 𝐥𝐨𝐠 [𝐇𝐀] Henderson-Hasselbalch Equation Buffers Buffer Solution ❖A solution of (a) a weak acid or weak base and (b) its salt Example: 1. Acetic acid (CH3COOH) and its salt sodium acetate (CH3COONa) 2. Ammonia (NH3) and its salt ammonium chloride (NH4Cl) ❖The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Buffers Unbuffered H2O Adding HCl: HCl(aq) → H+ + Cl- pH of the solution becomes acidic Adding NaOH: NaOH(aq) → Na+ + OH- pH of the solution becomes basic Buffer Solution: HA / A- Adding HCl: HCl(aq) → H+ + Cl- If you add a strong acid (HCl) to a buffer solution, H+ + A- → HA the conjugate base (A-) in the solution consumes the hydrogen ions. Adding NaOH: NaOH(aq) → Na+ + OH- If you add a strong base (NaOH) to a buffer solution, the acid (HA) in OH- + HA → A- + H2O the solution consumes the hydroxide ions. Buffers The Henderson-Hasselbalch equation is useful for estimating the pH of a buffer solution. Henderson-Hasselbalch Equation [A− ] 𝒎𝒐𝒍 𝑨− 𝒎𝒐𝒍 𝑯𝑨 pH = pKa + log Since, [𝑨− ]=𝑽𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 (𝑳) and [𝐇𝐀] = 𝑽𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 (𝑳) [HA] mol A− pH = pKa + log Vsolution (L) Summary mol HA Vsolution(L) In terms of concentration: [𝑨− ] It can also be expressed in terms of number of moles: pH = pKa + 𝐥𝐨𝐠 [𝐇𝐀] mol A− pH = pKa + log In terms of number of moles: mol HA 𝒎𝒐𝒍 𝑨− pH = pKa + 𝐥𝐨𝐠 𝐦𝐨𝐥 𝐇𝐀 Buffers Calculation: 1. You need to produce a buffer solution that has a pH of 5.27. You already have a solution that contains 10.0 mmol of acetic acid. How many mmol of sodium acetate will you need to add to this solution? The pKa of acetic acid is 4.75. Buffer solution: Acetic acid (CH3COOH) and Sodium acetate (CH3COO-Na+) Acid (HA) Conjugate base (A-) Given: pH of buffer solution = 5.27 pka = 4.75 # moles of CH3COOH = 10.0 mmol ______________________________________________________________________________________ mol A− pH = pKa + log # mol CH3COO−Na+ mol HA = 100.52 10.0 mmol # mol CH3COO−Na+ 5.27 = 4.75 + log 10.0 mmol # mol CH3COO−Na+ = (100.52 )(10.0mmol) # mol CH3COO−Na+ log = 5.27 - 4.75 10.0 mmol # 𝐦𝐨𝐥 CH3COO−Na+ = 33.11 𝐦𝐦𝐨𝐥 # mol CH3COO−Na+ log = 0.52 10.0 mmol Buffers Calculation: 2. Aspirin has a pKa of 3.4. What is the ratio of A¯ to HA in: ______________________________________________________________________________________ (a) the blood (pH = 7.4) (b) the stomach (pH = 1.4) [A− ] [A− ] pH = pKa + log pH = pKa + log [HA] [HA] [A− ] [A− ] 7.4 = 3.4 + log 1.4 = 3.4 + log [HA] [HA] [A− ] [A− ] log = 7.4 – 3.4 log = 1.4 – 3.4 [HA] [HA] [A− ] [A− ] log = 4 log = -2 [HA] [HA] [A− ] [A− ] = 104 = 10−2 [HA] [HA] ❖ There are 10,000 𝐀− for every 1 HA ❖ There are 1 𝐀− for every 100 HA Buffers Biological Buffer System Bicarbonate Buffer: Carbonic acid (H2CO3) and bicarbonate ion (HCO3-) ❖ maintenance of blood pH is regulated. ❖ At acidic environment, buffer acts to form carbon dioxide gas. The lungs H+ + HCO3- H2CO3 H2O + CO2 expel this gas out of the body during the process of respiration. ❖ At basic environment, pH back to neutral by causing excretion of the bicarbonate ions through the urine. OH- + H2CO3 HCO3- + H2O Buffers Biological Buffer System Phosphate Buffer: dihydrogen phosphate ion (H2PO4-) and hydrogen phosphate ion (HPO4-2) H+ + HPO4-2 H2PO4- ❖ internal environment of all cells contains this buffer comprising hydrogen phosphate ions and OH- + H2PO4- HPO4-2 + H2O dihydrogen phosphate ions. ❖ When excess hydrogen enters the cell, it reacts with the hydrogen phosphate ions. ❖ Under alkaline conditions, the dihydrogen phosphate ions accept the excess hydroxide ions that enter the cell. END