Chapter 2 Water: The Solvent for Biochemical Reactions PDF

Chapter 2 Water: The Solvent for Biochemical Reactions © 2018 Cengage Learning. All Rights Reserved. Chapter Outline Part 1 (S3-S32) (2-1) Water and polarity (2-2) Hydrogen bonds (2-3) Acids, bases, and pH Part 2...

Chapter 2 Water: The Solvent for Biochemical Reactions © 2018 Cengage Learning. All Rights Reserved. Chapter Outline Part 1 (S3-S32) (2-1) Water and polarity (2-2) Hydrogen bonds (2-3) Acids, bases, and pH Part 2 (S33-S53) (2-4) Titration curves (2-5) Buffers © 2018 Cengage Learning. All Rights Reserved. 2-1 Water and Polarity Water is the main component of most cells. The geometry of water molecule and its properties as a solvent play major roles in determining the properties of living systems. The tendency of an atom to attract electrons to itself in a chemical bond is called Electronegativity Note: Oxygen and nitrogen are more electronegative than carbon and hydrogen. Table 2.1 © 2018 Cengage Learning. All Rights Reserved. Polar Bonds. What is polarity? Electrons are unequally Figure 2.1 - The Structure of Water shared between O & H. The dipole moment in this figure More negative charge is points in the direction from found closer to one of the positive to negative. atoms. Water is a bent molecule. Polar bond is formed because of the difference in electronegativity of atoms involved in the bond. © 2018 Cengage Learning. All Rights Reserved. Polar Bonds and Molecules Bonds in a molecule may be polar, but the molecule itself can be nonpolar because of its geometry Molecules such as CO2 have polar bonds but given their geometry, are nonpolar Nonpolar: Bond in which two atoms share electrons evenly like between H & H. © 2018 Cengage Learning. All Rights Reserved. Solvent Properties of Water Some important noncovalent bonds: 1. Ionic bonds (Fig. 2.2) Held together by positive and negative ions 2. Salt bridge Interaction that depends on the attraction that occurs when oppositely charged molecules are near one another (like between COO- and – NH3+). 3. Ion–dipole interactions Occur when ions in solution interact with molecules that have dipoles (Figures 2.2 and 2.3a). Example - NaCl dissolved in H2O © 2018 Cengage Learning. All Rights Reserved. Figure 2.2 - Ionic Bonds Become Replaced by Ion–Dipole Interactions In ionic solids, ionic bonds hold the cations and anions together. In aqueous solution, these ionic bonds are replaced by ion–dipole interactions. The negatively charged chloride ions are attracted to the partial positive charges on water. The positively charged sodium ions are attracted to the partial negative charges on the water Water surrounding ions of this type are called hydration shells. © 2018 Cengage Learning. All Rights Reserved. Figure 2.3 - Ion–Dipole and Dipole–Dipole Interactions Ion–dipole and dipole–dipole interactions help ionic and polar compounds dissolve in water. © 2018 Cengage Learning. All Rights Reserved. Solvent Properties of Water (continued 1) 4. Van der Waals forces: Noncovalent associations based on weak attractions of transient dipoles for one another Called van der Waals interactions or van der Waals bonds Included in these forces are 3-noncovalent bonds that do not involve electrostatic interactions of a fully charged ions. 4.1 Dipole–dipole interactions (Fig. 2.3b) Forces that occur between molecules that are dipoles Partial positive side of a molecule attracts the partial negative side of another molecule © 2018 Cengage Learning. All Rights Reserved. Solvent Properties of Water (continued 2) 4.2 Dipole–induced dipole interactions (Fig. 2.4). Permanent dipole in one molecule can induce a transient dipole in another molecule through momentary distortion of electron clouds Weak and generally do not lead to solubility in water 4.3 Induced dipole–induced dipole interactions Fig. 2.5 This attractive force is often referred to as London dispersion force. (i.e., Attraction between transient induced dipoles) © 2018 Cengage Learning. All Rights Reserved. Figure 2.4 - Dipole-Induced Dipole Bonds Polar water can distort the electron cloud of a nonpolar molecule, like oxygen, creating a momentary dipole. Once the dipole is created, water is attracted to it. © 2018 Cengage Learning. All Rights Reserved. Figure 2.5 - Induced Dipole-Induced Dipole Interactions Also called London dispersion forces, these attractions arise when otherwise nonpolar molecules bump into each other and distort each other’s electron shells, forming induced dipoles. Once the dipoles are formed, the molecules are momentarily attracted to one another. © 2018 Cengage Learning. All Rights Reserved. Table 2.2 - Strengths of Bonds Found in Biochemistry © 2018 Cengage Learning. All Rights Reserved. Solvent Properties of Water (continued 3) Hydrophobic Interactions Hydrophobic interactions: Attractions between molecules that are nonpolar; also called hydrophobic bonds. Hydrophilic Substances (water loving) Tending to dissolve in water. See Table 2.3 Examples - Ionic and polar substances Hydrophobic Substances (water-hating) Tending not to dissolve in water. Amphipathic Substances Molecules that contain both hydrophobic and hydrophilic regions Example - Sodium palmitate (Fig. 2.6) © 2018 Cengage Learning. All Rights Reserved. Table 2.3 - Examples of Hydrophobic and Hydrophilic Substances © 2018 Cengage Learning. All Rights Reserved. Figure 2.6 - Sodium Palmitate, an Amphipathic Molecule Amphiphilic molecules are frequently symbolized by a ball and zigzag line structure where the ball represents the hydrophilic polar head, and the zigzag line represents the nonpolar hydrophobic hydrocarbon tail. © 2018 Cengage Learning. All Rights Reserved. Fig. 2.7 Micelle Formation by Amphipathic Molecules Spherical arrangement When micelles form, the of organic molecules in ionized polar groups are in contact with the water, and water solution clustered the nonpolar parts of the so that their: molecule are protected from Hydrophobic parts are contact with the water. buried inside the sphere Hydrophilic parts are on the surface of the sphere and in contact with the water environment Formation depends on the attraction between temporary induced dipoles © 2018 Cengage Learning. All Rights Reserved. 2-2 Hydrogen Bonds Another type of noncovalent attractive interactions between dipoles when the: Positive end of one dipole is a hydrogen atom bonded to a highly electronegative atom (hydrogen-bond donor) Negative end of the other dipole is an atom with a lone pair of electrons (hydrogen-bond acceptor). Fig. 2.8 A comparison of linear and nonlinear H-bonds. Nonlinear bonds are weaker than bonds in which all three atoms lie in a straight line. © 2018 Cengage Learning. All Rights Reserved. Figure 2.9 - Hydrogen-Bonding Sites A comparison of the numbers of hydrogen-bonding sites in HF, H2O, and NH3. (Actual geometries are not shown.) Each HF molecule has one hydrogen-bond donor and three hydrogen-bond acceptors. Each H2O molecule has two donors and two acceptors. Each NH3 molecule has three donors and one acceptor. © 2018 Cengage Learning. All Rights Reserved. Interesting and Unique Properties of Water Each water (ice) molecule is hydrogen bonded to four others. Acts as a donor in 2 and as an acceptor in the other 2 Enabled by the tetrahedral arrangement of the water molecule with bond angles of 104.3° Fig 2.10 shows: Tetrahedral H-bonding in water. In an array of H2O molecules in an ice crystal, each H2O molecule is hydrogen-bonded to four others. © 2018 Cengage Learning. All Rights Reserved. Table 2.4 - Comparison of Properties of Water, Ammonia, and Methane Even though hydrogen bonds are weaker than covalent bonds, they have a significant effect on the physical properties of hydrogen-bonded compounds © 2018 Cengage Learning. All Rights Reserved. Figure 2.11 Hydrogen Bonding between Polar Groups and Water If a polar solute can serve as a donor or an acceptor of hydrogen bonds, it can: Form hydrogen bonds with water Participate in nonspecific dipole–dipole interactions R-OH, R-NH2, R-COOH, R-CHO, & ketones can form H-bonds with water, so they are soluble in water. © 2018 Cengage Learning. All Rights Reserved. Table 2.5 - Examples of Major Types of Hydrogen Bonds Found in Biologically Important Molecules Hydrogen bonding is important in the stabilization of 3-D structures of biological molecules, such as DNA, RNA, and proteins. © 2018 Cengage Learning. All Rights Reserved. 2-3 Acids, Bases, and pH Acids and Bases Acid Molecule that behaves as a proton donor. Commonly known as a Brønsted acid. Base Molecule that behaves as a proton acceptor. Commonly known as a Brønsted base. © 2018 Cengage Learning. All Rights Reserved. Acid Strength Tendency of an acid to dissociate to a hydrogen ion and its conjugate base Characterized by acid dissociation constant (Ka) The greater the value of Ka, the stronger is the acid. The correct form of the equation: © 2018 Cengage Learning. All Rights Reserved. Figure 2.13 – The Ionization of Water Let us quantitatively examine the dissociation of water:  H +  OH -  Ka =     H 2O Molar concentration of pure water, [H2O], is equal to 1000/18 = 55.5 M. Thus, + -  H  OH  Ka = 55.5 K a  55.5 =  H +  OH -  = K w © 2018 Cengage Learning. All Rights Reserved. Ion Product Constant for Water (Kw) The value of Kw increases with the increase of temperature, i.e., the concentration of H+ and OH- ions increases with increase in temperature. A measure of the tendency of water to dissociate to give H+ and OH– ions The numerical value can be determined experimentally by measuring [H+] of pure water In monoprotic acids, [H+] = [OH–] At 25°C in pure water,  H +  = 10−7 M = OH −  At 25°C, the numerical value of Kw is given by: K w =  H +  OH −  = (10−7 )(10−7 ) = 10−14 © 2018 Cengage Learning. All Rights Reserved. pH and pKa Quantity for expressing [H+] and [OH–] in aqueous solutions pH = -log10  H  + Difference of one pH unit implies a tenfold difference in [H+] Neutral solutions have pH = 7, acidic solutions have pH < 7, and basic solutions have pH > 7 pKa Is a convenient numerical measure of acid strength pK a = − log10 K a The smaller the value, the stronger the acid © 2018 Cengage Learning. All Rights Reserved. Example 2.1 - pH Calculations In pure water, [H+] = 1×10–7 M and pH = 7.0 Calculate the pH of the following aqueous solutions: a. 1×10–3 M HCl b. 1×10–4 M NaOH Assume that the self-ionization of water makes a negligible contribution to the concentrations of hydronium ions and of hydroxide ions, which will typically be true unless the solutions are extremely dilute. © 2018 Cengage Learning. All Rights Reserved. Example 2.1 - Solution The key points in the approach to this problem are the definition of pH, which needs to be used in both parts, and the self-dissociation of water, needed in the second part a. For 1×10–3 M HCl, [H3O+] = 1×10–3 M Therefore, pH = 3 b. For 1×10–4 M NaOH, [OH–] = 1×10–4 M Because [OH–] [H3O+] = 1×10–14, [H3O+] = 1×10–10 M Therefore, pH = 10.0 © 2018 Cengage Learning. All Rights Reserved. Why do we want to know the pH? Figure 2.15 - pH versus Enzymatic Activity Pepsin, trypsin, and lysozyme all have steep pH optimum curves. Pepsin has maximum activity under very acidic conditions, as would be expected for a digestive enzyme that is found in the stomach. Lysozyme has its maximum activity near pH 5, while trypsin is most active near pH 6. Note how the activities of three enzymes can rapidly change as pH leaves optimum range © 2018 Cengage Learning. All Rights Reserved. Henderson–Hasselbalch Equation Mathematical  H +   A −  relationship between the Ka = pKa of an acid and the  HA  pH of a solution  A −  containing the acid and − log K a = log  H  + log +  its conjugate base  HA  When concentrations of  A −  a weak acid and its − log  H  = − log K a + log +  conjugate base are  HA  equal, the pH of the solution equals the pKa  A −  pH = pK a + log of the weak acid  HA  © 2018 Cengage Learning. All Rights Reserved. 2-4 Titration Curves Titration: Is an experiment in which measured amounts of standard base are added to measured amounts of an acid. Equivalence point: Point in an acid–base titration at which enough base has been added to neutralize the acid. It coincides with the end point. An inflection point is reached when the pH equals the pKa of acetic acid (Fig. 2.16). This is for a monoprotic weak acid. © 2018 Cengage Learning. All Rights Reserved. Figure 2.16 - Titration Curve for Acetic Acid Note that there is a region near the pKa at which the titration curve is relatively flat. In other words, the pH changes very little as base is added in this region of the titration curve. © 2018 Cengage Learning. All Rights Reserved. Example 2.2 - Calculating pH Values for Weak Acids and Bases Calculate the relative amounts of acetic acid and acetate ion present at the following points when 1 mol of acetic acid (pKa=4.76) is titrated with sodium hydroxide NaOH + CH3COOH → CH3COONa + H2O 1 mol 1 mol 1 mol 1 mol Also use the Henderson–Hasselbalch equation to calculate the values of the pH at these points: a) 0.1 mol of NaOH is added b) 0.3 mol of NaOH is added c) 0.5 mol of NaOH is added © 2018 Cengage Learning. All Rights Reserved. Example 2.2 - Solution Approach this problem as an exercise in stoichiometry There is a 1:1 ratio of moles of acid reacted to moles of base added Difference between the original number of moles of acid and the number reacted is the number of moles of acid remaining These are the values to be used in the numerator and denominator, respectively, of the Henderson–Hasselbalch equation © 2018 Cengage Learning. All Rights Reserved. Example 2.2 - Solution (continued 1) a. When 0.1 mol of NaOH is added, 0.1 mol of acetic acid reacts with it to form 0.1 mol of acetate ion, leaving 0.9 mol of acetic acid (1.0- 0.1). Composition is 90% acetic acid and 10% acetate ion 0.1 pH = pK a + log 0.9 0.1 pH = 4.76 + log 0.9 pH = 4.76 − 0.95 pH = 3.81 © 2018 Cengage Learning. All Rights Reserved. Example 2.2 - Solution (continued 2) b. When 0.3 mol of NaOH is added, 0.3 mol of acetic acid reacts with it to form 0.3 mol of acetate ion, leaving 0.7 mol of acetic acid (1.0- 0.3). Composition is 70% acetic acid and 30% acetate ion 0.3 pH = pK a + log pH = 4.39 0.7 c. When 0.5 mol of NaOH is added, 0.5 mol of acetic acid reacts with it to form 0.5 mol of acetate ion, leaving 0.5 mol of acetic acid (1.0- 0.5). Composition is 50% acetic acid and 50% acetate ion © 2018 Cengage Learning. All Rights Reserved. Example 2.2 - Solution (continued 3) 0.5 pH = pK a + log pH = 4.76 0.5 Note that this one is possible without doing much math We know that when the [HA] = [A–], the pH = pKa Therefore, the minute we saw that we added 0.5 mol of NaOH to 1 mol of acetic acid, we knew that we had added enough NaOH to convert half the acid to the conjugate base form Thus, the pH has to be equal to the pKa © 2018 Cengage Learning. All Rights Reserved. Titration Curves (continued) Acid dissociation constants for three groups of acids Monoprotic acids (e.g., Pyruvic acid) release one H+ ion and have a single Ka and pKa Diprotic acids release two H+ ions and have two Ka values and two pKa values (e.g., Oxalic acid). Polyprotic acids release more than two H+ ions ( e.g., Phosphoric acid). Here is a way to keep track of protonated and deprotonated forms of acids and their conjugate bases: When pH < pKa, the weak acid predominates H+ on, substance protonated When pH > pKa, the conjugate base predominates H+ off, substance deprotonated © 2018 Cengage Learning. All Rights Reserved. 2-5 Buffer Solutions Tend to resist changes in pH when small to moderate amounts of a strong acid or strong base are added. A buffer solution consists of a mixture of a weak acid and its conjugate base. How do buffers work? See Fig. 2.17 Examples of acid–base buffers are solutions containing: (acetate) CH3COOH and CH3COONa (bicarbonate) H2CO3 and NaHCO3 © 2018 Cengage Learning. All Rights Reserved. Figure 2.17 - Buffering Acid is added to the two beakers on the left. The pH of unbuffered H2O drops dramatically while that of the buffer remains stable. Base is added to the two beakers on the right. The pH of the unbuffered water rises drastically while that of the buffer remains stable. © 2018 Cengage Learning. All Rights Reserved. Example 2.3 Using the Henderson- Hasselbalch Equation. For illustration consider a solution in which the concentrations are [HPO42–] = 0.063 M and [H2PO4–] = 0.10 M Thus, the conjugate base/weak acid ratio is 0.63 If 1.0 mL of 0.10 M HCl is added to 99.0 mL of the buffer, the following reaction takes place and almost all the added H+ is used up 2− + − HPO 4 +H H 2 PO 4 The concentrations of [HPO42–] and [H2PO4–] will change, and the new concentrations can be calculated © 2018 Cengage Learning. All Rights Reserved. Mechanism of Buffers - Example (continued) The new pH can then be calculated using the Henderson–Hasselbalch equation and the phosphate ion concentrations. The appropriate pKa is 7.20 (Table 2.6).  HPO 24−  0.062 pH = pK a + log pH = 7.20 + log = 6.99  H 2 PO 4  − 0.101 © 2018 Cengage Learning. All Rights Reserved. Mechanism of Buffers Buffers follow Le Chatelier’s principle If stress is applied to a system in equilibrium, the equilibrium will shift in the direction that relieves the stress When H+ ion is added to a buffer system, stress is added to the reaction HA H+ + A− To relieve the stress, H+ reacts with A– to maintain equilibrium. See example 2.4 on p.50 © 2018 Cengage Learning. All Rights Reserved. Buffer Selection A consideration of titration curves can give insight into how buffers work (Fig. 2.18a) next slide. pH changes little in the vicinity of the inflection point of a titration curve At the inflection point, half the amount of acid originally present has been converted to the conjugate base The second stage of ionization of phosphoric acid, was the basis of the buffer used in this example. 2− + − HPO + H4 H 2 PO 4 When the pH is one unit higher than pKa, the ratio of conjugate base form to the conjugate acid form is 10 When pH is two units higher than pKa, the ratio is 100, and so on. Table 2.7 shows this. © 2018 Cengage Learning. All Rights Reserved. Figure 2.18 - Relationship between the Titration Curve and Buffering Action in H2PO4– © 2018 Cengage Learning. All Rights Reserved. Table 2.7 - pH Values and Base/Acid Ratios for Buffers © 2018 Cengage Learning. All Rights Reserved. Buffer Selection A buffer solution can maintain a relatively constant pH value, which is met at values at or near the pKa of the acid. The buffer is effective within a range of about 2 pH units (Fig. 2.18b). The buffer is effective at range of pH = pKa ±1. Buffering Capacity is a measure of the amount of acid or base that can be absorbed by a given buffer solution Related to the concentrations of the weak acid and its conjugate base. The greater the concentration of the weak acid and its conjugate base, the greater the buffering capacity.. © 2018 Cengage Learning. All Rights Reserved. How do we make buffers in the laboratory? Start with the HA form and add NaOH until the pH is correct, as determined by a pH meter or start with A– and add HCl until the pH is correct. Figure 2.19 - Two Ways of Looking at Buffers. In the titration curve, we see that the pH varies only slightly near the region in which [HA] = [A–]. In the circle of buffers, we see that adding OH– to the buffer converts HA to A−. Adding H+ converts A− to HA. © 2018 Cengage Learning. All Rights Reserved. Biological Buffers What naturally occurring pH buffers are present in living organisms? The H2PO4–/HPO42– pair is the principal buffer in cells. The buffering system in blood is based on the dissociation of carbonic acid (H2CO3) H2CO3 HCO3– + H+ Where the pKa = 6.37. The pH of the blood, 7.4, is near the end of the buffering range of this system, but another factor enters the situation. CO2 can dissolve in water and water-based fluids such as blood. The dissolved CO2 forms H2CO3, which reacts to produce bicarbonate ion. The conversion is catalyzed by carbonic anhydrase. This is one of the most efficient enzymes in biochemistry. CO 2 ( g ) CO 2 ( aq) CO 2 ( g ) + H 2O(l ) H 2CO3 ( aq) H 2 CO3 ( aq) H + ( aq) + HCO3- ( aq) Net equation: CO 2 ( g ) + H 2O(l ) H + ( aq) + HCO 3− ( aq) © 2018 Cengage Learning. All Rights Reserved. Naturally Occurring Buffers (continued) The phosphate buffer system is common in the lab. It can be: In vitro (outside the living body) Example - Buffer system based on tris (hydroxymethyl) aminomethane (TRIS) In vivo (inside the living body) Example - Phosphate buffer system Other buffers that are used widely are zwitterions, Which are:. compounds that have both a positive and negative charge.. considered less likely to interfere with biochemical reactions than some of the earlier buffers. See (Table 2.8). © 2018 Cengage Learning. All Rights Reserved. Table 2.8 Acid and Base Form of Some Useful Biochemical Buffers © 2018 Cengage Learning. All Rights Reserved.