BIOC 460 General Chemistry Review: Chemical Equilibrium, Ionization of H 2 O, ph, pk a General Equilibrium: What are the UNITS of K eq? Example reactions: A --> B units of K eq? A --> B + C units of K eq? A + B --> C units of K eq? A + B --> C + D units of K eq? 2A --> B units of K eq? 2A + B --> C + D units of K eq? Units of K eq depend on the reaction! Dissociation of H 2 O: Bioc460 General Chemistry Review (2006) p. 1
Table 2-6, from Nelson & Cox: Lehninger Principles of Biochemistry, 4th edition (2004) NOTE: If ph > 7, [H + ] < 10 7 M (solution is "basic") If ph < 7, [H + ] > 10 7 M (solution is "acidic") If ph = 7, [H + ] = 10 7 M (solution is "neutral") ph scale is logarithmic: If change in ph (ΔpH) = 1, Δ[H + ] = 10 1 (10-fold). If change in ph (ΔpH) = 2, Δ[H + ] = 10 2 (100-fold). Bioc460 General Chemistry Review (2006) p. 2
Ionization of weak acids and bases: Brοnsted definitions: acid = a proton donor base = a proton acceptor (Some of this material is reviewed on pp. 73-74 of the textbook Biochemistry, 5th ed., by Berg, Tymoczko & Stryer.) Definition of pk a : Below is a table of some weak acids, with their pk a values (25 C). Explain the differences in pk a within the 3 groups below. Why does substitution of each successive H in the methyl group of acetic acid with a Cl decrease the pk a? Why does substitution of each additional methyl or methylene group for the H in formic acid or in methylamine increase the pk a? Why does addition of the OH group to ethylamine to make ethanolamine decrease the pk a? Bioc460 General Chemistry Review (2006) p. 3
Acid pk a Acid pk a acetic acid 4.76 ammonium ion 9.25 + CH 3 -COOH NH 4 chloroacetic acid 2.85 methylamine 10.6 + ClCH 2 -COOH CH 3 -NH 3 dichloroacetic acid 1.48 ethylamine 10.8 + (Cl) 2 CH-COOH CH 3 -CH 2 -NH 3 trichloroacetic acid 0.70 ethanolamine 9.5 + (Cl) 3 C-COOH HO-CH 2 -CH 2 -NH 3 formic acid H-COOH acetic acid CH 3 -COOH propionic acid CH 3 -CH 2 -COOH 3.75 4.76 4.87 Fig. 2-16: Proton dissociation reactions of some mono-, di-, and triprotic acids, with their pk a s arranged on a ph scale (from Nelson & Cox: Lehninger Principles of Biochemistry, 4th edition (2004) Bioc460 General Chemistry Review (2006) p. 4
Quantitative relationship between pk a and ph: Look again at the equilibrium dissociation reaction and remember LeChatelier's Principle: HA H + + A The higher the [H + ] (i.e., the lower the ph), the more the equilibrium shifts to the left, so the more conjugate acid will be present. Conversely, the lower the [H + ] (i.e., the higher the ph), the more the equilibrium shifts to the right, so the more conjugate base will be present. The exact ratio of base to acid depends on the ph and the pk a (Henderson-Hasselbalch Equation). Note that when ph = pk a, [A ] = [HA], so [base] /[ acid] ratio = 1/1 ([base] = [acid]). When ph = pk a, the buffering capacity of the mixture will be the greatest. That is, a given change in concentration of base (OH ) or acid (H + ) will result in the smallest change in ph. If the ph is ABOVE the pk a, there will be more base than acid in the solution, i.e., the [base] /[ acid] ratio will be greater than 1/1. If the ph is BELOW the pk a, there will be more acid than base, i.e., the [base] /[ acid] ratio will be less than 1/1. When you do calculations, always stop at the end to think about whether your answer makes sense. Work the posted practice problems at the end of this review, on ph, pk a, and [base]/[acid] ratio, and learn how to convert a [base]/[acid] ratio to the FRACTION of the total ([base] + [acid]) that's in the form of the conjugate base (or the conjugate acid). Bioc460 General Chemistry Review (2006) p. 5
Titration Curves: add measured volume of solution of strong base (NaOH) of known concentration measure the ph, e.g. with a ph meter can be used to determine amount of acid in the solution (from mols of OH required to completely deprotonate each acidic group), which permits conversion of volume of standard OH added (which was measured) to equivalents OH added (mols OH /mol acidic group) pka values of acidic groups (from ph at which an acidic group is exactly half titrated) example: titration of acetic acid, CH3-COOH involves 2 reversible equilibria: Fig. 2-17: Titration curve of acetic acid (from Nelson & Cox: Lehninger Principles of Biochemistry, 4th edition (2004). When no OH has been added, predominant form is HAc; at the endpoint, when 1 equivalent OH has been added, all the HAc has been converted to Ac. When exactly 0.5 equivalent of OH has been added, exactly half the original HAc has dissociated, so [Ac ] = [HAc], i.e., [base] = [acid]), and ph = pka. Shaded zone shows the ph range of useful buffering power, from about 1 ph unit below to about 1 ph unit above the pka, i.e. between about 10% and 90% titration of the weak acid. Bioc460 General Chemistry Review (2006) p. 6 Fig. 2-18: Titration curves of acetic acid, H2PO4, and ammonium ion (from Nelson & Cox: Lehninger Principles of Biochemistry, 4th edition (2004).
Fig. 2.13 from Garrett & Grisham, Biochemistry, 3rd ed., 2004. For a polyprotic acid (more than one acidic group, so more than one pk a on titration curve), it takes 1 equivalent of OH to titrate each acidic group, so a triprotic acid like phosphoric acid requires 3 equivalents. The pk a values are ph values at each half equivalent added. Summary: 1) H 2 O and acids in aqueous solution dissociate to yield protons (H + ), which in actuality would be hydrated to form H 3 O +. 2) The tendency of an acid to donate its proton to H 2 O (dissociate the proton) can be quantitatively described by its dissociation equilibrium constant K a (or its pk a, which = logk a ). 3) pk a values can be very accurately measured by titration curves, as the ph at half equivalence points. 4) Relationship between ph, pk a, and ratio of [base]/[acid] can be described by the Henderson- Hasselbalch Equation: Buffers: homeostasis: the maintenance of constant conditions in internal environment Fluids in living systems have a characteristic, almost constant, ph. How is ph controlled in living systems? By buffer systems. Buffer: aqueous system that tends to resist changes in ph when small amounts of acid or base are added. Buffer system: aqueous solution of a weak acid and its conjugate base Buffer range of a weak acid: near its pk a, about ±1 ph unit from the pk a. Maximum buffering capacity is at the pk a. Look again at the equilibrium dissociation reaction and remember LeChatelier's Principle: HA H + + A Bioc460 General Chemistry Review (2006) p. 7
The higher the [H + ] (i.e., the lower the ph), the more the equilibrium shifts to the left, so the more conjugate acid will be present. Conversely, the lower the [H + ] (i.e., the higher the ph), the more the equilibrium shifts to the right, so the more conjugate base will be present. The exact ratio of base to acid depends on the ph and the pk a (Henderson-Hasselbalch Equation). Remember that when ph = pk a, [A ] = [HA], so [base] /[ acid] ratio = 1/1 ([base] = [acid]). When ph = pk a, the buffering capacity of the mixture will be the greatest. That is, a given change in concentration of base (OH ) or acid (H + ) will result in the smallest change in ph. 2 physiologically important buffer systems: 1) INTRACELLULAR: inorganic phosphate ion (H 2 PO 4 <--> HPO 4 2, pk a ~ 7) and organic phosphates, e.g. phosphomonoesters (R-OPO 2 OH <--> R-OPO 3 2, pk a ~ 7) 2) EXTRACELLULAR (blood plasma of mammals): bicarbonate buffer system (CO 2 + H 2 O <--> H 2 CO 3 <--> H + + HCO 3 ) 3 linked equilibria: How can the [HCO 3 ] / [H2 CO 3 ] system effectively buffer the blood plasma at ph ~7.4, with a pk a well below 7? clinical considerations of bicarbonate buffer system: a) What happens physiologically if the blood ph (i.e. [H + ]? (acidosis) You exhale excess CO 2, so [CO 2 ], so (Le Chatelier's Principle) [H + ], i.e. ph goes back up. b) What happens physiologically if the blood ph (i.e. [H + ]? (alkalosis) You exhale less CO 2, so [CO 2 ], so (Le Chatelier's Principle) [H + ], i.e. ph goes back down. c) What if you have difficulty exhaling (respiratory failure)? Blood [CO 2 ], so [H + ] (acidosis) and you can't get rid of the excess [CO 2 ] by exhaling. Treatment: inject bicarbonate (HCO 3 ) to convert some of the excess H + to H2 CO 3, which would bring ph back up -- and get the person breathing again to get rid of the excess CO 2! For a little online quiz on acids, bases, ph & pk a, and buffers, try the Biology Project website at http://www.biology.arizona.edu/biochemistry/problem_sets/ph/ph.html Click on "tutorial" at the bottom of each question for the tutorial that goes with that question. Bioc460 General Chemistry Review (2006) p. 8
For a little online quiz on clinical correlates of blood ph, try the Biology Project website at http://www.biology.arizona.edu/biochemistry/problem_sets/medph/medph.html Click on "tutorial" at the bottom of each question for the tutorial that goes with that question. PRACTICE PROBLEMS Work these without looking at the answers, which are on the last page.) NOTE ESPECIALLY PROBLEM 2B, how to go from the ratio of base/acid (Henderson-Hasselbalch calculation) to the fraction of the total (acid + base) that's in the form of the acid (or what fraction is in the form of the base). In fact, anytime you calculate a RATIO and get a number, STATE the (understood) "1" in the denominator. If ratio is 0.4, state it as 0.4 / 1. Also, start by writing chemical equations to describe acid dissociation reactions, with CHARGE BALANCE. 1. For a solution whose ph is 6.0, what is [H + ]? If [H + ] is 5 x 10 7 M, what is the ph? (Use a simple scientific calculator if necessary to do log problems.) 2. A. For a weak acid such as the R group carboxyl group of Glu or Asp in a protein, if the pk a of that specific residue's carboxyl group is exactly 4.0, in an environment in which the ph = 5.0, what would be the ratio of base to acid ( [COO ]/[COOH] )? B. For that same carboxyl group, what fraction, or what percentage, of the total (population of all the molecules in solution) is present in the form of the ACID (COOH) at ph 5.0? What fraction of the total is present in the form of the BASE (COO ) at ph 5? 3. Suppose that about 1% of the molecules of a particular protein in solution have the imidazole group of a specific His residue (say it's residue #20 in the amino acid sequence of that protein) in the uncharged (neutral) form at ph 4.5. What is the pk a of that specific His residue in that protein? (You need to know the ionization properties of the functional group of His to answer this question -- look at the ionization equilibrium for His in Fig. 3-15 (p. 48) and in Table 3-1 (p. 50), of Berg, Tymoczko & Stryer, Biochemistry, 5th ed. All that concerns you is the ionization of the R group (side chain), the imidazole group. The pk a of the imidazole group of the free amino acid histidine is about 6.0, but the exact pk a values of amino acid functional groups in proteins vary somewhat, depending on the specific environment around that residue in that protein. You're being asked to calculate this specific His residue's pk a.) Bioc460 General Chemistry Review (2006) p. 9
SOLUTIONS TO PRACTICE PROBLEMS 1. ph = log[h + ] (Remember, the log is the exponent.) If ph = 6.0, [H + ] =10 ph = 10 6.0 M (Use a calculator for this.) If [H + ] = 5 x 10 7 M, ph = log(5 x 10 7 ) = 6.3 2. acid dissocation reaction for a carboxyl group: (R)-COOH <--> (R)-COO + H + A. Use the Henderson-Hasselbalch Equation whenever you see a problem involving the relationship of ph, pk a, and base/acid ratio (or fraction of the total group that's in form of the base, or of the acid). ph = pk a + log([base]/[acid]) log([base]/[acid]) = ph pk a = 5.0-4.0 = 1.0 ([base]/[acid]) = 10 1 = 10/1 (There's 10 times as much of the carboxyl group in the form of the base as in the form of the acid when the ph is 1 unit above the pk a.) B. At ph 5.0, the COO /COOH ratio is 10/1. We can do this problem most simply by using proportions; the total of the carboxyl group is COOH + COO = 1 + 10 = 11. The fraction of the total in the form of the acid (COOH) is (COOH) / total = (COOH) / (COOH + COO ) = 1/11 = 0.091 9.1% of the molecules in solution have that carboxyl group in its acid form. COO / total = COO / (COOH + COO ) = 10/11 = 0.909 Alternatively, if you've already calculated that acid/total = 0.091, then base/total = 1 (acid/total) = 1 (0.091) = 0.909. Either way, 90.9% of the molecules in solution have that carboxyl group in its base form. 3. We'll abbreviate the imidazole group in question as Im, so neutral form is Im, the conjugate base, and charged form is Im +, the conjugate acid. Im + <--> Im + H + Use the Henderson-Hasselbalch Equation again. "About 1% neutral" means that base (Im) / total = about 0.01 or 1/100. Thus base/acid ratio would be base/(total base) = 1 / (100 1) = 1/99. ph = pk a + log([base]/[acid]) pk a = ph log([base]/[acid]) = 4.5 log (1/99) = 4.5 ( 2.0) = 4.5 + 2.0 = 6.5 The pk a of that specific His residue in that protein must be about 6.5. (Note that this is slightly different from the pk a of the R group of the free amino acid histidine.) Bioc460 General Chemistry Review (2006) p. 10