1. Acids, Bases and pH Student Edition 5/23/13 Version Pharm. 304 Biochemistry Fall 2014 Dr. Brad Chazotte 213 Maddox Hall chazotte@campbell.edu Web Site: http://www.campbell.edu/faculty/chazotte http://www.campbell.edu/faculty/chazotte Original material only ©2004-14 B. Chazotte

2. Goals Review the ionization of water, K eq and K w Review the concept of an acid dissociation constant, K a Understand the concept of pH and a pH scale Review the Henderson-Hasselbalch equation & its use Be able to calculate the pH of a weak acid Understand buffers Review titration curves for monoprotic & polyprotic acids Review the effect of pH on protein solubility and enzyme function Review the concept of, and the calculation of, ionic strength

3. Ionization of Water Water has a slight tendency to undergo a reversible ionization. H 2 O  H + + OH - Since free protons do not exist in solution one writes: H 2 O  H 3 O + + OH - (H 3 O +, a hydronium ion) Proton jumping: the proton can rapidly jump from one water molecule to the next in one of the fastest reactions in solutions.

4. K eq and K w A + B  C + D K eq = [C] [D] [A][B] [] concentration approximates the activity coefficient FOR WATER: H 2 O  H + + OH - K eq = [H + ] [OH - ] [H 2 O] In pure water at 25 ºC [H 2 O] =55.5 M i.e., essentially constant compared to ions Therefore we can write : (55.5 M) (K eq ) = [H + ] [OH - ] = K w (ion product of water) K w = 1 * 10 -14 @ 25.0 ºC

5. Lehninger 2001 Table 4.2 Table: The pH Scale As defined by Sorensen pH = -log [H + ] pH + pOH = 14 pH is a shorthand way of designating the hydrogen ion activity of a solution The “p” in pH designates the negative logarithm The ion product of water is the basis for the pH scale Sum of pH and pOH always =14 Voet, Voet & Pratt 2013 Fig. 2-16

6. pH’s of Some Aqueous Fluids Lehninger 2000, Figure 4.13 >pH 7 basic [H + ] < [OH - ] pH 7 neutral [H + ] = [OH - ] <pH 7 acidic [H + ] > [OH - ]

7. Acids & Bases - Definitions Brønsted and Lowry: Brønsted Acid - a substance that donates protons (hydrogen ions) Brønsted Base - a substance that accepts protons (hydrogen ions) When a Brønsted acid loses a proton a Brønsted base is produced. In this context the original acid loses a proton and produces a base, These are referred to as a conjugate pair, e.g. HA and A -. A strong acid or base is one that ionizes almost 100% in aqueous solution.

8. K a (Dissociation Constant) K a numerically describes the strength of an acid. K = [H 3 O + ] [A - ] [HA] [H 2 O] since [water] = 55.5 M in dilute solution, [H 2 O] ≈ constant K a = K [H 2 O] = [H+] [A-] [HA] Strong acids K a >> 1 Weak acids K a < 1

9. Henderson- Hasselbalch Equation Shows the relationship of a solution’s pH and the concentration of an acid and its conjugate base in solution. [H + ] = K a ([HA] / [A - ]) (1) pH = - log K a - log ([HA] / [A - ]) since pH = -log [H + ] (2) pH = pK a + log ([A - ] / [HA]) since pK = -log K (3) HA = proton donor A - = proton acceptor When [HA] = [A - ] then log ([A - ] / [HA]) = 0 And pH = pK a ! But the H-H equation cannot be used for strong acid or base.

10. Henderson-Hasselbalch Calculation Calculate the pK a of lactic acid given that the concentration of lactic acid is 0.010 M, the concentration of lactate is 0.087 M and the pH is 4.80. [lactate] pH = pK a + log [lactic acid] [lactate] pK a = pH - log [lactic acid] 0.087 M pK a = 4.80 - log 0.010 M = 4.80 - log 8.7 = 4.80 - 0.94 pK a =3.9

11. Extent of Ionization Relationship of pH and pK a Weak acids % Compound Ionized Weak Bases pH = pK a ~50% pH = pK a + 1~90% pH = pK a + 2~99% pH = pK a + 3~99.9% pH = pK a + 4~99.99% pH = pK a ~50% pH = pK a - 1~90% pH = pK a - 2~99% pH = pK a - 3~99.9% pH = pK a - 4~99.99% Cairns “Essentials of Pharmaceutical Chemistry”2008 p.22

12. pH of a Weak Acid Solution Calculation I Problem: A weak acid HA is 0.1% ionized (dissociated) in a 0.2 M solution. A)What is the equilibrium constant for the acid’s dissociation? B)What is the pH of the solution?

13. pH of a Weak Acid Solution Calculation II “A”HA  H+A- Start0.2M00 Change-(0.1% of 0.2M) =-(2 x 10 -4 M) +(2 x 10 -4 M) +(2 x 10 -4 M) Equilibrium 0.2M -(2 x 10 -4 M) (2 x 10 -4 M) (2 x 10 -4 M) [H+] [A-] (2 x 10 -4 M) x (2 x 10 -4 M) K a = [HA]= (0.2 M - 2 x 10 -4 M) 4 x 10 -8 K a = 1.998 x 10 -1 = 2 x 10 -7 Segal 1975

14. pH of a Weak Acid Solution Calculation III “B” pH = log (1 / [H + ]) = log (1 / [2 x 10 -4 ]) = log (5000) = 3.7

15. Acid-Base Neutralization Calculation Problem How many ml of 0.025 M H 2 SO 4 are required to neutralize 525 ml of 0.06 M KOH? Setup: 1# moles H + (equivalents) req. = # moles OH - (equivalents) present 2Liters x normality (N) = # equivalents 3Liters acid x N acid = liters base x N base 1 H 2 SO 4 = 0.025 M = 0.05 N (two hydrogens per mole sulfuric) Liters acid x 0.05 M = 0.525 liters base x 0.06 M 0.63 liters acid = (0.525 liters x 0.06 M) / 0.05 M

16. Acids, Bases, Salts & Solutions Types of salts and solutions formed when acid and base combine Strong acid + Strong base →Neutral salt HCl + NaOH → Na + Cl - + H 2 O Strong acid + Weak base →Acidic salt HCl + NH 3 → NH 4 + Cl - Weak acid + Strong base →Basic salt CH 3 COOH+ NaOH → CH 3 COO - Na + + H 2 O Weak acid + Weak base →Neutral salt CH 3 COOH + NH 4 OH → NH 4 + CH 3 COO - + H 2 O Cairns “Essentials of Pharmaceutical Chemistry”2008 p11 NH 4 + Cl - ↔ NH 4 + Cl - NH 4 + + H 2 O ↔ NH 3 + H 3 O + Example of strong acid and weak base from above → acidic salt This is why: Some Examples of Drugs Formulated as Salts Diphenhydramine hydrochloride (Benadryl) Naproxen sodium (Aleve) Cetirizine hydrochloride (Zyrtec) Morphine sulfate Oxycodone Hydrochloride (Oxycontin)

17. Buffers The maintenance of a relatively constant pH is critically important to most biological systems. pH changes can effect the structure, function, and/or reactivity of biomolecules. One pH unit is equivalent to a 10-fold change in H + ion concentration. A buffer (system) resists changes in solution pH by changes in the concentration of the buffers’ acid and conjugate base (HA and A - ). This can be illustrated in the titration curve of a weak acid. A system’s maximum buffering “capacity” is at or near the pK a of the molecule with a range typically ± 1 pH.

18. Titration Curve Examples Acetic Acid, Phosphate & Ammonia Voet. Voet & Pratt, 2013 Fig 2.17 HA  H + + A - K a = [H + ] [A - ] [HA] CH 3 COOH CH 3 COO - 5.76 3.76 4.76 050100% titrated

19. Titration of a Polyprotic Acid e.g., H 3 PO 4 Voet. Voet & Pratt, 2013 Fig 2.18 A polyprotic acid has a pKa for each ionization or ionizable group. The ionization of one group creates an electrostatic charge that raises the pK a of the subsequent ionization

20. A Buffer System: Acetic Acid / Acetate Lehninger 2000, Figure 4.17 Adding H + ions drives the equilibrium to acetic acid taking up the added H + ions while adding OH - ions drive the equilibrium to acetate. Sum of the buffer components does not change, only their ratio.

21. The Bicarbonate Buffer System (Lungs & Blood) Lehninger 2000, page 105 (Voet et al. Rx #1) (Voet et al. Rx #2)

22. pH and Solubility of a Protein Matthews et al 1999 Figure 2.21 At a protein’s isoelectric point, called pI, the sum of the positive charges equals that of the negative charges leaving no net charge At the pI a protein is least soluble.

23. pH Optima of Three Enzymes Lehninger 2000, Figure 4.19

24. Ionic Strength A measure of the total concentration of ions in solution. The more charged the ion the more it is counted μ = ½ Σ c i Z i 2 where c i = concentration of ith species and Z i = the charge on the ith species Why care? The greater the μ of a solution the higher the charge in the “ionic atmosphere” around an ion and the less net charge so the less attraction between any given cation and anion. Relates thermodynamic activity, a i, to concentration [A]. a A = [A] γ A (see also extended Debye-Hückel Equation)

25. Sample Calculation of Ionic Strength  = ½ ((0.090))  = 0.045 What is the ionic strength of a 0.015 M solution of CaCl 2 ? The equation is  = ½  c i Z i 2 The reaction is CaCl 2  Ca 2+ + 2 Cl - The concentrations are [0.015] + [0.030]  = ½ ((0.015 x 2 2 ) + (0.030 x -1 2 ))  = ½ ((0.060) + (0.030))

26. The Effect of Ionic Strength Matthews et al 1999 Figure 2.22

27. End of Lecture