1. Inter Molecular bonding High boiling Point Polarity (high dipole moment) Flickering High dielectric constant

2. Water molecules force lipids to associate and position polar heads outward

3. Water ionization 1. Structured water can ionize spontaneously 2. Only a very small fraction of the water molecules ionize. 3. Ionization gives rise to hydronium ions and hydroxide ions [H 3 O + ] [OH - ] One to one 4. Ionization gives rise to a neutral solution

4. Ionization equations Keq = [H 3 O + ][OH - ] [H 2 O] 2 H2OH2O H2OH2O Keq = [H + ][OH - ] [H 2 O] [H 2 O] = 1000 g / 18 g/gMwt = 55.5 M H 2 O + H 2 O H 3 O + + OH - See Biochemical Strategies, p. 20

5. Ionization (Continued) [H 2 O] x Keq = [H + ][OH - ] [55.5] x Keq = 1 x 10 -14 1 x 10 -14 = [H + ][OH - ] Since [H + ] = [OH - ] 1 x 10 -14 = [H + ] 2 1 x 10 -7 = [H + ] in a neutral solution See Biochemical Strategies, pp 20-21 See Biochemical Strategies, pp 20-21

6. Log Expressions pH = - log [H+] pOH = -log[OH - ] pKw = -log[10 -14 ] = 14 pK a = -log Ka Therefore: pH + pOH = 14 If [H+][OH-] = 1 x 10 -14 Then -log[H + ] + -log[OH - ] = -log 1 x 10 -14 See Biochemical Strategies, p. 21 See Biochemical Strategies, p. 21

7. K a = [H + ][A - ] [HA] [H + ] = [HA] [A - ] KaKa Take log of both sides Multiply by -1 log [H + ] = log K a + log [HA] [A - ] [HA] log [H + ] = log K a + log pH = pK a + log [A - ] [HA] See Biochemical Strategies, p23 See Biochemical Strategies, p23 HA H + + A - HENDERSON-HASSELBALCH EQUATION

8. 1. A shield 2. A weak acid and its conjugate base PRESENT IN NEARLY EQUAL PROPORTIONS 3. A buffer is made up of two components HA + OH - A - + H 2 O HA + H + NO REACTION A - + H+HA A - + OH - NO REACTION See Biochemical Strategies, pp 22-24 HA and A - See Buffer tutorial on web

9. HA + A - Add OH - HAA-A- Add H + HA A-A- HA + A - (Total buffer) does not change because HA and A - change to same amount but in opposite directions. ONLY, The ratio of A - /HA changes The addition of acid or base to a buffer changes the ratio of the conjugate acid and base without changing the total The addition of acid or base to a buffer changes the ratio of the conjugate acid and base without changing the total HA + A - HA A-A- A-A- A-A- A-A- A-A- At Neutrality

10. Titration Curve Plateau Region Half-way Point pKa HCOOH 00.51.0 pH HCOO -- 2 4 6 8 10 NH 4 + NH 3 NaOH equivalents HCOOH + OH - HCOO - + H 2 O Conjugate acid Conjugate base See Biochemical Strategies p. 25 [HCOOH] = [HCOO - ] Smallest change in pH

11. SOLVING BUFFER PROBLEMS H+H+ = Ka [HA] [A - ] pH = pKa + log [HA] [A - ] HA + A - + OH - HA + A - + H 2 O 55 Before2 3 7 2 After [A - ] [HA] = 1.0 [A - ] + [HA] = 10 [A - ] [HA] = 2.33 [A - ] + [HA] = 10 Independent variable Independent variable Dependent variable Dependent variable Constant

12. Buffer Problem (Continued) pH = pKa + log [A - ] [HA] New pH is: = pKa + log 7 3 = pKa + 0.37 pH has become 0.37 units more alkaline = pKa + log 2.36 A - = 7 mM, HA = 3 mM After Neutralization Note: HA + A - = 10 mM

13. Meaning of K a K a is numerically equal to the proton concentration when the acid is half ionized. K a is numerically equal to the proton concentration when the acid is half ionized. K a = [H + ][A - ] [HA] Rearranging [H + ] =KaKa [HA] [A - ] When [HA] = [A - ], the acid is half ionized and…. [H + ] = K a Adjusting the proton concentration to equal K a assures the acid will be half ionized. Adjusting the proton concentration to equal K a assures the acid will be half ionized. K a is a dissociation constant for an acid, but much more... K a is a dissociation constant for an acid, but much more... See Strategies p. 22

14. Significance of pK a 1. The pH at the point of half ionization 2. Point of maximum buffering capacity 3. Relative Acid strength 4. Order of proton dissociation from a polyprotic acid An acid with a pK a of 4 is 100 times stronger than one with a pK a of 6 A dissociable group with a pK a of 6 is a 1000 times stronger acid than one with a pK a of 9 pK a is the negative log of K a, i.e., -Log K a The numerical value for pK a allows one to determine

15. Problem: A 0.1 M acetate buffer has a pH of 5.0 How much NaOH must be added to raise the pH to 5.5? SOLUTION: 1. Calculate the A - /HA at the START 2. Calculate the A - /HA at the END 3. Change in either HA or A will determine base added pK a acetate = 4.7

16. pH = pK a + log [A - ] [HA] Start [A - ] [HA] log = pH - pK a [A - ] [HA] log = pH - pK a END = 5.0 - 4.7 = 0.3 [A - ] [HA] = 2.0/ 1.0 = 5.5 - 4.7 = 0.8 [A - ] [HA] = 6.3/1.0 A - = 0.067 mM HA = 0.033 mM A - = 0.086 mM HA = 0.014 mM 0.019 mmoles was absorbed