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

Water molecules force lipids to associate and position polar heads outward

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

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

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

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

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

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 HA and A - See Buffer tutorial on web

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

Titration Curve Plateau Region Half-way Point pKa HCOOH pH HCOO 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

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

Buffer Problem (Continued) pH = pKa + log [A - ] [HA] New pH is: = pKa + log 7 3 = pKa 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

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

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

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

pH = pK a + log [A - ] [HA] Start [A - ] [HA] log = pH - pK a [A - ] [HA] log = pH - pK a END = = 0.3 [A - ] [HA] = 2.0/ 1.0 = = 0.8 [A - ] [HA] = 6.3/1.0 A - = mM HA = mM A - = mM HA = mM mmoles was absorbed