1. Buffers and the Henderson-Hasselbalch Equation -many biological processes generate or use H + - the pH of the medium would change dramatically if it were not controlled (leading to unwanted effects) --biological reactions occur in a buffered medium where pH changes slightly upon addition of acid or base -most biologically relevant experiments are run in buffers how do buffered solutions maintain pH under varying conditions? to calculate the pH of a solution when acid/base ratio of weak acid is varied: Henderson-Hasselbalch equation comes from: K a = [H + ] [A – ] / [HA] take (– log) of each side and rearrange, yields: pH = pK a + log ( [A – ] / [HA] ) some examples using HH equation: what is the pH of a buffer that contains the following? 1 M acetic acid and 0.5 M sodium acetate

2. Titration example (similar one in text:) Consider the titration of a 2 M formic acid solution with NaOH. 1. What is the pH of a 2 M formic acid solution? use K a = [H + ] [A – ] / [HA] HCOOH H + + HCOO – let x = [H + ] = [HCOO – ] thenK a = 1.78 x 10 –4 = x 2 / (2 – x) for an exact answer, need the quadratic equation but since formic acid is a weak acid (Ka is small), x <<< [HCOOH] and equation becomes K a = 1.78 x 10 –4 = x 2 / 2 so x = [H + ] = [HCOO – ] = 0.019 and pH = 1.7 2. Now start the titration. As NaOH is added, what happens? NaOH is a strong base --- completely dissociates OH – is in equilibrium with H +, K w = [H + ] [OH – ] = 10 –14, K w is a very small number so virtually all [OH – ] added reacts with [H + ] to form water

3. Titration continued: - to satisfy the equilibrium relationship given by K a K a = [H + ] [HCOO – ] / [HCOOH] = 1.78 x 10 -4 more HCOOH dissociates to replace the reacted [H + ] and -applying HH, see that [HCOO – ] / [HCOOH] will increase pH = pK a + log ( [HCOO – ] / [HCOOH] ) -leading to a slow increase in pH as the titration proceeds _______________________________________________ consider midpoint of titration where half of the HCOOH has been neutralized by the NaOH [HCOO – ] / [HCOOH] = 1 HH becomes: pH = pK a + log 1 = pK a = 3.75 for HCOOH Titration curve: - within 1 pH unit of pK a over most of curve - so pK a defines the range where buffering capacity is maximum - curve is reversible

4. Simple problem: -have one liter of a weak acid (pK a = 5.00) at 0.1 M -measure the initial pH of the solution, pH = 5.00 -so it follows that initially, [A – ] = [HA] where pH = pK a -add 100mL of 0.1M NaOH, following occurs HA + OH – = A – + H 2 O 0.01moles -so, 0.01 moles of HA reacted and new [HA] = 0.1 – 0.01 = 0.09 new [A – ] = 0.11 -use HH to get new pH = 5 + log (0.11 / 0.09) = 5.087 _______________________________________________ now consider,100mL of 0.1 M NaOH added to 1 L without the weak acid to see how well the weak acid buffers 0.01 moles OH – / 1.1L = 9.09 x 10 -3 = [OH – ] use K w = [OH – ] [H + ] = 1 x 10 -14 to get pH = 11.96 _______________________________________________ what happens when 0.1 moles of base have been added? what happens when the next 1 mL of base is added? Known as overrunning the buffer

5. Sample Buffer Calculation (in text) -want to study a reaction at pH 4.00 -so to prevent the pH from drifting during the reaction, use weak acid with pK a close to 4.00 -- formic acid (3.75) -can use a solution of weak acid and its conjugate base -ratio of formate ion to formic acid required can be calculated from the Henderson - Hasselbalch equation: 4.00 = 3.75 + log [HCOO – ] / [HCOOH] [HCOO – ] / [HCOOH] = 10 0.25 = 1.78 -so can make a formate buffer at pH 4.0 by using equal volumes of 0.1 M formic acid and 0.178 M sodium formate -Alternatively, exactly the same solution could be prepared by titrating a 0.1 M solution of formic acid to pH 4.00 with sodium hydroxide. _______________________________________________ some buffer systems controlling biological pH: 1. dihydrogen phosphate-monohydrogen phosphate pKa = 6.86- involved in intracellular pH control where phosphate is abundant 2. carbonic acid-bicarbonate pKa = 6.37, blood pH control 3. Protein amino acid side chains with pKa near 7.0

6. Example of an ampholyte - molecule with both acidic and basic groups glycine: pH 1NH 3 + – CH 2 – COOH net charge +1 pH 6NH 3 + – CH 2 – COO – net charge 0 zwitterion pH 14 NH 2 – CH 2 – COO – net charge –1 pKa values carboxylate group2.3 amino group9.6 can serve as good buffer in 2 different pH ranges ______________________________________________ use glycine to define an important property isoelectric point (pI) - pH at which an ampholyte or polyampholyte has a net charge of zero. for glycine, pI is where: [NH 3 + – CH 2 – COOH] = [NH 2 – CH 2 – COO – ] can calculate pI by applying HH to both ionizing groups and summing (see text) yields: pI = {pK COOH + pK NH 3+ } / 2 = {2.3 + 9.6} / 2 = 5.95 pI is the simple average for two ionizable groups

7. polyampholytes are molecules that have more than 2 ionizable groups lysineNH 3 + - C- (CH 2 ) 4 - NH 3 + COOH titration of lysine shows 3 pKa’s: pH<2, exists in above form first pKa = 2.18, loss of carboxyl proton second at pH = 8.9 third at pH = 10.28 need model compounds to decide which amino group loses a proton first _____________________________________________ to determine pI experimentally use electrophoresis (see end of Chapter 2) 1. Gel electrophoresis-electric field is applied to solution of ions, positively charged ions migrate to cathode and negatively charged to anode, at it’s pI an ampholyte does not move because net charge = zero 2. Isoelectric focusing- charged species move through a pH gradient, each resting at it’s own isoelectric point _____________________________________________ Macromolecules with multiples of either only negatively or only positively charged groups are called polyelectrolytes polylysine is a weak polyelectrolyte - pKa of each group influenced by ionization state of other groups

8. Solubility of macroions (polyelectrolytes and polyampholytes, including nucleic acids and proteins) depends on pH. For polyampholytes: high or low pH leads to greater solubility (due to – or + charges on proteins, respectively) At the isoelectric pH although net charge is zero, there are + and – charges and precipitation occurs due to: - charge-charge intermolecular interaction - van der Waals interaction to minimize the electrostatic interaction, small ions (salts) are added to serve as counterions, they screen the macroions from one another Ionic Strength = I = ½  (M i Z i 2 ) (sum over all small ions)M is molarity Z is charge Consider the following 2 processes that can take place for protein solutions: 1. Salting in: increasing ionic strength up to a point (relatively low I ), proteins go into solution 2. Salting out: at high salt, water that would normally solvate the protein goes to solvate the ions and protein solubility decreases. Most experiments use buffers with NaCl or KCl