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Buffers and the Henderson-Hasselbalch Equation

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)

ardelle-york
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Buffers and the Henderson-Hasselbalch Equation

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  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: Ka = [H+] [A–] / [HA] take (– log) of each side and rearrange, yields: pH = pKa + 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 Ka = [H+] [A–] / [HA] • HCOOH H+ + HCOO– • let x = [H+] = [HCOO–] • then Ka = 1.78 x 10 –4 = x2 / (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 Ka = 1.78 x 10 –4 = x2 / 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+ , Kw = [H+] [OH–] = 10–14 , • Kw 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 Ka • Ka = [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 = pKa + 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 = pKa + log 1 = pKa = 3.75 for HCOOH Titration curve: - within 1 pH unit of pKa over most of curve - so pKa defines the range where buffering capacity is maximum - curve is reversible

  4. Simple problem: -have one liter of a weak acid (pKa = 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 = pKa -add 100mL of 0.1M NaOH, following occurs HA + OH– = A– + H2O 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 Kw = [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 pKa 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 1 NH3+ – CH2 – COOH net charge +1 pH 6 NH3+ – CH2 – COO– net charge 0 zwitterion pH 14 NH2 – CH2 – COO– net charge –1 pKa values carboxylate group 2.3 amino group 9.6 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: [NH3+ – CH2 – COOH] = [NH2 – CH2 – 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 • lysine NH3+- C- (CH2)4 - NH3+ • 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  (Mi Zi2) • (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

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