1 / 8

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)

sonel
Download Presentation

Buffers and the Henderson-Hasselbalch Equation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related