Buffers and the Henderson-Hasselbalch Equation
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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)

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Buffers and the Henderson-Hasselbalch Equation

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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)

--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


Buffers and the henderson hasselbalch equation

  • 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–]

  • thenKa = 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


Buffers and the henderson hasselbalch equation

  • 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


Buffers and the henderson hasselbalch equation

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


Buffers and the henderson hasselbalch equation

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


Buffers and the henderson hasselbalch equation

Example of an ampholyte - molecule with both acidic and basic groups

glycine: pH 1NH3+ – CH2 – COOH net charge +1

pH 6NH3+ – CH2 – COO– net charge 0

zwitterion

pH 14 NH2 – CH2 – 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:

[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


Buffers and the henderson hasselbalch equation

  • polyampholytes are molecules that have more than 2 ionizable groups

  • lysineNH3+- 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


Buffers and the henderson hasselbalch equation

  • 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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