Henderson Hasselbalch Equation

1
Henderson - Hasselbalch Equation

• If ignoring x in added acid base is valid, ie
  the extent of dissociation is small, then for
  solution of weak acid and its conjugate base
  (salt)

Convenient but should be checked if not valid,
we can always solve quadratic.
2
Buffers

• A solution containing both
• a weak acid its salt.
• OR
• a weak base its salt.
• withstands pH changes when (limited) amounts
  of acid or base are added.
• Reason Le Châteliers principle.

• vice-versa if add base

3
Example

• Find pH of solution which is 0.1 M in NH4 and
  0.2 M in NH3 where pKa of NH4 9.2

before 0.2 0.1
0 after 0.2 x 0.1 x
x
x 0.2/0.1 ? 10 4.8 (indeed x ltlt 0.1) pOH
log x 4.5 so pH 9.5
4
Effect of pH Change on Buffer

• Consider change in pH of pure water (pH 7) if
  we add 103 M HCl
• H 103 M (can neglect amount already
  present in water), so pH goes from 7 to 3
• Huge change.

5
Example

• Consider a buffer solution with 0.1 M each of
  sodium acetate (NaA) acetic acid (HA)
• found last week that pH 4.7
• What is the pH when 103 M HCl is added?

initially 0.1 103
0.1 eqm 0.1103x x 0.1103
x ie suppose all but x of the added H HA but
x will be very small...
6

• x 1.02 ? Ka 0.000020 ltlt 0.001
• pH log x 4.69
• the pH hardly changes from 4.7!
• Solution is buffered against pH change

CH3COOH OH-? H2O CH3COO-
H2O CH3COOH ? H3O CH3COO-
Silberberg Fig 19.2
7
Buffer Capacity
Buffer Preparation

• If pH of required buffer is pKa of available
  acid use equimolar amounts of acid and conjugate
  base
• If the required pH differs from the pKa use the
  Henderson-Hasselbalch equation

• Buffer capacity amount of added acid or base
  without significant pH change
• Depends on amount of acid conjugate base in
  solution highest when HA ? A

8
Buffers in Natural Systems

• Biological fluids, eg blood, contain buffers pH
  control essential because biochemical reactions
  are very sensitive to pH
• Human blood is slightly basic, pH ? 7.39 7.45
• In a healthy person, blood pH is never more than
  0.2 pH units from its average value
• pH lt 7.2, acidosis pH gt 7.6, alkalosis
• Death if pH lt 6.8 or gt 7.8

9
Buffer System in Blood

• extracellular buffer (outside cell)

• Removal of CO2 shifts equilibria to right,
  reducing H , ie, raising the pH
• The pH can be reduced by

10
Another Blood Buffer

• phosphate buffer, present inside cells
  (intracellular buffer)
• H2PO4 and HPO42

from H3PO4 , a tribasic (triprotic) acid
11
Polyprotic Acids

• removing more protons is harder increasing pKa
  decreasing Ka K a1 gt Ka2 gt Ka3
• reason harder to remove ve charge against
  increasing ve charge
• large difference in pKa values ? only need to
  consider one equilibrium at a time (simplifies
  maths)

12
Example

• Buffer 0.5 M NaH2PO4 0.4 M Na2HPO4
• Which Ka value is most important and what is the
  pH of the solution?
• Because we have approximately the same amounts of
  H2PO4 HPO42 we only need to consider pKa2
• Could go through whole procedure, or simply use
  Henderson-Hasselbalch equation

( just usual approximation of ignoring x ) pH
7.2 log(0.4/0.5) 7.1