Chem 142: Lecture 8B
Advanced concepts of Buffers (8.3)
Buffer Capacity (8.4)
Buffer Capacity independent of insult
Buffer before Insult Base is Added
pH of a solution that is 0.500 M in both acetic acid (HC2H3O2) and
sodium acetate (NaC2H3O2) with Ka = 1.8 x 10-5?
HAc(aq) Ac − (aq) + H + (aq )
[ Species ] [ Init ] [ ∆ ]
 H +   Ac −  X ( 0.5 + X )
K A = QC =
=
= 1.8 ⋅10−5
[ HAc ] = 0.50 − X
0.5 − X
[ HAc ]
 H +  =
+X
0
( 0.5 − X ) 1.8 ⋅10−5
 H +  =
( 0.5 + X )
 Ac −  = 0.50 + X
•


X =  H +  = 1.8 ⋅10−5 = K A
pH = pK A = 4.74
We will c compare with the exact treatment , including water ,
the Full Henderson Hasselbalch Equation. Not too different.
[ HAc ] = K
 H +  = K A
 Ac − 
(
(
)
)
C Acid −  H +  − OH − 
A
CSalt +  H +  − OH − 
Only term missing;
usually very small
Add an Insult base to a buffer
•
Add 0.010 mol of NaOH to 1.0 L of the acetic acid/acetate
solution. Adding NaOH as a solid avoids volume dilution.
What is the pH?
HAc(aq ) Ac − (aq) + H + (aq ) where Ac − ≡ C2 H 3O2 −
•
The base reduces the amount of acid, so (LeChatelier’s
principle) equilibrium is shifted to the right as H+
concentration goes down.
Concentration of acetic acid present initially? [ HAc ]o = 0.500 M
•
OH- then converts acetic acid to acetate!!
•
[ HAc ]o = ( 0.500 − 0.010 )
 Ac −  = ( 0.500 + 0.010 )
o
= 0.490 = C Acid − CInsult , Base
= 0.510 = CSalt + CInsult , Base
Buffer After Insult Base was Added
•
Use new initial concentrations and recalculate pH:
HAc(aq) Ac − (aq) + H + (aq )
[ Species ] [ Init ] [ ∆ ]
[ HAc ] = 0.49 − X
 H +  =
0
+X
 Ac −  =
0.51
+X
 H +   Ac −  X ( 0.51 + X )
K A = QC =
=
= 1.8 ⋅10−5
0.49 − X
[ HAc ]
( 0.49 − X ) 1.8 ⋅10−5 = 0.96 ⋅1.8 ⋅10−5 = 1.7 ⋅10−5
 H +  =
( 0.51 + X )
X =  H +  = 1.8 ⋅10−5 = 0.96 K A
pH = pK A − log ( 0.96 ) = 4.74 + 0.02 = 4.76
The pH did increase but very little. (increase is +0.02 pH units)
[ HAc ]0 = 0.50 − 0.01 = C Acid − CNaOH
 Ac −  = 0.50 + 0.01 = CSalt + CNaOH
0
How would you set up the initial acid and salt conditions if a
strong acid (0.01 M HCl) were added (instead of the NaOH)?
Compare effect of NaOH on water.
•
What if we just put 0.010 moles of NaOH into 1.0 L
of water?
•
Concentration of hydroxide ion is simply:
0.010 moles
= 0.010 M OH1.0 L
pOH = − log(0.010) = 2.0 ⇒ pH = 12.0
•
And the pH is:
•
In this case the pH moved from 7 to 12, or 5 units.
•
This is dramatically different than the value of 4.76
observed for the same amount of base the buffer,
and a change of 0.02 pH units.
HH Expression: Rearrange KA
−
H+ 

A
HAc]
[

+
KA =
or H  = KA −
 A 
[ HAc]
[ HAc] 
−log H+  = pH = pKA − log  − 
  A  
 [A − ] 
pH = pK a + log 

 [HA] 
Henderson-Hasselbalch Equation
The pH in a buffer is
determined by the equil.
between HA and A- when the
insult [H+] or [OH-] is much
less than [HA] and [A-].
Utility of HH Expression.
[ HA]
 H  = K A −
 A 
+
The reason the fully correct form of the KA in the HendersonHasselbalch form has utility:
The actual amounts of Acid and Anion can be replaced by the
initial amounts. This is the Simple Henderson-Hasselbalch
(SHH) form that is very accurate as long as the ratio is roughly
between 0.1 and 10. And concentrations are ‘large’ w.r.t. [H+].
 [ HA] 
HA
[
]
SHH :  H +  = K A − o or pH = pK A − log  − o 
 A 
  A  o 
o
[ HA]o C Acid
If we add a base: (Use LeC’s Principle)
=
The amount of A- will go up/down/same?
CSalt
 A− 
The amount of HA will go up/down/same?
o
The pH will go up/down?
HH: Insult Modifies HA/A- values.
with your new A-/HA
concentrations.
This is the The central Assumption! The amount
of A-/HA in your buffer will be reduced exactly in
the amount of H+/OH- you have added.
Add Insult Base
[ HA]o, New = [ HA]o − OH −  = C Acid − CBase, Insult
 A− 
=  A−  + OH −  = CSalt + CBase, Insult
o , New
o
Modify these equations when the insult is a strong acid instead of a strong base.
Summary: Features of Buffer
Buffers work when pH is near pKA of acid or pOH near pKB of base.
Works when both the acid concentration and the salt concentration are large
compared to the pH range applicable (so the weak limit CA,CS>> KA)
Go from the full or exact HH, FHH, form to the SHH form gives us a way to directly
compute the pH of a buffer. Easy to check if correct. Useful only if the pH is within
one unit of the pKA. So the ratio of Salt to buffer is between 0.1 and 10.0 SHH shows
this nicely. The working equation is the SHH. And it really is a rearrangement of the KA
expression.
Use the charge balance to show several ways to make a buffer. Add NaOH to HAc and
buffer at pH=pKA, because fA-= ½. This gets you to form a buffer. Fractional
dissociation can go between .1 and 10. and the buffer is still able to do its job.
Compare adding a bit of base (the Insult Base, Cinsult) to water vs adding the base to
the acetic acid buffer, and the pH moves from 7 to 12 without the buffer and barely
moves away from 5 with the buffer.
Now need to quantitatively define a buffer, B.C. = CI/deltapH. Large B.C. means the
pH moved very little when the insult acid/base was added. Get simple formula for B.C.
Buffer and Charge Balance
The Charge Balance a complementary approach:
• An alternative to the NICE table
• An alternative to thinking about how concentrations are
moved.
• Gives exact expressions and includes the insult acid/base.
Let’s review a problem we already did: To a buffer of 0.5 M acetic acid/ sodium acetate is
added 0.01 M NaOH. What is the old, and new pH and the change in the pH. What is the
buffer capacity. We found pH moved from 4.74 to 4.76 (note an increase).
List All species present.
From the list of all species (only) generate the (exact or full) charge balance expression.
Use the material (or mass) balance to relate actual concentration to initial (or total)
concentrations.
Buffer and Charge Balance
To a buffer of 0.5 M acetic acid/ sodium acetate is added 0.01 M NaOH. What is the old,
and new pH and the change in the pH. What is the buffer capacity. We found pH moved
from 4.74 to 4.76 (note an increase).
List All species present.
Na+ -- Spectator ion from two sources (Acid-Salt) and insult base.
H+ and OH- -- always there, the reactions are in water after all.
Weak Acid -- Generates both HA and A-.
From the list of all species (only) generate the (exact or full) charge balance expression.
 Na +  +  H +  = OH −  +  A− 
Material (or Mass) balance expressions:
 Na +  = CSalt + CInsult
 A−  + [ HA] =  A−  + [ HA]0 = C Acid + CSalt
0
K
+
−
+
δ =  H  − OH  =  H  − W
 H + 
HH and Charge Balance
Develop the HH expression:
The Charge Balance gives us an exact expression for [A-]
From the Charge balance add [HA} to both sides and use mass balance to obtain an exact
expression for [HA].
+
−
 Na  + δ =  A 
 A−  =  Na +  + δ = CSalt + CInsult + δ
Develop an equation of [HA] from Charge Balance (Add [HA] to both sides,
simplify)
[ HA] +  Na +  + δ =  A−  + [ HA] = CSalt + C Acid
[ HA] + CSalt + CInsult + δ =  A−  + [ HA] = CSalt + C Acid
[ HA] = C Acid − CInsult − δ
Insert these equations for [HA] and [A-] into the KA expression:
FHH
Expression
[ HA] = K
 H  = K A −
 A 
+
C Acid − CInsult − δ
A
CSalt + CInsult + δ
Drop the small corrections, called ‘delta’ from the r.h.s. and get SHH, a very usable
expression; same as the one we have been using all along. Now the limits on the
SHH equation can be discerned.
Adapt this expression for an insult Acid.
Buffer Capacity
•
•
•
Buffer capacity is defined as the amount of (insult) strong acid
or base, CI, that can be absorbed by the buffer per unit
change in the pH. Therefore the Buffer Capacity has the units
of concentration or Molarity.
The buffer capacity, B.C., depends on the concentrations (i.e.
Molarity) of the (weak) acid and conjugate base used.
CI .
B.C. ≡
∆pH
Back to the acetic acid/acetate buffer: To a buffer of 0.5 M
acetic acid/ sodium acetate is added 0.01 M NaOH. What is
the old, and new pH and the change in the pH? What is the
buffer capacity? We found pH moved from 4.74 to 4.76 (note
an increase). What is the Buffer Capacity:
CI .
0.01M
B.C. ≡
=
= 0.5M
∆pH
0.02
The amount of the weak acid and salt
defined the buffer capacity. The buffer can
absorb up to its amount.
Compare Buffer Capacity
A. What is the pH of a solution that is 0.50 M in both acetic acid
and acetate?
B. What is the pH of a solution that is 0.050M in both acetic acid
and acetate?
C. What is the new pH of these two solutions after 0.03M NaOH
has been added? Compare the capacity, B.C., of these two
buffers. (NaOH added as a solid; No Volume Dilution.)
• For all 4 problems use:
−
  Ac  

0 
and pK a = 4.74
SHH : pH = pK a + log 
 [ HAc ]0 


A) pH = pK a and B) pH = pK a
• The pH of both solutions is the same. This means dilution with
water will not move the pH of a buffered solution. But which
has the greater buffer capacity against an insult?
Comparing Buffer Change
•
Using SHH: The pH for solution after 0.03M NaOH is added to
the 0.050M Hac/Ac- Buffer.
  Ac −  
 0.080 

0 
SHH : pH = pK a + log 
= pK a + log 

 [ HAc ]0 
0.020 



C
0.03
pH = 4.74 + 0.60 = 5.34 ∆pH = 0.60 B.C. ≡ I . =
= 0.050M
∆pH 0.60
•
What if we did the same calculation for the 0.500 M buffer
solution?
−
  Ac  
 0.530 

0 
SHH : pH = pK a + log 
= pK a + log 

 [ HAc ]0 
0.470 



C
0.03M
pH = 4.74 + 0.05 = 4.79 ∆pH = 0.052 B.C. ≡ I . =
= 0.57 M
∆pH
0.052
•
•
NB: The change in the pH, is independent of the pKA.
The higher concentration buffer solution has a greater buffer
capacity; as one would expect. B.C. ~ Concentration Acid.
What pH to Buffer?
•
Buffer solutions are used to keep the pH balanced at a target
pH.
•
Buffers are most effective with the concentrations of weak
acid and conjugate base are similar (within a factor of 10).
•
This can be accomplished by picking a weak acid where the
pKa of the acid is close to the desired pH.
•
Example: Calculate the ratio of weak acid to conjugate base
needed to make a 4.30 pH solution using chloroacetic acid (Ka
= 1.35 x 10-3) and benzoic acid (Ka = 6.4 x 10-5)
Setting the pH with the ratio
•
Using the Henderson-Hasselbalch Eq. for the ClAc buffer:
 ClAc −  
 ClAc −  



 ⇒ pH − pK a = log  
HH : pH = pK a + log 
 [ HClAc] 
 [ HClAc] 




−
 ClAc−  


ClAc





4.30 − 2.87 = 1.43 = log
∴
= 101.43 = 26.9
 [ HClAc]  [ HClAc]


•
Repeating for BzA:
  BzA −  
  BzA −  



 ⇒ pH − pK a = log  
pH = pK a + log 
 [ HBzA ] 
 [ HBzA ] 




  BzA −    BzA − 
 ∴ 
 = 100.1 = 1.26
4.30 − 4.20 = 0.10 = log  
 [ HBzA ]  [ HBzA ]


Benzoic Acid-based buffer has a higher B.C. because the ratio is closer to 1. The ClAc
will buffer well against insult acid but not against base because it is mostly made of
anion; and the pH is far more basic than the pKA.
CInsult
B.C.
  H + ( CI )  


+
+
The definition of the change in pH: ∆pH = − log  H ( CI )  − − log  H ( CI = 0 )  = − log  +



H
0
(
)
 
 
Develop the expression for the Change in pH:
A general approach to pH Change.
(
)
C
[ HA]
 H + ( CI = 0 )  = K A − = K A Acid
CSalt
 A 
[ HA]
C − CInsult
 H + ( CI )  = K A − = K A Acid
CSalt + CInsult
 A 
The SHH before and after adding insult base.
The SHH after adding base:
∆pH =
SHH
Expressions
The Ratio of the two (no KA left). The ratio of H+ is in terms of change of acid times change
of salt concentration ratios.  H + ( CI )   C − C  

C

=

+
 H ( 0 )  
Acid
C Acid
Insult
Salt


C
+
C
Insult 
  Salt
Ratio converted into the pH change ( using the above definition )
  H + ( CI )  
  = − log  C Acid − CInsult
∆pH = − log   +

  H ( 0) 
C Acid

 
 
 C

 C

∆pH = − log 1 − Insult  + log 1 + Insult 
CSalt 
 C Acid 


 CSalt + CInsult 
 + log 

C
Salt



This is exactly the same form we used to compute the pH change in the previous
examples, but there we computed the pH before and after. We did notice the change
was independent of the KA, and this fits that. Here the change is the sum of the two
terms, one for the acid and the other for the salt change.
Redo the pH Change and compare with Algebraic Reduction
 CInsult
∆pH = − log 1 −
 C Acid

 CInsult 
 + log 1 +

C
Salt 


What is the change in pH of the Acetic Acid/Acetate buffer at 0.050M each, when
0.030 M NaOH is added (no dilution, magically). NB: We do not need the KA to
solve this problem. This could be any acid/salt buffer system. (Later: Do not
need the Insult Concentration to get the buffer capacity.)
The change in pH before was 0.60 and 0.052 For the 0.05 and 0.5 Acetate buffers.
Now , with the above expression, we get the same answer.
How to make a useful approximation.
∆pH ≈
1  CInsult CInsult
⋅
+
2.303  CSalt C Acid
log (1 + x ) =

1  1
1
=
⋅
+


 2.303  CSalt C Acid
ln (1 + x )
x
≈
2.303
2.303

 CInsult

Let’s compare this approximation with the exact answer on the previous slide. For the
0.050M buffer, the change in pH was 0.60 vs 0.52 here (not bad) and for the 0.50 the
change in pH was 0.052 vs 0.052 here (the same). So it is a very useful approximation.
1  0.03
 1
∆pH ≈ 
+
= 0.52

 0.05 0.05  2.303
 CSalt ⋅ C Acid 
B.C. = 2.303 ⋅ 

+
C
C
Acid 
 Salt
Ways to Make an optimal Buffer (SHH)
An optimal buffer occurs when the salt concentration equals the acid concentration.
Or when the exact concentrations of anion and un-dissociated acid are equal.
Then pH=pKA.
Look at the SHH formula when pH=pKA
HA] C Acid − CNaOH
[ HA]
C − CInsult
[
 H +  = K A − = K A Acid
@ OPT ⇒ 1 = − =
CSalt + CInsult
 A 
 A  CSalt + CNaOH
Many ways to make an optimum buffer. The simplest is add no NaOH, then
CSalt = C Acid
This is the typical condition we have had all along. But there are more
possibilities, we can include the NaOH in the buffer as long as
CSalt + CNaOH = C Acid − CNaOH
So if we set the Csalt=0, we can make a buffer if we add ½ as much NaOH as
weak acid. This makes EXACTLY the same buffer as if we mixed weak acid/salt 1:1.
Conversely we can make a buffer with the salt of the weak acid and a strong acid
like HCl. But you can’t make a buffer from the weak acid and a strong acid, nor
can you make a buffer from the salt of a weak acid and a strong base.
Ways to Make an Optimal Buffer (Charge Balance)
An optimal buffer occurs when the salt concentration equals the acid concentration.
Or when the exact concentrations of anion and un-dissociated acid are equal.
Then pH=pKA.
What is the dissociation fraction when pH=pK?
KA
1
1
−


A
HA
=
⇒
=
=
( CSalt + C Acid )
[
]


+
2
2


KA + H 
Let’s review the buffer where pH=pKA. We added equimolar amounts of acid and
salt of the acid. The Charge Balance is:
+
+
−
−
f A− =
 Na  +  H  = OH  +  A 
And the mass balance requires:
Substitute the mass balance into
The charge balance gives:
CSalt
CSalt
1
Na+  = CSalt and  A−  = ( CAcid + CSalt )
2
1
+  H  = OH  + ( C Acid + CSalt )
2
+ 2  H +  = 2 OH −  + C Acid
+
CSalt ≈ C Acid
−
The charge balance shows that
the optimal buffer is found
when the salt and acid are
nearly in equimolar amounts.
There are other ways to get to
exactly this same set of
concentrations.
Charge Balance and the optimal Buffer
Let’s develop other methods of moving the pH so that pH=pKA. Consider the Acid
only, and add NaOH so that pH=pK. How much strong base (NaOH) do you need
to add to get the same pH as the optimum buffer?
NB: We are not adding the salt of the weak acid; Just the weak acid.
Use the Charge Balance, and the mass balance (parallel to previous work):
1
Na+  = CNaOH and  A−  = CAcid
2
 Na +  +  H +  = OH −  +  A− 
1
+
−
CNaOH +  H  = OH  + C Acid
2
1
CNaOH ≈ C Acid
2
If we add half as much strong base
as weak acid present, we will have
exactly the same concentration of
all ions present as the classic buffer
we have been working with. Need
to double the initial weak acid C.
This expression also contains the
very slight correction for the pH, but
we are in the weak acid limit where
Cacid >> KA= [H+].
Design a buffer where you add just the salt of a weak acid (eg NaAc) and the strong
acid HCl. What condition is there on the amount of HCl needed to get pH=pKA in this
case? [Hint: You need to include the chloride ion in the charge balance.]
Charge Balance and the optimal Buffer
Describe a way to make a buffer with a 0.02 M HCN solution and either NaOH or
HCl, that buffers optimally. What is its buffer capacity?
Compare the buffer capacity of this solution wit that of a solution that has 0.01M
HCN and 0.010 M NaCN.
If you have solution that contains .01 M NaAc and .04 M HAc, how much NaOH
would you add to make this an optimal buffer. What is its buffer capacity when it is
optimal? (Neglect any volume dilution.)
Explain why adding HCl will work to make a solution that is .01M HAc and 0.040 M
NaAc will make an optimal buffer but NaOH will not.