Chem 142: Lecture 8A
Solutions (8.1) Containing a Common Ion
Develop Buffer Rules (8.2)
An alternative strategy to Buffers (8.3)
Buffer Capacity (8.4)
Solutions that contain a Common Ion
•
To this point we have calculated the pH of solutions of weak
acids and weak bases and the salts of these acids/bases.
What happens in a solution of a weak acid and its conjugate
base…the common-ion effect.
We will discover that solutions of this type are able to resist
large changes in pH…they are buffered solutions.
To Illustrate this: Find the pH of the following solutions...
1.
2.
3.
0.500 M hydrocyanic acid (HCN, Ka = 4.9 x 10-10)
0.500 M sodium cyanide (NaCN) – Salt of the acid
0.500 M hydrocyanic acid and 0.50 M sodium cyanide 50:50 mix of both.
•
•
•
pH of a weak acid
pH of a solution that is C=0.500 M in HCN (Ka) = 4.9 x 10-10
•
[ Species ] [ Init ] [ ∆ ]
[ HCN ] = 0.50 − X
+
 H  =
0
+X
CN −  =
0
+X
HCN (aq) CN − (aq) + H + (aq)
 H +  CN − 
X2
=
= 4.9 ⋅10−10
K A = QC =
0.5 − X
[ HCN ]
Exact Treatment, Includes water
X =  H +  = 1.6 ×10−5
pH = − log(1.6 ×10−5 ) = 4.8
 H +  = CN −  + OH − 
Use Fraction dissociation, multiply by [H+], take square root
 H +  =
KA
+

 C + KW
H
+ 
K A +  H 
X =  H +  Same as Simple Form K AC
To 4 sig figs, same.
Dilute acid by 100 fold and pH
increase by exactly 1 unit, to 5.8.
pH of a Salt of Weak Acid
•
pH of a solution that is CSalt=0.500 M in HCN (Ka) = 4.9 x 10-10
[ Species ] [ Init ] [ ∆ ]
CN −  =
0.50
−X
OH −  =
[ HCN ] =
0
+X
0
+X
Includes water,
CN − (aq) HCN (aq ) + OH − (aq )
K B, A
OH −  [ HCN ]
X2
1 ⋅10−14
−5
= QC =
=
= 2.0 ⋅10 =
−
0.5 − X
4.9 ⋅10−10
CN 
Exact Treatment
X = OH −  = 3.2 ×10−3
K
 = 11.50
pH = − log  W

3.2 ×10−3 

 Na +  +  H +  = CN −  + OH − 
Use Fraction dissociation, regroup charge, multiply by [OH-], take square
root
K B, A
−
Dilute acid by 100 fold and pH
OH −  =

 CSalt + KW
OH
− 
K B , A + OH 
decrease by exactly 1 unit, to 10.50.
X = OH −  Same as Simple Form K BACSalt
To 4 sig figs, same.
pH of a Salt with its Weak Acid
pH of a solution that is CSalt=0.500 M in HCN and CA=0.50M in NaCN
(KA = 4.9 x 10-10)
[ Species ] [ Init ] [ ∆ ]
HCN (aq) CN − (aq ) + H + (aq)
[ HCN ] =
0.50
−X
+
 H  =
0
+X
CN −  =
0.50
+X
 H +  CN −  X ( 0.5 + X )
=
= 4.9 ⋅10−10
K A = QC =
0.5 − X
[ HCN ]
Exact Treatment Includes water (now it must)
X =  H +  = 4.9 ×10−10 = K A
pH = − log(4.9 ×10−10 ) = pK A = 9.31
 Na +  +  H +  = CN −  + OH − 
 Na +  = CSalt and CN −  + [ HCN ] = CSalt + C Acid
CN −  = CSalt +  H +  − OH − 
Dilute acid by 100 fold and pH does
NOT change. , H moved to 4.94e-10.
[ HCN ] = C Acid + OH −  −  H + 
[ HCN ] = K
 H  = K A
CN − 
+
(
(
)
)
C Acid −  H +  − OH − 
A
CSalt +  H +  − OH − 
Gives Same answer as
simple form
pH of a Salt with its Weak Acid
•
•
pH of a solution that is CSalt=0.500 M in HCN (Ka) = 4.9 x 10-10
Problem can be solved with either reaction (the acid) or the salt of
the acid.
CN − (aq) HCN (aq ) + OH − (aq )
[ Species ] [ Init ] [ ∆ ]
CN −  =
0.50
OH −  =
[ HCN ] =
0
+X
0.50
+X
−X
NB: Algebraically the two
problems are equivalent
OH −  [ HCN ]
K [ HCN ]
 H +  = W −
K B , A  H  =
−
CN 
CN 
 H +  CN −  KW
=
= KA
K B, A
[ HCN ]
+
K B, A
K B, A
K B, A
OH −  [ HCN ]
= QC =
CN − 
X ( 0.5 + X )
1 ⋅10−14
−5
=
= 2.0 ⋅10 =
0.5 − X
4.9 ⋅10−10
X ( 0.5 + X )
( 0.5 − X ) = OH − 
=
or X = K B , A

0.5 − X
( 0.5 + X ) 
X = OH −  = 2.04 ×10−5
K

pH = − log  W
−5  = 9.31
2.04 ×10 

Buffer can be made with a weak acid or a weak base; here
HCN and NH3 buffer at almost the same pH (based on pK).
Make a Buffer from a weak base
pH of a solution that is 0.500 M in both NH4OH (base) and NH4CL
(Salt of the base) KB = 1.8 x 10-5
NH 3 (aq ) NH 4 + (aq ) + OH − (aq )
[ Species ] [ Init ] [ ∆ ]
 NH 4 +  OH −  X ( 0.5 + X )
K B = QC =
=
= 1.8 ⋅10−5
[ NH 3 ] = 0.50 − X
0.5 − X
[ NH 3 ]
OH −  =
+X
0
( 0.5 − X ) 1.8 ⋅10−5
OH −  = X =
( 0.5 + X )
 NH +  = 0.50 + X
•

4

NB: The problem is the same answer as
an acid buffer but you solve for OH- or
pOH and then convert to pH at the end.
The balance works the same way, more
base, CB increases the pH (makes the
solution more basic)
X = OH −  = 1.8 ⋅10−5 = K B
pOH = pK B = 4.74
pH = pK A, B = 14 − 4.74 = 9.26
CB
−


SHH : OH  = K B
CSalt
Buffer can be made with a weak acid or a weak base; here
HCN and NH3 buffer at almost the same pH (based on pK).
What will the addition of an insult acid (or base), CInsult, do to the pH? Use LeCh’s
Prin. to get the direction of the change correct.
Later , Compare with the exact Treatment , including water , the
Full Henderson Hasselbalch, FHH, Equation
Summary of a Buffer (Common Ion Effect)
•
•
•
•
•
•
Presence of the weak acid and its conjugate base changes the
initial concentration of both species. (Also true for solution of
a weak base and its conjugate acid.)
Notice that the pH of the 0.500 M HCN solution was less (pH =
4.8) than when CN- was present (9.3).
Conversely, the pH of a CN- solution (pH = 11.5) is greater than
when the acid, HCN, is present (pH = 9.3).
Both perspectives demonstrate that the presence of the
conjugate “inhibits” the dissociation of the acid / and proton
association of the base.
A weak acid and a weak base are mixed, the result (i.e. pH) is
in between.
A buffer is best described as the point where pH=pKA (or
pOH=pKB). At this point the dissociation fraction is ½.
Buffer Solutions: So What?
• Buffers are solutions of a weak acid and its conjugate base (or
a weak base and its conjugate acid).
• Buffers are able to absorb a certain quantity of OH- or H+ ions
(the insult to the buffer)without undergoing a strong variation
in pH.
• This works because...
– a weak acid HA is a “hidden” proton source (better than
water), so OH- is neutralized and replaced with A-.
– the conjugate base A- wants an extra proton (more than
water) so H+ is neutralized and replaced with HA.
• This works when...
– the concentration of added OH- or H+ ions is less than the
concentration of HA or A- needed to react.
Insult to a Buffer Solution
Buffer response to
added OH- ions:
The strong base OH- reacts
completely with weak acid HA,
which is the best available
proton source.
Buffer response to
increased H+ ions:
The strong acid H+ reacts
completely with weak base A-,
which is the best available
proton acceptor.
pH Up
slightly
increase. H+
Increased H3O+ ions
replaced by HA
pH Down
slightly
NB: We specify “increased H+“ rather than “added H+” because a strong acid added to solution
will ionize completely to give its conj. base and H+, the strongest acid that can exist in aqueous
solution.
Buffer: New initial conditions
with your new A-/HA
concentrations.
This is the KEY!
Add Insult Base
The amount of A-/HA in your buffer will
be reduced exactly in the amount of
H+/OH- you have added.
[ HA]o, New = [ HA]o − CB.I .
and  A− 
=  A−  + CB. I .
o , New
o
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
= 0.510
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
+
Ka =
−
[H 3O ][A ]
[HA]
The pH in a buffer is
[HA]
+
determined by the equil.
[H 3O ] = K a
−
[A ] between HA and A- when the
insult [H+] or [OH-] is much
less than [HA] and [A-].
 [HA] 
− log[H3O + ] = − log ( K a ) − log  − 
 [A ] 
 [HA] 
pH = pK a − log  − 
 [A ] 
 [A − ] 
pH = pK a + log 

 [HA] 
Henderson-Hasselbalch Equation
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?