pH of Polybasic acid buffers

pH of Polybasic bases and
their buffers
1:1 buffers and non-1:1 buffers
plus the pH of the first weak base proton.

pH of Polybasic acid buffers
The concept of the equilibriums involve with polyprotic acids
and amphoteric bases has been presented in a previous slide
show titled “Amphoteric (amphiprotic) acids and bases”
Please review this before continuing if you have not mastered
the subject.

pH of Polybasic acid buffers
Here are some examples (from the previous slide show) of
polyprotic acids along with their Ks.
H 2 NNH 2 + H 2 O º H 2 NNH 3 + + OH ! K b1 = 1.2 × 10 –6
H 2 NNH 3 + + H 2 O º H 3 NNH 3 2+ + OH ! K b2 = 8.9 × 10 -16
H 2 N(CH 2 ) 6 NH 2 + H 2 O º H 2 N(CH 2 ) 6 NH 3 + + OH ! K b1 = 7.2 × 10 –3
H 2 N(CH 2 ) 6 NH 3 + + H 2 O º H 3 N(CH 2 ) 6 NH 3 2+ + OH ! K b2 = 5.8 × 10 -4
H 2 N(CH 2 ) 2 NH 2 + H 2 O º H 2 N(CH 2 ) 2 NH 3 + + OH ! K b1 = 5.2 × 10 –4
H 2 N(CH 2 ) 2 NH 3 + + H 2 O º H 3 N(CH 2 ) 2 NH 3 2+ + OH ! K b2 = 3.7 × 10 -7
H 2 N(CH 2 ) 3 NH 2 + H 2 O º H 2 N(CH 2 ) 3 NH 3 + + OH ! K b1 = 8.7 × 10 –4
H 2 N(CH 2 ) 3 NH 3 + + H 2 O º H 3 N(CH 2 ) 3 NH 3 2+ + OH ! K b2 = 1.1 × 10 -5
Given these data, calculate the pH of a 0.10 M solution of each
of the pure bases. Answers are on the next slide.

pH of Polybasic acid buffers
Answers to the question, “What is the pH of each of the
bases if they are made up to be 0.10 M?”
H 2 NNH 2 pOH = 3.46 Y pH =10.54
H 2 N(CH 2 ) 6 NH 2 pOH = 1.57 Y pH =12.43
H 2 N(CH 2 ) 2 NH 2 pOH = 2.14 Y pH =11.86
H 2 N(CH 2 ) 3 NH 2 pOH = 2.03 Y pH =11.97

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pOH .pK b

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pH .pK a
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in H 2 NNH 2 and 0.10 M in H 2 NNH 3 Cl.

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pH .pK a
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in H 2 NNH 2 and 0.10 M in H 2 NNH 3 Cl.
The two reactions associated with H 2 NNH 2 are:
H 2 NNH 2 + H 2 O º H 2 NNH 3 + + OH !
and H 2 NNH 3 + + H 2 O º H 3 NNH 3 2+ + OH !
Which one should be used?

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pH .pK a
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in H 2 NNH 2 and 0.10 M in H 2 NNH 3 Cl.
The two reactions associated with H 2 NNH 2 are:
H 2 NNH 2 + H 2 O º H 2 NNH 3 + + OH !
and H 2 NNH 3 + + H 2 O º H 3 NNH 3 2+ + OH !
Which one should be used?
Notice that H 2 NNH 2 and H 2 NNH 3 + are relevant.
(Cl – ions are spectator ions. If this is mysterious to you (tish,
tish), see the CHEM 1110 slide show “Overall versus Net
Ionic Reactions” for a review!)

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pH .pK a
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in H 2 NNH 2 and 0.10 M in H 2 NNH 3 Cl.
The two reactions associated with H 2 NNH 2 are:
H 2 NNH 2 + H 2 O º H 2 NNH 3 + + OH !
and H 2 NNH 3 + + H 2 O º H 3 NNH 3 2+ + OH !
Which one should be used?
Notice that H 2 NNH 2 and H 2 NNH 3 + are relevant.
Since this is a 1:1 buffer, one need only set the pOH to the
pK b
Kb1 = 1.2 × 10 –6
ˆ pOH = 5.92 Y pH = 8.08

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pH .pK a
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in H 2 NNH 2 and 0.10 M in H 2 NNH 3 Cl.
The two reactions associated with H 2 NNH 2 are:
H 2 NNH 2 + H 2 O º H 2 NNH 3 + + OH !
and H 2 NNH 3 + + H 2 O º H 3 NNH 3 2+ + OH !
Which one should be used?
Notice that H 2 NNH 2 and H 2 NNH 3 + are relevant.
Since this is a 1:1 buffer, one need only set the pOH to the
pK b
Kb1 = 1.2 × 10 –6

pH of Polybasic acid buffers
Recall that for the 1:1 buffer, the pH .pK a
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in H 2 NNH 2 and 0.10 M in H 2 NNH 3 Cl.
The two reactions associated with H 2 NNH 2 are:
H 2 NNH 2 + H 2 O º H 2 NNH 3 + + OH !
and H 2 NNH 3 + + H 2 O º H 3 NNH 3 2+ + OH !
Which one should be used?
Notice that H 2 NNH 2 and H 2 NNH 3 + are relevant.
Since this is a 1:1 buffer, one need only set the pOH to the
pK b
Kb1 = 1.2 × 10 –6
ˆ pOH = 5.92 Y pH = 8.08

pH of Polybasic acid buffers
Here’s some more examples, determine the answers:
Example 2: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 6 NH 3 Cl and 0.10 M in H 3 N(CH 2 ) 6 NH 3 Cl 2 .
Example 3: Calculate the pH of a solution made up to be
0.50 M in H 2 N(CH 2 ) 2 NH 2 and 0.50 M in H 2 N(CH 2 ) 2 NH 3 Cl .
Example 4: Calculate the pH of a solution made up to be
0.35 M in H 2 N(CH 2 ) 3 NH 3 Br and 0.35 M in H 3 N(CH 2 ) 3 NH 3 Br 2 .
The answers are on the next slide.

pH of Polybasic acid buffers
Here’s some more examples, determine the answers:
Example 2: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 6 NH 3 Cl and 0.10 M in H 3 N(CH 2 ) 6 NH 3 Cl 2 .
Answer: pH = pK b2 ˆ pOH = 3.24 Y pH = 10.76
Example 3: Calculate the pH of a solution made up to be
0.50 M in H 2 N(CH 2 ) 2 NH 2 and 0.50 M in H 2 N(CH 2 ) 2 NH 3 Cl .
Answer: pH = pK b1 ˆ pOH = 3.28 Y pH = 10.72
Example 4: Calculate the pH of a solution made up to be
0.35 M in H 2 N(CH 2 ) 3 NH 3 Br and 0.35 M in H 3 N(CH 2 ) 3 NH 3 Br 2 .
Answer: pH = pK b2 ˆ pOH = 4.96 Y pH = 9.04
These are all 1:1 buffer so pH .pK a is OK.

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
So, which of these reactions is relevant?
H 2 N(CH 2 ) 2 NH 2 + H 2 O º H 2 N(CH 2 ) 2 NH 3 + + OH ! K b1 = 5.2 × 10 –4
H 2 N(CH 2 ) 2 NH 3 + + H 2 O º H 3 N(CH 2 ) 2 NH 3 2+ + OH ! K b2 = 3.7 × 10 -7

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
So, which of these reactions is relevant?
H 2 N(CH 2 ) 2 NH 2 + H 2 O º H 2 N(CH 2 ) 2 NH 3 + + OH ! K b1 = 5.2 × 10 –4
H 2 N(CH 2 ) 2 NH 3 + + H 2 O º H 3 N(CH 2 ) 2 NH 3 2+ + OH ! K b2 = 3.7 × 10 -7
It is the second reaction.

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
H 2 N(CH 2 ) 2 NH 3 + + H 2 O º H 3 N(CH 2 ) 2 NH 3 2+ + OH ! K b2 = 3.7 × 10 -7

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH
x 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
H 2 N(CH 2 ) 2 NH + 3 + H 2 O º H 3 N(CH 2 ) 2 NH 2+ 3 + OH ! K b2 = 3.7 × 10 -7
This then is a normal type II buffer problem. Set up a table:
[H 2 N(CH 2 ) 2 NH + 3 ]
[OH – ]
[H 3 N(CH 2 ) 2 NH 2+ 3 ]
before equilibrium after equilibrium
0.10
0.0
0.10 ! x
x
0.50
0.50 + x

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
x
H 2 N(CH 2 ) 2 NH + 3 + H 2 O º H 3 N(CH 2 ) 2 NH 2+ 3 + OH ! K b2 = 3.7 × 10 -7
This then is a normal type II buffer problem. Set up a table:
[H 2 N(CH 2 ) 2 NH + 3 ]
[OH – ]
[H 3 N(CH 2 ) 2 NH 2+ 3 ]
before equilibrium
0.10
0.0
0.50
after equilibrium
0.10 ! x
x
0.50 + x
Fill in the
equilibrium
expression:
K b
− 2+
⎡OH
⎤⎡H3N(CH 2) 2NH
⎤
3
=
⎣ ⎦⎣ ⎦
+
⎡
⎣HN(CH)NH
⎤
2 2 2 3⎦

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
x
H 2 N(CH 2 ) 2 NH + 3 + H 2 O º H 3 N(CH 2 ) 2 NH 2+ 3 + OH ! K b2 = 3.7 × 10 -7
This then is a normal type II buffer problem. Set up a table:
[H 2 N(CH 2 ) 2 NH + 3 ]
[OH – ]
[H 3 N(CH 2 ) 2 NH 2+ 3 ]
before equilibrium after equilibrium
0.10
0.0
0.10 ! x
x
0.50
0.50 + x
Fill in the
equilibrium
expression:
( x)( x)
( 0.10 − x)
− 7
0.50 +
3.5× 10 =

pH of Polybasic acid buffers
What if the mixture is not a 1:1 buffer?
Example 5: Calculate the pH of a solution made up to be
0.10 M in H 2 N(CH 2 ) 2 NH 3 Cl and 0.50 M in H 3 N(CH 2 ) 2 NH 3 Cl 2 .
x
H 2 N(CH 2 ) 2 NH + 3 + H 2 O º H 3 N(CH 2 ) 2 NH 2+ 3 + OH ! K b2 = 3.7 × 10 -7
This then is a normal type II buffer problem. Set up a table:
[H 2 N(CH 2 ) 2 NH + 3 ]
[OH – ]
[H 3 N(CH 2 ) 2 NH 2+ 3 ]
before equilibrium after equilibrium
0.10
0.0
0.10 ! x
x
0.50
0.50 + x
Fill in the
equilibrium
expression:
( x)( x)
( 0.10 − x)
− 7
0.50 +
3.5× 10 =
and solve*: x = 7.0 × 10 -8
or: pOH = 7.15 Y pH = 6.85
* the approximation ±x works here.

pH of Polybasic bases and
their buffers
1:1 buffers and non-1:1 buffers
plus the pH of the first weak base proton.
THE END