pH of Polyprotic acids and
their buffers
1:1 buffers and non-1:1 buffers
plus the pH of the first weak acid proton.
Review the slideshow on Amphoteric Acids and
Bases in preparation.
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pH of Polyprotic acid buffers
The concept of the equilibria involved 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 Polyprotic acid buffers
Here are some examples (from the previous slide show) of
polyprotic acids along with their Ks.
H2SO4 + H2O º H3O+ + HSO4!
HSO4! + H2O º H3O+ + SO42!
Ka1 = very large
Ka2 = 2.0 × 10-2
H2SO3 + H2O º H3O+ + HSO3!
HSO3! + H2O º H3O+ + SO32!
Ka1 = 1.7 × 10–2
Ka2 = 6.24 × 10–8
H3PO4 + H2O º H3O+ + H2PO4!
H2PO4! + H2O º H3O+ + HPO42!
HPO4! + H2O º H3O+ + PO43!
Ka1 = 1.1 × 10–2
Ka2 = 7.5 × 10–8
Ka2 = 4.8 × 10–12
H2CO3 + H2O º H3O+ + HCO3!
HCO3! + H2O º H3O+ + CO32!
Ka1 = 4.31 × 10–7
Ka2 = 4.4 × 10–11
Given these data, calculate the pH of a 0.10 M solution of each
of the pure acids. Answers are on the next slide.
pH of Polyprotic acid buffers
Answers to the question, “What is the pH of each of the acids
if they are made up to be 0.10 M?”
H2SO4
pH = 1.00
(Recall it’s a strong acid.)
H2SO3
pH = 1.52
(Missed this one? Note x is not valid)
H3PO4
pH = 1.55
(""""""")
H2CO3
pH = 3.68
pH of Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
pH of Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in NaHSO3 and 0.10 M in Na2SO3.
pH of Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in NaHSO3 and 0.10 M in Na2SO3.
The two reactions associated with H2SO3 are:
H2SO3 + H2O º H3O+ + HSO3–
and HSO3– + H2O º H3O+ + SO32–
Which one should be used?
pH of Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in NaHSO3 and 0.10 M in Na2SO3.
The two reactions associated with H2SO3 are:
H2SO3 + H2O º H3O+ + HSO3–
and HSO3– + H2O º H3O+ + SO32–
Which one should be used?
Notice that it is the HSO3– and the SO32– that are relevant.
(Na+ 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 Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in NaHSO3 and 0.10 M in Na2SO3.
The two reactions associated with H2SO3 are:
H2SO3 + H2O º H3O+ + HSO3–
and HSO3– + H2O º H3O+ + SO32–
Which one should be used?
Notice that it is the HSO3– and the SO32– that are relevant.
Since this is a 1:1 buffer, one need only set the pH to the pKa
pH of Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in NaHSO3 and 0.10 M in Na2SO3.
The two reactions associated with H2SO3 are:
H2SO3 + H2O º H3O+ + HSO3–
and HSO3– + H2O º H3O+ + SO32–
Which one should be used?
Notice that it is the HSO3– and the SO32– that are relevant.
Since this is a 1:1 buffer, one need only set the pH to the pKa
Ka2 = 6.24 × 10–8
pH of Polyprotic acid buffers
Recall that for the 1:1 buffer, the pH .pKa
Example 1: Calculate the pH of the buffer solution made up to
be 0.10 M in NaHSO3 and 0.10 M in Na2SO3.
The two reactions associated with H2SO3 are:
H2SO3 + H2O º H3O+ + HSO3–
and HSO3– + H2O º H3O+ + SO32–
Which one should be used?
Notice that it is the HSO3– and the SO32– that are relevant.
Since this is a 1:1 buffer, one need only set the pH to the pKa
Ka2 = 6.24 × 10–8
ˆ
pH = 7.205
pH of Polyprotic 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 KH2PO4 and 0.10 M in K2HPO4.
Example 3: Calculate the pH of a solution made up to be
0.50 M in K2HPO4 and 0.50 M in K3PO4.
Example 4: Calculate the pH of a solution made up to be
0.35 M in Li2CO3 and 0.35 M in LiHCO3.
The answers are on the next slide.
pH of Polyprotic 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 KH2PO4 and 0.10 M in K2HPO4.
Answer: pH = pKa2 ˆ pH = 7.205
Example 3: Calculate the pH of a solution made up to be
0.50 M in K2HPO4 and 0.50 M in K3PO4.
Answer: pH = pKa3 ˆ pH = 11.32
Example 4: Calculate the pH of a solution made up to be
0.35 M in Li2CO3 and 0.35 M in LiHCO3.
Answer: pH = pKa2 ˆ pH = 10.36
These are all 1:1 buffer so pH .pKa is OK.
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
So, which of these reactions is relevant?
H3PO4 + H2O º H3O+ + H2PO4!
H2PO4! + H2O º H3O+ + HPO42!
HPO4! + H2O º H3O+ + PO43!
Ka1 = 1.1 × 10–2
Ka2 = 7.5 × 10–8
Ka2 = 4.8 × 10–12
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
So, which of these reactions is relevant?
H3PO4 + H2O º H3O+ + H2PO4!
H2PO4! + H2O º H3O+ + HPO42!
HPO4! + H2O º H3O+ + PO43!
It is the second one.
Ka1 = 1.1 × 10–2
Ka2 = 7.5 × 10–8
Ka2 = 4.8 × 10–12
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
H2PO4! + H2O º H3O+ + HPO42!
Ka2 = 7.5 × 10–8
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
x
!
Ka2 = 7.5 × 10–8
H2PO4 + H2O º H3O+ + HPO42!
This then is a normal type II buffer problem. Set up a table:
[H2PO4–]
[H3O+]
HPO42–]
before equilibrium
0.10
0.0
0.50
after equilibrium
0.10 ! x
x
0.50 + x
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
x
!
Ka2 = 7.5 × 10–8
H2PO4 + H2O º H3O+ + HPO42!
This then is a normal type II buffer problem. Set up a table:
[H2PO4–]
[H3O+]
[HPO42–]
Fill in the
equilibrium
expression:
before equilibrium
0.10
0.0
0.50
⎡⎣H3O+ ⎤⎦ ⎡⎣HPO24− ⎤⎦
Ka 2 =
⎡⎣H2 PO4− ⎤⎦
after equilibrium
0.10 ! x
x
0.50 + x
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
x
!
H2PO4 + H2O º H3O+ + HPO42!
Ka2 = 7.5 × 10–8
This then is a normal type II buffer problem. Set up a table:
[H2PO4–]
[H3O+]
[HPO42–]
Fill in the
equilibrium
expression:
before equilibrium
0.10
0.0
0.50
7.5 ×10
−8
x )( 0.50 + x )
(
=
( 0.10 − x )
after equilibrium
0.10 ! x
x
0.50 + x
pH of Polyprotic 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 KH2PO4 and 0.50 M in K2HPO4.
x
!
Ka2 = 7.5 × 10–8
H2PO4 + H2O º H3O+ + HPO42!
This then is a normal type II buffer problem. Set up a table:
[H2PO4–]
[H3O+]
[HPO42–]
Fill in the
equilibrium
expression:
before equilibrium
0.10
0.0
0.50
after equilibrium
0.10 ! x
x
0.50 + x
-8
and
solve*:
x
=
1.5
×
10
x )( 0.50 + x )
(
−8
7.5 ×10 =
( 0.10 − x ) or: pH = 7.82
* the approximation ±x works here.
pH of Polyprotic acid and
their buffers
1:1 buffers and non-1:1 buffers
plus the pH of the first weak acid proton.
THE END