Diprotic Acids

1
Diprotic Acids
Finish reading Chapter 11.

• Today we will extend what we learned about
  buffers and the titrations of weak acids and
  bases to the analogous problems involving
  diprotic acids.
• We will follow the same approach as was used for
  weak acids, making minor modifications, i.e. we
  will
• write down relevant equilbiria,
• write down the CB and MB equations,
• write CB eqn as a function of H,
• solve for pH numerically using mathcad or
• solve analytically by neglecting certain
  terms.

2
Comparison of Mono- and Diprotic Weak Acids
Ka1
Ka1Ka2
Kw
The MB and CB equations are
3
Rewrite CB Eqn as a function of H
Monoprotic
Diprotic
In both mono- and diprotic acids we now have one
equation and one unknown.
4
Solution of Equations
We cant solve for H as a function of CHA, so
we do the reverse and solve for CHA as a function
of H.
Monoprotic
Diprotic
We will review the as then compare titrations
with NaOH.
5
Review Expressions for as
Start with general formula for a0 and plug in
above expressions for HA- and A2-.
6
Titration of HA with NaOH
Rewrite CB Eq.
versus
Titration of H2A with NaOH
Rewrite CB Eq.
7
Titration of HA with NaOH
Now solve for Vt to find.
Titration of H2A with NaOH
Now solve for Vt to find
This Excel document uses the above Vt expression.
8
Titration of weak diprotic acid with a strong base
The key to solving weak for the pH is to know
what region you are in. Why is this easy to
do?
pH
3
1
2



9
Titration of weak diprotic acid with a strong base
1
2
3



Vt
10
Example 1
If 15 mL of .500M NaOH is added to 100 mL of
.200M H2A. What is the pH if pKa14.00 and
pKa210.00?
We have initially .0075 moles NaOH and .020 moles
H2A, so we are in region 1.
Which terms can be neglected?
11
In region 1 we can neglect a2.
1
2
3
I



12
In region 1 a2 ltlta1 so pHltltpKa2 and
simplifies to
This is just the same formula for the monoprotic
acid. Diprotic and monoprotic acids also have
the same formulas for a0 in region 1.
13
If 15 mL of .500M NaOH is added to 100 mL of
.200M H2A. What is the pH if pKa14.00 and
pKa210.00? Begin by determining which terms in
CB eqn below can be neglected.
Now rewrite in terms of H.
Now do a little algebra to solve for H.
14
We have initially .0075 moles NaOH and .020 moles
H2A, so we are in region 1.
Was it OK to neglect the above terms? We need to
check.
Excel
15
Example 2
If 56 mL of .500M NaOH is added to 100 mL of
.200M H2A. What is the pH if pKa14.00 and
pKa210.00?
We have initially .023 moles NaOH and .020 moles
H2A, so we are in region 2.
Which terms can be neglected in CB equation below?
We will now look at the expressions for ai in
region 2.
16
In region 2 a0 ltlta1 so pHgtgtpKa1 and
simplifies to
How does this formula compare to that of a
monoprotic acid? What is the formula for a2?
17
Example 2 Continued
We have .023 moles NaOH and .020 moles H2A.
18
Example 2 Continued
If 56 mL of .500M NaOH is added to 100 mL of
.200M H2A. What is the pH if pKa14.00 and
pKa210.00?
We have initially .028 moles NaOH and .020 moles
H2A, so we are in region 2.
Mathcad
Lets compare the pH formulas for regions 1and 2.
19
Diprotic Region 1 or Monoprotic Region 1
Diprotic Region 2
20
Diprotic Acids
We have developed approximate methods to
calculate the pH of mixtures of diprotic acids
and weak bases. An important step is to
recognize that it is easy to determine which
region you are in. You can then make the
appropriate approximations.
21
Problems
Since you have the Excel program, you can create
your own problems, solving for the pH in
different regions. You can change the Ka1 and
Ka2 values as well. You should be able to
calculate the pH in any region of the curve.