Acids

1
Acids Bases

• Weak Acids

2
Weak acids and weak bases do not completely
ionize in water. The reaction is also reversible.
When the ionization reaches equilibrium (almost
instantaneously gt very fast), the solution
contains ionized and unionized materials. The
position of equilibrium is described by an
equilibrium constant.
3
For a weak acid HA HA H2O ? H3O A-
Ka KeqH2O H3OA- / HA Note
concentrations are equilibrium values.
CHA analytical concentration of HA HA
A-.
4
The analytical concentration of a substance
refers to the total solute concentration of a
compound in solution how many moles are present
for example independent of the form or species
actually present in solution. For example, if 1
mole acetic acid is dissolved in water, there is
less than 1 mole CH3COOH present the rest is
CH3COO- H. The total amount
5
of material containing carbon derived from acetic
acid in solution remains at 1 mole. Weak acids
can be titrated successfully with strong bases
because the strong base pulls the neutralization
to completion. A weak base weak acid cant be
used, since the reaction is not complete.
6
General shape of the weak acid-strong base
titration curve
A before base added B buffer region C
equivalence pt D post equiv pt
pH
mL strong base added
7

• pH calculation at regions A, B, C, D is different
  gt different processes control H3O.
• Before any base added pH is controlled only by
  the H3O arising from dissociation of the weak
  acid HA.
• Relevant reaction is HA H2O ?
  H3O A-

8
General approach to solving problem Need to know
Ka and analytical concentration of HA. Must find
equilibrium value of H3O to calculate
pH. Note balanced equation shows 1 mole A- and 1
mole H3O form for each 1 mole HA that
dissociates.
9
If CHA is the analytical concentration of the
acid ( starting value), the equilibrium value
of HA CHA H3O, and A- H3O. Ka
H3OA- / HA can be written as Ka
H3O2 / (CHA H3O). So the only unknown is
H3O.
10
Note that this is a quadratic (second order)
equation Ka(CHA H3O) H3O2, or H3O2
Ka H3O - KaCHA 0. How to
solve? Quadratic equation, successive
approximation, or simple approximation.
11
An exact value for H3O can be found using the
quadratic equation, ax2 bx c 0, with x
H3O. Thus x -b SQRT(b2 4ac)
2a
In some cases, it may be possible to find H3O
more simply.
12
On slide 10 we saw that the equilibrium
calculation led to Ka(CHA H3O) H3O2
. If the weak acid has not dissociated very much,
then H3O is much smaller than CHA. In turn
this means the equilibrium concentration of weak
acid HA, HA ? CHA. Thus, the
13
equation becomes KaCHA H3O2 . So, H3O
SQRT(KaCHA), which is much simpler to
compute. The question then becomes when is it
appropriate to use this approximation? The answer
is it depends on how large the error in the
calculated H3O is acceptable for a given
purpose.
14
As a routine matter, an error of lt 5 is often
acceptable because of the uncertainty in
measuring the H3O. This level of uncertainty
can be obtained when CHA gtgt Ka100. The error
drops to lt 2 when CHA gtgt Ka1000. Unless
otherwise specified, use the 100Ka guideline.
15
Specific Example What is the pH of a 0.055 M
solution of acetic acid, CH3COOH, if Ka 1.75 x
10-5 ?