A Look at Buffers
Note that in the model, Ac = acetate ion
Initially:
Beaker A: 100.0 ml 1.00 x 10-5 M HCl
To describe what is happening before anything is added to beaker A, we can write:
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
Beaker B: 100.0 ml of a buffer solution that is 0.295 M HAc and 0.537 M NaAc
To describe what is happening before anything is added to beaker B, we can write:
HAc (aq) + H2O (l)  H3O+ (aq) + Ac- (aq)
The picture shows that a base, NaOH, will be added to beaker A and beaker B. We will
explore how this alters the composition of the solutions in each beaker, and ultimately,
calculate the pH of each new solution.
Questions:
1) Even though the sample in beaker A is described as 1.00 x 10-5 M, why are there
no HCl molecules in solution A? (Hint, what kind of acid is HCl, strong or
weak?)
2) Since HAc is a weak acid, how does the [H3O+] compare to the [HAc] in solution
B? Explain.
3) Why does the Na+ appear in beaker B, but not in the reaction equation describing
beaker B?
4) How do the moles of each of the following change (increase, decrease, no
change) when OH- is added to each solution? Explain.
In Beaker A:
H3O+
Cl-
In Beaker B:
H3O+
Na+
Ac-
HAc
Time to Calculate:
1)
HCl
Show with calculations that the pH of each solution is initially 5.00. The Ka of
HAc is 1.8 x 10-5
HAc
2) Determine the moles of OH- that is added to each solution if 50.0 ml of 0.10 M
NaOH is added. This would be the amount of OH- initially present for each of the
tables on the next page.
3) Complete the tables for each solution showing the stoichiometry of the reactions
that occur when the base is added.
Solution A: H3O+ (aq) + OH-(aq)  2 H2O (l)
(moles) H3O+
OH-
initially
change
final
Solution B: HAc (aq) + OH- (aq)  H2O (l) + Ac- (aq)
(moles) HAc
OH-
Ac-
initially
change
final
For the table above:
The initial moles of OH-is the same as for beaker A since you are adding the same
volume of solution to beaker B as you did in beaker A.
Both HAc and Ac- are present in the solution initially so use the volume of solution in
beaker B and the molarities given on the first page describing beaker B, to find the moles
of each ion initially present.
Since these tables describe a reaction that goes to completion examine the reactions next
to the tables and use the appropriate mole ratios to complete the change row. Hint: one
of the reactants must run out, the other will be in excess.
4) Calculate each using the final moles from the tables and the other given
information. Be sure to account for the fact that the volume of the final solution
has increased.
Solution A:
[OH-]
[Cl-] (Hint: the Cl- does not react with the base.)
Solution B:
[Na+] (Hint: Na+ coming both from the original solution in beaker B, and
the base that you are adding.
[Ac-]
[HAc]
5) Find the final [H3O+] for each solution (after the addition of OH-). It is not
possible to find the [H3O+] in solution B using stoichiometry because this is an
equilibrium system. You must use an ICE table and the given equilibrium
constant. Find [H3O+] for solution B. In beaker A, you will need to use the [OH-]
and the fact that Kw = [H3O+] [OH-] to find [H3O+]
6) Calculate the pH of each solution after the base has been added. Why does the
pH change much more for solution A (in fact, this solution ends up basic), even
though solution A contains a strong acid, than it does for solution B?