A titration is performed using by adding 0.100 M NaOH to 40.0 mL of 0.1 M
HA (Ka is 1.0 x 10-5.).
e) Calculate the pH after addition of 10.0 mL of 0.100 M NaOH past the endpoint.
Answer:
When a strong base like NaOH is added to a weak acid like HA a neutralization
reaction occurs,
NaOH(aq) + HA(aq) ---> NaA(aq) + H2O(aq)
The K for this reaction is very large,
so the reaction goes to completion. To determine how much reaction occurs we
need to set up an ICE 'like' table.
|
NaOH(aq)
|
+ HA(aq)
|
--->
|
NaA(aq)
|
+ H2O(aq)
|
I
|
|
|
|
|
|
C
|
|
|
|
|
|
F
|
|
|
|
|
|
Determine the moles of NaOH (that were added) and the moles of HA (present
initially);
The moles of NaOH added from the buret are;
The volume of 50.0 mLs was obtained by adding 10.0 mLs (volume
past the equivalence point) plus the 40.0 mLs of NaOH required to reach the
equivalence point.
The moles of HA in the flask initially are;
So the 'ICE table' now looks like;
|
NaOH(aq)
|
+ HA(aq)
|
--->
|
NaA(aq)
|
+ H2O(aq)
|
I
|
0.00500 mol
|
0.00400 mol
|
|
0
|
-
|
C
|
|
|
|
|
|
F
|
|
|
|
|
|
Since K for the reaction is large, the reaction will go to
completion. This means that the reactant in the smallest amount will completely
react. At this point in the titration there are fewer moles of base compared
to acid. So all the base reacts. Adding this to the 'ICE table' we have;
|
NaOH(aq)
|
+ HA(aq)
|
--->
|
NaA(aq)
|
+ H2O(aq)
|
I
|
0.00500 mol
|
0.00400 mol
|
|
0
|
-
|
C
|
-0.00400 mol
|
-0.00400 mol
|
|
+0.00400 mol
|
-
|
F
|
|
|
|
|
|
The final amount of each species can be obtained just as it
always has been in the other ICE tables, by adding the change to the initial
amount.
|
NaOH(aq)
|
+ HA(aq)
|
--->
|
NaA(aq)
|
+ H2O(aq)
|
I
|
0.00500 mol
|
0.00400 mol
|
|
0
|
-
|
C
|
-0.00400 mol
|
-0.00400 mol
|
|
+0.00400 mol
|
-
|
F
|
0.00100 mol
|
0
|
|
+0.00400 mol
|
|
Focusing on the Final row there is base remaining and all the
acid has reacted. The salt has been formed.
To calculate the pH we must recognize the type of system we
have. To do that we look at the final row and see what species are present.
In this case there are moles of a strong base (NaOH) and the salt (NaA) which
is also a base. So we have a solution with a strong base and a weak base.
This is a type of common ion system.
To calculate the pH of the solution we must determine the concentration
of NaOH and the salt NaA in the mixture. The concentration is obtained by
dividing the volume of the solution into the moles of NaOH and the moles of
NaA. The volume of the solution is 90.0 mLs (40.0 mLs of HCl and 50.0 mLs
of NaOH).
So the [NaOH] is;
and the [A-] is;
To determine the pH of the solution we must decide on the chemical reaction
that best desribes the system. Since the system is a strong base and a weak
base, we must use the equation describing the weak base to determine the pH.
The equation along with the ICE table is;
|
A-(aq)
|
+ H2O(aq)
|
--->
|
HA(aq)
|
+ OH-(aq)
|
I
|
|
|
|
|
|
C
|
|
|
|
|
|
F
|
|
|
|
|
|
Substituting into the ICE table and completing the ICE table;
|
A-(aq)
|
+ H2O(aq)
|
--->
|
HA(aq)
|
+ OH-(aq)
|
I
|
0.0444 M
|
-
|
|
0
|
0.0111
|
C
|
- x
|
-
|
|
+ x
|
+ x
|
F
|
0.0444 - x
|
-
|
|
+ x
|
0.0111 + x
|
The important point to make from these calculations is that the pH of a solution
containing a strong base and a weak base is defined by the strong base. The
weak base does not contribute significantly to the [OH-].
We can see this point on the titration curve (as point 1e).
