Thank you for completing this ACA. You should print this page out and bring it to class so if you have any questions you can ask them.
ACA Response Page
This is ACA # 24. It is
OK to use your textbook, but if you can answers the questions without it that
is OK too.
I recommend you print out this page and bring it to class. Click
here to show a set of five student responses, randomly selected from
all of the student responses thus far, in a new window.
, here are your responses
to the ACA and the Expert's response.
Our discussion of acids and bases has prepared you to calculate the pH of
the following solutions,
strong acids
weak acids
strong bases
weak bases
the salt of a strong acid and a strong base
the salt of a strong acid and a weak base
the salt of aweak acid and a strong base
common ion - weak acid and its conjugate base
common ion - weak base and its conjugate acid
A secret for calculating the pH of neutralization reactions/titration curves
is recognizing which of the above solution is formed after mixing an acid and
a base.
Here is an example of what I mean;
Predict the type of solution formed when 30.0 mL of 0.450 M NaOH
is added to 15.0 mL of 0.800 M HCl.
To answer this question we write the neutralization equation
and set up an ICF table. In an ICF table we have to work in moles of each component.
So we must determine the moles of NaOH and HCl we have initially before any
reaction occurs. Below is the completed ICF table.
|
NaOH
|
+
|
HCl
|
|
NaCl
|
+
|
H2O
|
Initial |
0.0135 mol
|
|
0.012 mol
|
|
0
|
|
-
|
Change |
-.012
|
|
-.012
|
|
.012
|
|
-
|
Final |
0.0013
|
|
0
|
|
.012
|
|
-
|
Looking at the the Final row we can make the following observations....
The final condition after the reaction occurs shows moles NaOH and the salt
NaCl are present. Since the salt NaCl does not effect the pH of the solution
only strong base is important in determining the pH of the solution. So mixing
30.0 mL of 0.450 M NaOH is added to 15.0 mL of 0.800 M HCl yields a solution
of a strong base. We can calculate the pH of a strong base by taking the moles
of the NaOH in the Final row and dividing by the total volume of the mixture
and then doing the -log thing.
So now you try it....
1. Indicate what substances are present in the solution (Final row) when,
(NOTE: you must follow the procedure described above)
a) 30.0 mL of 0.100 M NaOH are mixed with 25.0 mL of 0.100 M HBr
The solution contains moles of NaOH and moles of NaBr
after mixing. This is a solution of a strong base.
Below is how the expert arrived at this answer....
|
NaOH(aq)
+ |
HBr(aq) |
|
NaBr(aq)
+ |
H2O(l) |
I
|
0.00300 mol |
0.00250 mol |
|
0 |
- |
C
|
-0.00250 mol |
-0.00250 mol |
|
+0.00250 mol |
- |
E |
0.00050 mol |
0 |
|
+0.00250 mol |
- |
The solution contains moles of NaOH and moles of NaBr
after mixing.
b) 20.0 mL of 0.0372 M NaOH are mixed with 34.0 mL 0.0520 M HC2H3O2
The solution contains moles of HC2H3O2 and moles of NaC2H3O2 after mixing. This is
a solution of a weak acid and its salt.
Below is how the expert arrived at this answer....
|
NaOH(aq)
+ |
HC2H3O2(aq) |
|
NaC2H3O2(aq)
+ |
H2O(l) |
I
|
0.000744 mol |
0.00177 mol |
|
0 |
- |
C
|
-0.000744 mol |
-0.000744 mol |
|
+0.000744 mol |
- |
E |
0 |
0.00103 mol |
|
+0.000744 mol |
- |
The solution contains moles of HC2H3O2 and moles of NaC2H3O2 after mixing.
c) 40.0 mL of 0.234 M NH3 are mixed with 34.0 mL 0.135 M HCl
The solution contains moles of NH3 and moles
of NH4Cl after mixing. This is a solution of a weak base and its
salt.
Below is how the expert arrived at this answer....
|
NH3(aq)
+ |
HCl(aq) |
|
NH4Cl(aq) |
I
|
0.00936 mol |
0.00459 mol |
|
0 |
C
|
-0.00459 mol |
-0.00459 mol |
|
+0.00459 mol |
E |
0.00477 mol |
0 |
|
+0.00459 mol |
The solution contains moles of NH3 and moles
of NH4Cl after mixing.
2. For each case in Q1 after the mixtures have been prepared, indicate whether
the solution is acidic, basic or neutral.
a) The
solution in 1a is basic.
b) The
solution in 1b is acidic. (Note: Ka for HC2H3O2 is greater than Kb for C2H3O2-.)
c) The
solution in 1b is basic. (Note: Kb for NH3 is greater
than Ka for NH4+.)
3. Below is a figure showing two titration curves,
List some similarities and some differences in the solutions
used to generate these titration curves. Use a) and b) to differentiate between
the two systems.
Similarities :
In both titrations a strong base is added to a
strong acid. The end point/equivalence point is the
same for both acids (7). The same volume of base is required for each titration.
Differences :
The initial concentration of the acid
is different, and the initial concentration of the base in each titration is
different. Solution a) is a higher concentration of the strong acid compared to Solution b)
4. Below is a titration curve. What can you tell me about the solutions used
to produce this titration curve?
This a titration of a weak acid and a strong
base. The end point/equivalence point is NOT 7, but is basic. This is characteristic
of a weak acid/strong base titration. Notice the pH of the solution in the early
part of the titration is very different compared to the the pH of the solution
in the early part of the titration in the curves of the SA/SB titration. Looks like the concentration
of the base in this titration is the same as the concentration in curve a) in
Q3.
4. Is there anything about the questions that you feel you do not understand?
List your concerns/questions.
5. If there is one question you would like to have answered in lecture,
what would that question be?