
To find:
a) The
b) The suitable indicator for the titration referring figure 16.5
c) The
d) Sketch the titration curve

Answer to Problem 16.126QA
Solution:
a) The calculated
b) The suitable indicator for the titration, referring figure 16.5, is Alizarin Yellow R.
c) The
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
d) In the sketched titration curve
The major species at the equivalence point is
The major species at the end of titration is
Explanation of Solution
1) Concept:
To calculate the pH of a weak acid with a strong base, we need to use reaction stoichiometry and RICE table. First, we will calculate the equivalence point from the volume and molarity of base and acid. We will calculate the pH at equivalence point by using the RICE table. At the equivalence point, moles of phenol and
2) Formulae:
i)
ii)
iii)
iv)
3) Given:
i) Volume of phenol =
ii) Molarity of phenol
iii)
iv) Molarity of
4) Calculations:
a) Calculating
Step 1)
Finding the volume of
At the equivalence point, moles of phenol = moles of
At the equivalence point, moles of phenol and
Calculating the volume of
So, at the equivalence point, the volume of
So, the total volume of solution is
At the equivalence point, sodium salt of phenol is formed, and its moles will be equal to moles of phenol and
Calculating the molarity of sodium phenoxide:
Step 2)
Creating an RICE table for the dissociation of sodium phenoxide
RICE | |||
Initial (M) | |||
Change (M) | |||
Equilibrium (M) |
Writing the
The calculated
b) Choosing the indicator, which is suitable for the titration referring to figure 16.5:
A
c) Calculating the
The calculated volume of
1) Addition of 0.0 mL NaOH:
At 0.0mL NaOH added, the
RICE | |||
Initial (M) | |||
Change (M) | |||
Equilibrium (M) |
So, the
2) Addition of 2.5 mL NaOH:
Moles of phenol:
Moles of
We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
The total sample volume is
Applying the Henderson-Hasselbalch equation for the above buffer system formed as
3) Addition of 5.0 mL NaOH:
Moles of phenol:
Moles of
We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
The total sample volume is
Applying the Henderson-Hasselbalch equation for the above buffer system formed as
4) Addition of 7.5 mL NaOH:
Moles of phenol:
Moles of
We create a modified RICE table to determine how many moles of acid remains and how many moles of conjugate base have been produced.
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
The total sample volume is
Applying the Henderson-Hasselbalch equation for the above buffer system formed as
5) Addition of 9.8 mL NaOH:
Moles of phenol:
Moles of
We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
The total sample volume is
Applying the Henderson-Hasselbalch equation for the above buffer system formed as
6) Addition of 10 mL NaOH:
Moles of phenol:
Moles of
We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
The total sample volume is
Applying the Henderson-Hasselbalch equation for the above buffer system formed as
7) Addition of 10.2 mL NaOH:
Moles of phenol:
Moles of
We create a modified RICE table to determine how many moles of acid remain and how many moles of conjugate base have been produced:
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
The total sample volume is
Applying the Henderson-Hasselbalch equation for the above buffer system formed as
8) Addition of 10.6 mL NaOH:
The contribution of
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
9) Addition of 10.8 mL NaOH
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
10) Addition of 11 mL NaOH
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
11) Addition of 12.5 mL NaOH
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
12) Addition of 15 mL NaOH
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
13) Addition of 17.5 mL NaOH:
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
14) Addition of
Moles of
Initial (mol) | |||
Change (mol) | |||
Final (mol) |
Total volume of solution =
Molarity of
d) Graph of
The above curve is a plot of
The volume of
So, at half equivalence point, the major species would be
At the equivalence point, moles of acid initially present are equal to moles of added base. So, there will not be phenol or
At the end of the titration, the only major species present will be
Conclusion:
In the titration of a weak acid and strong base, the resultant salt formed is a basic salt. Hence, the
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