The Images provided are the way they want the problem to be solved, it's weird and I'm sorry about that. The one I'm trying to wrap my head around goes as follows; When a 16.3 mL sample of a 0.394 M aqueous hydrofluoric acid solution is titrated with a 0.415 M aqueous potassium hydroxide solution, (1) What is the pH at the midpoint in the titration? (2) What is the pH at the equivalence point of the titration? (3) What is the pH after 23.2 mL of potassium hydroxide have been added?
The Images provided are the way they want the problem to be solved, it's weird and I'm sorry about that. The one I'm trying to wrap my head around goes as follows; When a 16.3 mL sample of a 0.394 M aqueous hydrofluoric acid solution is titrated with a 0.415 M aqueous potassium hydroxide solution, (1) What is the pH at the midpoint in the titration? (2) What is the pH at the equivalence point of the titration? (3) What is the pH after 23.2 mL of potassium hydroxide have been added?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
The Images provided are the way they want the problem to be solved, it's weird and I'm sorry about that. The one I'm trying to wrap my head around goes as follows;
When a 16.3 mL sample of a 0.394 M aqueous hydrofluoric acid solution is titrated with a 0.415 M aqueous potassium hydroxide solution,
(1) What is the pH at the midpoint in the titration?
(2) What is the pH at the equivalence point of the titration?
(3) What is the pH after 23.2 mL of potassium hydroxide have been added?
![Use the References to access important values if needed for this question.
When a 22.4 mL sample of a 0.397 M aqueous hypochlorous acid solution is
titrated with a 0.402 M aqueous barium hydroxide solution,
(1) What is the pH at the midpoint in the titration? 1
(2) What is the pH at the equivalence point of the titration?
(3) What is the pH after 16.6 mL of barium hydroxide have been added?
Incorrect
Please note that while the results of the intermediate calculations below are
displayed rounded to 3 significant digits, the actual calculations are done
internally without rounding.
When the strong base, Ba(OH)2, is added, it reacts to neutralize the HCIO. The
balanced equation is:
2HCIO + Ba(oн),—Ва(ClO), + 2 H20
The tabulated value of K, for HClo is 3.50x10-8.
(1) pH at the midpoint of the titration
At the midpoint of the titration, half of the HCIO has been converted
to clo", and the ratio [HCIO] / [Clo] = 1.
This is a buffer solution, and the pH can be calculated using the
Henderson-Hasselbalch equation:
[Clo]
pH = pk, + log
[HCIO).
pH = -log(3.50×10-8) + log(1) = 7.46 + 0 = 7.46
Note that the pH at the midpoint of the titration is equal to the pk3a
of the weak acid, HCIO.
(2) pH at the equivalence point: 22.4 mL of 0.397 M HClo
Ba(Он)2
0.402 M
The number of moles of the weak acid HCIO is given by 0.397M X
2.24x10-2 L = 8.89×10-3 mol.
The number of moles of Ba(OH)2 required to neutralize the acid is
1 mol
Ba(OH)2
8.89x10-3 mol
4.45x10-3 mol
HCIO X
-Ba(ОН)2
2 mol HCIO
The volume of Ba(OH), required is thus 4.45x10-3 mol / 0.402 M
= 1.11x10-2 L
and the total volume at the equivalence point is 2.24x10-2 L +
1.11x10-2 L = 3.35×10-2 L
The number of moles of Clo" that is formed is
1 mol Clo-
8.89x10-3 mol HClo
= 8.89x10-3 mol
1 mol
clo
HCIO
and the concentration of Clo" at the equivalence point is 8.89x10-3
mol Clo / 3.35×102L = 0.266 M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3e94073-e2a1-4a71-89ba-10c58e3f4fb1%2F55418137-35be-41e0-b588-970c5030ddcb%2Fx3wwjbn_processed.png&w=3840&q=75)
Transcribed Image Text:Use the References to access important values if needed for this question.
When a 22.4 mL sample of a 0.397 M aqueous hypochlorous acid solution is
titrated with a 0.402 M aqueous barium hydroxide solution,
(1) What is the pH at the midpoint in the titration? 1
(2) What is the pH at the equivalence point of the titration?
(3) What is the pH after 16.6 mL of barium hydroxide have been added?
Incorrect
Please note that while the results of the intermediate calculations below are
displayed rounded to 3 significant digits, the actual calculations are done
internally without rounding.
When the strong base, Ba(OH)2, is added, it reacts to neutralize the HCIO. The
balanced equation is:
2HCIO + Ba(oн),—Ва(ClO), + 2 H20
The tabulated value of K, for HClo is 3.50x10-8.
(1) pH at the midpoint of the titration
At the midpoint of the titration, half of the HCIO has been converted
to clo", and the ratio [HCIO] / [Clo] = 1.
This is a buffer solution, and the pH can be calculated using the
Henderson-Hasselbalch equation:
[Clo]
pH = pk, + log
[HCIO).
pH = -log(3.50×10-8) + log(1) = 7.46 + 0 = 7.46
Note that the pH at the midpoint of the titration is equal to the pk3a
of the weak acid, HCIO.
(2) pH at the equivalence point: 22.4 mL of 0.397 M HClo
Ba(Он)2
0.402 M
The number of moles of the weak acid HCIO is given by 0.397M X
2.24x10-2 L = 8.89×10-3 mol.
The number of moles of Ba(OH)2 required to neutralize the acid is
1 mol
Ba(OH)2
8.89x10-3 mol
4.45x10-3 mol
HCIO X
-Ba(ОН)2
2 mol HCIO
The volume of Ba(OH), required is thus 4.45x10-3 mol / 0.402 M
= 1.11x10-2 L
and the total volume at the equivalence point is 2.24x10-2 L +
1.11x10-2 L = 3.35×10-2 L
The number of moles of Clo" that is formed is
1 mol Clo-
8.89x10-3 mol HClo
= 8.89x10-3 mol
1 mol
clo
HCIO
and the concentration of Clo" at the equivalence point is 8.89x10-3
mol Clo / 3.35×102L = 0.266 M
![HClo
and the concentration of Clo at the equivalence point is 8.89x10-3
mol Clo / 3.35x10-2 L = 0.266 M
The value of Kh for the weak base Clo can be calculated from the
tabulated value of K, for its conjugate acid HClo.
1.0x10-14
Kh =
2.86x10-7
3.50x10-8
Solve for the [OH] in a 0.266 M solution of Clo using the ICE
method:
Clo
+ H20
OH + HClo
Initial (M)
0.266
Change (M)
-x
+X
+X
Equilibrium (M)) 0.266 -x
[OH"][HCI0]
x2
K = 2.86x10-7 -
[Clo]
0.266 - x
Assume that x is small relative to 0.266. This is acceptable if
[lo), > 100-Kg: 0.266 > 100-2.86x10-7.
Then:
2.86×10-7 = x² / (0.266)
x = [(2.86x10-7)(0.266)]1/2 = 2.76×10-4 M = [OH"]
[H3o+] = 1.0x10-14 / 2.76×10-4 = 3.63×10-11 M
pH = -log [H3o+] = 10.44
(3) pH after 16.6 mL 0.402 M Ba(OH), have been added to
22.4 ml 0.397 M HCIO
From part (2), the number of moles of Ba(OH)2 required
to neutralize the acid is 4.45×10-3 mol Ba(OH)2.
2 mol OH"
Moles OH" added = 0.402 M
1.33x10
=
Вa(он)2 X 1.66х10°2 L X
mol
2 mol
Ba(OH)2
There is, therefore, excess OH".
Moles OH" excess =
1.33x10-2 - 8.89×10-3 =
4.45×10-3 mol
Total volume = 22.4 mL + 16.6 mL = 39.0 mL =
3.90x10-2 L
[OH] = 4.45x10-3 mol / 3.90x10-2 L = 0.114 M
from the excess Ba(OH)2
The weak base clo will have only a negligible effect on
the hydroxide ion concentration, so the pOH can be
calculated from the molarity of the excess OH":
pOH = - log(0.114) = 0.94
pH = 14 - 0.94 = 13.06
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Transcribed Image Text:HClo
and the concentration of Clo at the equivalence point is 8.89x10-3
mol Clo / 3.35x10-2 L = 0.266 M
The value of Kh for the weak base Clo can be calculated from the
tabulated value of K, for its conjugate acid HClo.
1.0x10-14
Kh =
2.86x10-7
3.50x10-8
Solve for the [OH] in a 0.266 M solution of Clo using the ICE
method:
Clo
+ H20
OH + HClo
Initial (M)
0.266
Change (M)
-x
+X
+X
Equilibrium (M)) 0.266 -x
[OH"][HCI0]
x2
K = 2.86x10-7 -
[Clo]
0.266 - x
Assume that x is small relative to 0.266. This is acceptable if
[lo), > 100-Kg: 0.266 > 100-2.86x10-7.
Then:
2.86×10-7 = x² / (0.266)
x = [(2.86x10-7)(0.266)]1/2 = 2.76×10-4 M = [OH"]
[H3o+] = 1.0x10-14 / 2.76×10-4 = 3.63×10-11 M
pH = -log [H3o+] = 10.44
(3) pH after 16.6 mL 0.402 M Ba(OH), have been added to
22.4 ml 0.397 M HCIO
From part (2), the number of moles of Ba(OH)2 required
to neutralize the acid is 4.45×10-3 mol Ba(OH)2.
2 mol OH"
Moles OH" added = 0.402 M
1.33x10
=
Вa(он)2 X 1.66х10°2 L X
mol
2 mol
Ba(OH)2
There is, therefore, excess OH".
Moles OH" excess =
1.33x10-2 - 8.89×10-3 =
4.45×10-3 mol
Total volume = 22.4 mL + 16.6 mL = 39.0 mL =
3.90x10-2 L
[OH] = 4.45x10-3 mol / 3.90x10-2 L = 0.114 M
from the excess Ba(OH)2
The weak base clo will have only a negligible effect on
the hydroxide ion concentration, so the pOH can be
calculated from the molarity of the excess OH":
pOH = - log(0.114) = 0.94
pH = 14 - 0.94 = 13.06
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