What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8 x 10-5 Quantity = _______ mol What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka = 1.8 x 10-5 Quantity = _______ mol 3. What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = 5.00? Ka = 1.8 x 10-5 Quantity = _______ mol
What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8 x 10-5 Quantity = _______ mol What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka = 1.8 x 10-5 Quantity = _______ mol 3. What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = 5.00? Ka = 1.8 x 10-5 Quantity = _______ mol
Chemistry
10th Edition
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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- What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = pKa?
Ka = 1.8 x 10-5
Quantity = _______ mol - What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = 4.00? Ka = 1.8 x 10-5
Quantity = _______ mol
3. What quantity (moles) of NaOH must be added to 1.0 L of 1.5 M HC2H3O2 to produce a solution buffered at pH = 5.00?
Ka = 1.8 x 10-5
Quantity = _______ mol
(A similar solution is attached for reference)
![Added OH converts HC2 H3 02 into C2H3 02:
HC,H3O2 + OH C,H3O2 + H20
From this reaction, the moles of C2H3O2 produced equal the moles of OH added. Also, the total concentration of acetic acid plus acetate ion must equal 1.7 M (assuming no volume change on addition of NaOH).
[C,H3O2 ]+ [HC, H3 O2] = 1.7 M and
[C,HO2 ] produced = [OH ] added
[CH3 O2 ]
[HC,H3O]
Гон ] added
[C2H3 O2 ]
[HC,H,O]
pH = pK, + log
; for pH = pK,, log
a.
Therefore,
[CH,O2]
[HC,HO]
= 1.0 and [C,H3 O2 ] = [HC;H3O2]-
Because [C2 H3 02 ]+ [HC,H3 O2] = 1.7 M :
[C,H3O2] = [HC,H;O2] = 0.85 M = [OH ] added
To produce a 1.0 M C2H3O2 solution, we need to add 0.85 mol of NaOH to 1.0 L of the 1.7 M HC2H3O2 solution. The resulting solution will have pH = pK = 4.74.
[C,H3O2-] [CH3O2]
[HC, H3O]
HC,H3O]
b.
4.00 = 4.74 + log
10-0.74 = 0.18
%3D
[C,HO2 ] = 0.18[HC,H;O2] or
%3D
[HC,H,O2] = 5.5(C,H;O2]
Because (C2H3 O2 ]+ [HC,H3O2] = 1.7 M :
[CH3O2¯]+ 5.5(C,H,O2¯] = 1.7 M
1.7
[C,H3O2 ] =
0.26 M [OH] added
6.5
We need to add 0.26 mol of NaOH to 1.0 L of 1.7 M HC2H3O2 solution to produce 0.26 M C2H3O2.
The resulting solution will have pH = 4.00.
Email](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c6da06a-bab2-438d-8ae5-b09a280a4515%2Ff0b53ec1-b28b-471e-88c2-625c61495e48%2Fq5bw2nf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Added OH converts HC2 H3 02 into C2H3 02:
HC,H3O2 + OH C,H3O2 + H20
From this reaction, the moles of C2H3O2 produced equal the moles of OH added. Also, the total concentration of acetic acid plus acetate ion must equal 1.7 M (assuming no volume change on addition of NaOH).
[C,H3O2 ]+ [HC, H3 O2] = 1.7 M and
[C,HO2 ] produced = [OH ] added
[CH3 O2 ]
[HC,H3O]
Гон ] added
[C2H3 O2 ]
[HC,H,O]
pH = pK, + log
; for pH = pK,, log
a.
Therefore,
[CH,O2]
[HC,HO]
= 1.0 and [C,H3 O2 ] = [HC;H3O2]-
Because [C2 H3 02 ]+ [HC,H3 O2] = 1.7 M :
[C,H3O2] = [HC,H;O2] = 0.85 M = [OH ] added
To produce a 1.0 M C2H3O2 solution, we need to add 0.85 mol of NaOH to 1.0 L of the 1.7 M HC2H3O2 solution. The resulting solution will have pH = pK = 4.74.
[C,H3O2-] [CH3O2]
[HC, H3O]
HC,H3O]
b.
4.00 = 4.74 + log
10-0.74 = 0.18
%3D
[C,HO2 ] = 0.18[HC,H;O2] or
%3D
[HC,H,O2] = 5.5(C,H;O2]
Because (C2H3 O2 ]+ [HC,H3O2] = 1.7 M :
[CH3O2¯]+ 5.5(C,H,O2¯] = 1.7 M
1.7
[C,H3O2 ] =
0.26 M [OH] added
6.5
We need to add 0.26 mol of NaOH to 1.0 L of 1.7 M HC2H3O2 solution to produce 0.26 M C2H3O2.
The resulting solution will have pH = 4.00.
Email
![[C,H3 O2 ] [C,H3O2¯]
10 0.74 0.18
b.
4.00 = 4.74 + log
[HC, H3 O]
HC, H; O]
[C,H3O2 ] = 0.18(HC,H;O2] or
(HC,H3 O2] = 5.5[C,H3 O2]
Весause [C> Hз02] + (НС, Нз О2] 3 1.7 М:
[C,H; O2 ]+ 5.5[C,H3O2 ] = 1.7 M
1.7
[C,H3O2 ] =
= 0.26 M = [OH ] added
6.5
We need to add 0.26 mol of NaOH to 1.0 L of 1.7 M HC,H3 O2 solution to produce 0.26 M C2H3 O2.
The resulting solution will have pH = 4.00.
[C,H3 O2]
HC, H3 O]
[C,H3O2]
[HC,H3O]
5.00 = 4.74 + log
100.26 = 1.8
C.
1.8 HC, H;O2] =[C2H3O2 ] or
[HC,H3O2] = 0.56[C,H3O2 ]
1.56[C2H3O2 ] = 1.7 M
[C,H3O2-1 = 1.1 M = [OH ] added
We need to add 1.1 mol of NaOH to 1.0L of 1.7 M HC2H3O2 solution to produce a solution with pH = 5.00.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c6da06a-bab2-438d-8ae5-b09a280a4515%2Ff0b53ec1-b28b-471e-88c2-625c61495e48%2Fxkocl7b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[C,H3 O2 ] [C,H3O2¯]
10 0.74 0.18
b.
4.00 = 4.74 + log
[HC, H3 O]
HC, H; O]
[C,H3O2 ] = 0.18(HC,H;O2] or
(HC,H3 O2] = 5.5[C,H3 O2]
Весause [C> Hз02] + (НС, Нз О2] 3 1.7 М:
[C,H; O2 ]+ 5.5[C,H3O2 ] = 1.7 M
1.7
[C,H3O2 ] =
= 0.26 M = [OH ] added
6.5
We need to add 0.26 mol of NaOH to 1.0 L of 1.7 M HC,H3 O2 solution to produce 0.26 M C2H3 O2.
The resulting solution will have pH = 4.00.
[C,H3 O2]
HC, H3 O]
[C,H3O2]
[HC,H3O]
5.00 = 4.74 + log
100.26 = 1.8
C.
1.8 HC, H;O2] =[C2H3O2 ] or
[HC,H3O2] = 0.56[C,H3O2 ]
1.56[C2H3O2 ] = 1.7 M
[C,H3O2-1 = 1.1 M = [OH ] added
We need to add 1.1 mol of NaOH to 1.0L of 1.7 M HC2H3O2 solution to produce a solution with pH = 5.00.
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