suppose there is 1.00L of an aqueous buffer containing 600.0 moral & acetic acid (pk. 4.76) and 40.0 ml of acetate. Calculate the pott of this buffer. pH = pka + log [#] and [HA]- acetic acid 60 moal [A-] acetate = 40 mmal pka = 4.76 PH= 4.76 log (488) = 4,58² # PH What volume of 3.50 M NaOH would be required to increase the PH to 4.93?
suppose there is 1.00L of an aqueous buffer containing 600.0 moral & acetic acid (pk. 4.76) and 40.0 ml of acetate. Calculate the pott of this buffer. pH = pka + log [#] and [HA]- acetic acid 60 moal [A-] acetate = 40 mmal pka = 4.76 PH= 4.76 log (488) = 4,58² # PH What volume of 3.50 M NaOH would be required to increase the PH to 4.93?
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![Suppose there is look of an aqueous buffer containing
600.0 moral & acetic acid (pk. - 4.76) and 40.0 mmol of
acetate. Calculate the pt of this buffer.
pH = pka + log [#]
acid
[HA]- acetic acid = 60 mmol
[A-] acetate = 40 mmal
pka = 4.76
PH² 4.76 log (128)-14.58
B What volume of 3.50 M NaOH would be required to increase.
the pH to 4.93?
Required ph= 4.93
I
e
E
0.17 =
10
Substitute into Henderson pH = pka + log [CH₂COO]
[CH3COOL+]
4.93 = 4.76 + log (40+ 3.
-4.76
-4.76
60-
1.47
TO
CH3COOH + OH → CH ₂ COO + H₂O
3.50x
40
1
-3.50x
(60-3.50x) D
40
-3.50x
0.1740+3.50x
60-2x
X = 20.35
+3.50X
40+ 8.50x1
60-2x
PH
40+ 3.50 x
40-2x)
-3.50 x
+3.50x/
(40+3,50x
کو](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7588948d-a8a4-4b5d-b85a-786e24f76953%2F9f873b52-035e-4162-963e-38536cfeb355%2Fnw5ksn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose there is look of an aqueous buffer containing
600.0 moral & acetic acid (pk. - 4.76) and 40.0 mmol of
acetate. Calculate the pt of this buffer.
pH = pka + log [#]
acid
[HA]- acetic acid = 60 mmol
[A-] acetate = 40 mmal
pka = 4.76
PH² 4.76 log (128)-14.58
B What volume of 3.50 M NaOH would be required to increase.
the pH to 4.93?
Required ph= 4.93
I
e
E
0.17 =
10
Substitute into Henderson pH = pka + log [CH₂COO]
[CH3COOL+]
4.93 = 4.76 + log (40+ 3.
-4.76
-4.76
60-
1.47
TO
CH3COOH + OH → CH ₂ COO + H₂O
3.50x
40
1
-3.50x
(60-3.50x) D
40
-3.50x
0.1740+3.50x
60-2x
X = 20.35
+3.50X
40+ 8.50x1
60-2x
PH
40+ 3.50 x
40-2x)
-3.50 x
+3.50x/
(40+3,50x
کو
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