Sample solution in the image. Compute for the pH, pOH, and [OH-] of a buffer solution that is 0.20 M in acetic acid and contains sufficient sodium acetate to make the acetate ion equal to 0.05 M. Ka = 1.85 x 10^-5.
Sample solution in the image. Compute for the pH, pOH, and [OH-] of a buffer solution that is 0.20 M in acetic acid and contains sufficient sodium acetate to make the acetate ion equal to 0.05 M. Ka = 1.85 x 10^-5.
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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Sample solution in the image.
Compute for the pH, pOH, and [OH-] of a buffer solution that is 0.20 M in
acetic acid and contains sufficient sodium acetate to make the acetate ion equal to 0.05 M. Ka = 1.85 x 10^-5.
![Henderson- Hasselbalch Equation
-use without using equilibrium equations, [H°] [OH]
-use only for initial concentration w/ ka| kb & Ca & Cb
In buffer pair of weak acid and its corresponding base, ionization takes place. The pH
of solution consisting weak acid is related to its dissociation constant (k).
HA +H* + A
Weak Acid CA
CB
@equilibrium Ka = [H*] [A*]
[HAJ
Log Ka = log (H°] [A"]
[HA]
Log Ka = log[H*] + log [A^)_
[HA]
- Log = log (H*] + log [A"]
(HA]
pka = pH - log base/ acid
pH = pKa - log acidic component
basic component
Module II
82
MODULE III
CHEMICAL EQUILIBRIUM AND IONIC EQUILIBRIUM
or pH = 14- pkb -log acidic component
basic component
pH = 14- pKb -log acidic component
basic component
pH = pkw- pKb + log base/salt
Sample problems:
1. Compute for the pH, pOH, and [OH-] of a buffer solution that is 0.20 M in acetic
acid and contains sufficient sodium acetate to make the acetate ion equal to 0.1 M.
Ka - 1.85 x 105
Solution:
HAC++ H*+ AC
НАС
Ac
Initial
0.2
0.1
Change
+x
+X
Equilibrium
0.2 -x
0.1+x
Ka = 0.1 x + X = 1.85 x 105
0.2 -x
[H'] = 3.7 x 10-5
pH =4.4
[OH] -2.7 x 10 10
Another solution
HHE
pH = pka + log (A]
[HAJ
pH = -log 1.85 x10-5 + log 0.1 - 4.4
0.2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F411cc000-d55d-48d1-92c8-6416a2ad5fb8%2Fdb2a7e59-1ebb-4612-9fe2-b325ff10431b%2Fwyxpwwa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Henderson- Hasselbalch Equation
-use without using equilibrium equations, [H°] [OH]
-use only for initial concentration w/ ka| kb & Ca & Cb
In buffer pair of weak acid and its corresponding base, ionization takes place. The pH
of solution consisting weak acid is related to its dissociation constant (k).
HA +H* + A
Weak Acid CA
CB
@equilibrium Ka = [H*] [A*]
[HAJ
Log Ka = log (H°] [A"]
[HA]
Log Ka = log[H*] + log [A^)_
[HA]
- Log = log (H*] + log [A"]
(HA]
pka = pH - log base/ acid
pH = pKa - log acidic component
basic component
Module II
82
MODULE III
CHEMICAL EQUILIBRIUM AND IONIC EQUILIBRIUM
or pH = 14- pkb -log acidic component
basic component
pH = 14- pKb -log acidic component
basic component
pH = pkw- pKb + log base/salt
Sample problems:
1. Compute for the pH, pOH, and [OH-] of a buffer solution that is 0.20 M in acetic
acid and contains sufficient sodium acetate to make the acetate ion equal to 0.1 M.
Ka - 1.85 x 105
Solution:
HAC++ H*+ AC
НАС
Ac
Initial
0.2
0.1
Change
+x
+X
Equilibrium
0.2 -x
0.1+x
Ka = 0.1 x + X = 1.85 x 105
0.2 -x
[H'] = 3.7 x 10-5
pH =4.4
[OH] -2.7 x 10 10
Another solution
HHE
pH = pka + log (A]
[HAJ
pH = -log 1.85 x10-5 + log 0.1 - 4.4
0.2
Transcribed Image Text:Sample Problems
1.Calculate the hydroxide ion concentration of a solution that is 0.1 F with NH3 and
0.25F with respect to (NH,)2SO4.
Reaction
NH3 + NH, + OH
NH3
NH4 *
OH-
initial
0.1
0.5
change
equilibrium
-X
+X
+X
0.1 - x
0.5 + X
Solution:
If (NH4)2SO4
0.25
- 2 NH4 *
2(0.25)
SO, 2
Ка 3D (0.5 + х) (х) %3D1.75 х 10 -5
(0.1 + x)
* x is negligible, Use Rule of 100
*If C/K is greater than 100 then x is negligible
(x)
3.5 x 10-6 = (OH1)
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