For the dipiotic weak acid H₂A, Kai = 3.3 X 10 8 az 1.7 X101, what is the pH of a 0.0800 M solution of H₂ A ? *For a diprofic acid, if kazka, then most of the. [Hot] ions come from the 1st dissociation step. pH = H₂A HA+Hts Kai HA A₂A + H² → Kaz [H₂A] 0.0800-X M [H+] = x [HA] XAN kaz Kai -? 2.3 X10 7.7 X10 K₁₁ = [H+] [HA] So: 3,3 × 10 = x² [H₂A] x² + (313X 10)- (0.0800-x)
For the dipiotic weak acid H₂A, Kai = 3.3 X 10 8 az 1.7 X101, what is the pH of a 0.0800 M solution of H₂ A ? *For a diprofic acid, if kazka, then most of the. [Hot] ions come from the 1st dissociation step. pH = H₂A HA+Hts Kai HA A₂A + H² → Kaz [H₂A] 0.0800-X M [H+] = x [HA] XAN kaz Kai -? 2.3 X10 7.7 X10 K₁₁ = [H+] [HA] So: 3,3 × 10 = x² [H₂A] x² + (313X 10)- (0.0800-x)
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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Is there a faster way to solve instead of plugging into the quadratic formula?
![170/400-
Faz
H₂A?
se
².
Byggtong
***
sape-Appry
For the dipretic weak acid H₂A, Ka, 3.3 X10-b
= 7.7 X 109, what is the pH of.
8
a 0.0800 M solution of
*For a diprotic acid, if Kancka, then most of "
[Hot] ions come from the 1st dissociation step.
the
PH= ?
H₂A HA+H²> Ka,
HA
A₂A + H² → Kaz
[H₂A] = 0.0800-X M
[HA] = XA
K₁₁ = [H+] [HA] So: 3,3 × 10 = x²
CH₂A]
x² + (313X 10)-
Kai
3.3 X105
(0,0800-x)
kaz
>7.7.X10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7588948d-a8a4-4b5d-b85a-786e24f76953%2F3d7bd8a4-a8d5-4d0a-aac9-3039af871bbe%2Fri5fp7g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:170/400-
Faz
H₂A?
se
².
Byggtong
***
sape-Appry
For the dipretic weak acid H₂A, Ka, 3.3 X10-b
= 7.7 X 109, what is the pH of.
8
a 0.0800 M solution of
*For a diprotic acid, if Kancka, then most of "
[Hot] ions come from the 1st dissociation step.
the
PH= ?
H₂A HA+H²> Ka,
HA
A₂A + H² → Kaz
[H₂A] = 0.0800-X M
[HA] = XA
K₁₁ = [H+] [HA] So: 3,3 × 10 = x²
CH₂A]
x² + (313X 10)-
Kai
3.3 X105
(0,0800-x)
kaz
>7.7.X10
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