The inverse logarithm, or -log, of [H3O*l][OH*l] = 10*14 is pH + pOH = 14 1. What happens to the [H3O*'] when acid is added to neutral water ? the [H3O*']goes 2. What happens to the pH when acid is added to neutral water ? an example of an increased [H3O*l] is [H3O*l] = 1 x 10-6. Thus pH = -log[10-6] = 6 thus, adding acid makes the pH go 3. What happens to the [OH-'] when acid is added to neutral water ? in the equation [H3O*'][OHl] = 10-14, when [H3O*l] increases, the [OH'] goes 4. What happens to the pOH when acid is added to neutral water ? when [OH] decreases, the pOH goes since the pOH = -log[OH¨'] this can also be seen using pH + pOH = 14 5. What happens to the Kw when acid is added to neutral water ? because it's a constant. a. down b. up c. no change
The inverse logarithm, or -log, of [H3O*l][OH*l] = 10*14 is pH + pOH = 14 1. What happens to the [H3O*'] when acid is added to neutral water ? the [H3O*']goes 2. What happens to the pH when acid is added to neutral water ? an example of an increased [H3O*l] is [H3O*l] = 1 x 10-6. Thus pH = -log[10-6] = 6 thus, adding acid makes the pH go 3. What happens to the [OH-'] when acid is added to neutral water ? in the equation [H3O*'][OHl] = 10-14, when [H3O*l] increases, the [OH'] goes 4. What happens to the pOH when acid is added to neutral water ? when [OH] decreases, the pOH goes since the pOH = -log[OH¨'] this can also be seen using pH + pOH = 14 5. What happens to the Kw when acid is added to neutral water ? because it's a constant. a. down b. up c. no change
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
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The Keq for the reaction of water with water is called Kw
H2O + H2O ⇔ H3O+1 + OH-1
[H3O+1][OH-1]
Kw = ––––––––––––– = [H3O+1][OH-1] = (1 x 10-7)2 = 1 x 10-14
[H2O]2
![The inverse logarithm, or -log, of [H3O*][OH1] = 10-14 is pH + pOH = 14
1. What happens to the [H30*1] when acid is added to neutral water ?
the [H3O*'] goes
2. What happens to the pH when acid is added to neutral water ?
an example of an increased [H30*l] is [H30*l]=1 x 10-6. Thus pH = -log[10ʻ] = 6
thus, adding acid makes the pH go
3. What happens to the [OH'] when acid is added to neutral water ?
in the equation [H3O+'][OH´1] = 10-14, when [H3O*1] increases, the [OH1] goes
4. What happens to the pOH when acid is added to neutral water ?
when [OH-'] decreases, the pOH goes
since the pOH = -log[OH1]
this can also be seen using pH + pOH = 14
5. What happens to the Kw when acid is added to neutral water ?
because it's a constant.
a. down
b. up
c. no change](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a9cf46f-1923-4110-8bfb-3c5048be5f9e%2F76f10bc4-42a1-408d-b2df-4c9af43f8aa0%2F118rqv_processed.png&w=3840&q=75)
Transcribed Image Text:The inverse logarithm, or -log, of [H3O*][OH1] = 10-14 is pH + pOH = 14
1. What happens to the [H30*1] when acid is added to neutral water ?
the [H3O*'] goes
2. What happens to the pH when acid is added to neutral water ?
an example of an increased [H30*l] is [H30*l]=1 x 10-6. Thus pH = -log[10ʻ] = 6
thus, adding acid makes the pH go
3. What happens to the [OH'] when acid is added to neutral water ?
in the equation [H3O+'][OH´1] = 10-14, when [H3O*1] increases, the [OH1] goes
4. What happens to the pOH when acid is added to neutral water ?
when [OH-'] decreases, the pOH goes
since the pOH = -log[OH1]
this can also be seen using pH + pOH = 14
5. What happens to the Kw when acid is added to neutral water ?
because it's a constant.
a. down
b. up
c. no change
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