レ」 Practice 13) Use an Initial/Change/Equilibrium table and the Ka for benzoic acid (C&H;COOH) to estimate the equilibrium concentrations of unreacted benzoic acid, benzoate (CHSCOO') and H3O* in solution if the initial concentration of benzoic acid was 0.100 M. (Hint-first write the balanced reaction for benzoic acid reacting with water and the expression for K, for benzoic acid). 3.796107- x2 C 6 Hs COOH CHs COO 0.100 Hg O+ kb= ih.tial 1.60 H30+] CIHSO6- change 一X o l00-X equilibrism とオ Kaニ 3.74x0~9 + C.Hs 00 CGHS COOH+H2O )-7 H30* + CHs 00 equilibrium
レ」 Practice 13) Use an Initial/Change/Equilibrium table and the Ka for benzoic acid (C&H;COOH) to estimate the equilibrium concentrations of unreacted benzoic acid, benzoate (CHSCOO') and H3O* in solution if the initial concentration of benzoic acid was 0.100 M. (Hint-first write the balanced reaction for benzoic acid reacting with water and the expression for K, for benzoic acid). 3.796107- x2 C 6 Hs COOH CHs COO 0.100 Hg O+ kb= ih.tial 1.60 H30+] CIHSO6- change 一X o l00-X equilibrism とオ Kaニ 3.74x0~9 + C.Hs 00 CGHS COOH+H2O )-7 H30* + CHs 00 equilibrium
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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Question
I need help with ICE Chart. No matter what I do I don’t get it right, I will show what I have. This is ungraded study guide.
![where HB* represents the conjugate acid of the base.
Questions/Exercises:
12) Write the general Kp expression for a base reacting in water based on the reaction above.
kb= [HBt ] LOH
[B]
Practice
In
13) Use an Initial/Change/Equilibrium table and the Ka for benzoic acid (C6H;COOH) to estimate the
S-
equilibrium concentrations of unreacted benzoic acid, benzoate (C6HSC00') and H3O* in solution if the
initial concentration of benzoic acid was 0.100 M. (Hint-first write the balanced reaction for benzoic acid
reacting with water and the expression for Ka for benzoic acid).
3.79K107- X2
C G Hs COOH
O.100
H3 O+
kb=
30+3 CLHS06-]
[Cしc00円T
in.tial
1.0
change
一X
o100-X
equilibrium
Kaニ 3.4x10-9
C6HG COOH +H0 1)-7 H30* + CHs 00
14) Use an Initial/Change/Equilibrium table and the Kp for aniline (C,H$NH2) to estimate the equilibrium
concentrations of unreacted aniline, its conjugate acid, and OH in solution if the initial concentration of
aniline was 0.230 M. (Hint-first write the balanced reaction equation for aniline reacting with water and
the expression for Kp for aniline).
CHS NH2 + Ha0-7 C6HS'NH3+ + 6H- kbz 1.76 x10-5
X 2
1.76:
0.230*X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F183f5d28-414b-4b67-9cc5-5966880e8282%2F91d9422a-1990-4396-893a-65138d4aabd6%2Fqw3nnjq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:where HB* represents the conjugate acid of the base.
Questions/Exercises:
12) Write the general Kp expression for a base reacting in water based on the reaction above.
kb= [HBt ] LOH
[B]
Practice
In
13) Use an Initial/Change/Equilibrium table and the Ka for benzoic acid (C6H;COOH) to estimate the
S-
equilibrium concentrations of unreacted benzoic acid, benzoate (C6HSC00') and H3O* in solution if the
initial concentration of benzoic acid was 0.100 M. (Hint-first write the balanced reaction for benzoic acid
reacting with water and the expression for Ka for benzoic acid).
3.79K107- X2
C G Hs COOH
O.100
H3 O+
kb=
30+3 CLHS06-]
[Cしc00円T
in.tial
1.0
change
一X
o100-X
equilibrium
Kaニ 3.4x10-9
C6HG COOH +H0 1)-7 H30* + CHs 00
14) Use an Initial/Change/Equilibrium table and the Kp for aniline (C,H$NH2) to estimate the equilibrium
concentrations of unreacted aniline, its conjugate acid, and OH in solution if the initial concentration of
aniline was 0.230 M. (Hint-first write the balanced reaction equation for aniline reacting with water and
the expression for Kp for aniline).
CHS NH2 + Ha0-7 C6HS'NH3+ + 6H- kbz 1.76 x10-5
X 2
1.76:
0.230*X
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