Calculate the pH of the buffered solution if at equilibrium [H2SO3] = 3.50 X 104 M, [HSO31] = 5.5 X 10-2 M, and Ka 1.75 x 10-6 ? everything is based upon the acid dissolving in water: H2SO3 + H20 <-> HSO31 + Hзо*1 I Ка solve for [H3O*1] then put in the numbers Ka [ [H3O+'] = [H3O*1] = = - log a. H30*1 b. Он-1 c. H2C6H606 d. HC6H6061 е. HSO;1 f. SO3? g. H2SO3 h. HC2H3O2 i. C2H3O21 j. H2S k. HS-1 I. s2 m. HX n. X-1 О. 1 р. 2 q. 3 r. 0.5000 s. 5.5 X 10-5 t. 3.5 X 10-4 u. 8.00 x 10-5 v. 1.75 x 10-6 w. 1.11 x 10-8 x. 5.50 x 102 у. О.7000 z. 1.60 x 10-5 аа. 1.143 х 10-5 bb. 0.4900 сс. 1.104 х 10-5 dd. 0.7100 ее. 1.26 х 10-5 ff. 4.901 gg. 7.93 hh. 4.942 ii. 4.957 j. 7.953
Calculate the pH of the buffered solution if at equilibrium [H2SO3] = 3.50 X 104 M, [HSO31] = 5.5 X 10-2 M, and Ka 1.75 x 10-6 ? everything is based upon the acid dissolving in water: H2SO3 + H20 <-> HSO31 + Hзо*1 I Ка solve for [H3O*1] then put in the numbers Ka [ [H3O+'] = [H3O*1] = = - log a. H30*1 b. Он-1 c. H2C6H606 d. HC6H6061 е. HSO;1 f. SO3? g. H2SO3 h. HC2H3O2 i. C2H3O21 j. H2S k. HS-1 I. s2 m. HX n. X-1 О. 1 р. 2 q. 3 r. 0.5000 s. 5.5 X 10-5 t. 3.5 X 10-4 u. 8.00 x 10-5 v. 1.75 x 10-6 w. 1.11 x 10-8 x. 5.50 x 102 у. О.7000 z. 1.60 x 10-5 аа. 1.143 х 10-5 bb. 0.4900 сс. 1.104 х 10-5 dd. 0.7100 ее. 1.26 х 10-5 ff. 4.901 gg. 7.93 hh. 4.942 ii. 4.957 j. 7.953
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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![Calculate the pH of the buffered solution if at equilibrium [H2SO3] = 3.50 X 104 M, [HSO31] = 5.5 X 102 M, and Ka =
1.75 x 10-6 ?
everything is based upon the acid dissolving in water:
H2SO3
+ H20
<-> HSO31 + H3O*1
Ка
[
solve for [H30*'] then put in the numbers
Ka [
[H3O*'] =
[
[H3O*1] =
%3D
pH = - log
a. H30+1
b. Он1
c. H2C6H606
d. HC6H6061
е. HSO31
f. SO3?
-2
g. H2SO3
h. HC2H3O2
i. C2H3O21
j. H2S
k. HS-1
I. s2
m. HX
n. X-1
О. 1
р. 2
q.
3
r. 0.5000
s. 5.5 X 10-5
t. 3.5 X 10-4
u. 8.00 x 10-5
v. 1.75 x 10-6
w. 1.11 x 10-8
х. 5.50 х 10-2
у. О.7000
z. 1.60 x 10-5
аа. 1.143 х 10-5
bb. 0.4900
СС. 1.104 х 10-5
dd. 0.7100
ee. 1.26 x 105 ff. 4.901
gg. 7.93
hh. 4.942
ii. 4.957
j. 7.953](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03121ea6-6cc9-419a-9866-a8a4918051de%2F9624921c-e9e7-48d0-a171-fa915681356f%2Fyouw8xk_processed.png&w=3840&q=75)
Transcribed Image Text:Calculate the pH of the buffered solution if at equilibrium [H2SO3] = 3.50 X 104 M, [HSO31] = 5.5 X 102 M, and Ka =
1.75 x 10-6 ?
everything is based upon the acid dissolving in water:
H2SO3
+ H20
<-> HSO31 + H3O*1
Ка
[
solve for [H30*'] then put in the numbers
Ka [
[H3O*'] =
[
[H3O*1] =
%3D
pH = - log
a. H30+1
b. Он1
c. H2C6H606
d. HC6H6061
е. HSO31
f. SO3?
-2
g. H2SO3
h. HC2H3O2
i. C2H3O21
j. H2S
k. HS-1
I. s2
m. HX
n. X-1
О. 1
р. 2
q.
3
r. 0.5000
s. 5.5 X 10-5
t. 3.5 X 10-4
u. 8.00 x 10-5
v. 1.75 x 10-6
w. 1.11 x 10-8
х. 5.50 х 10-2
у. О.7000
z. 1.60 x 10-5
аа. 1.143 х 10-5
bb. 0.4900
СС. 1.104 х 10-5
dd. 0.7100
ee. 1.26 x 105 ff. 4.901
gg. 7.93
hh. 4.942
ii. 4.957
j. 7.953
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