Use the "Square of Death" to determine the following for the solution in problem 6 where pH = 11.3010 [OH-¹] = [H3O+¹] = pOH = -log [OH^¹] = a. acetic b. hydrochloric g. sulfurous c. phosphoric d. perchloric h. hydrosulfuric i. OH-1 j. H30+1 k. SO4² f. sulfuric e. hydrofluoric 1. Mg+2 n. HPO3-² 0. H₂PO3-¹ p. CO3-² q. HCO3-1 r. H₂S S. HS-1 t. S-² u. 0.2500 v. 0.1000 w. 0.02500 z. 1x 10-14 x. 1.000 x 10-3 dd. 2 aa. 2.6990 ff. 2.000 x 10-3 gg. 5.000 x 10-12 mm. conjugate acid bb. 1.000 x 10-11 hh. 11.3010 oo. base CC. 11.0000 ii. 0.01250 pp. conjugate base y. 3.0000 ee. 1 jj. 0.22100 kk. 0.05656 m. Cl-1 II. acid
Use the "Square of Death" to determine the following for the solution in problem 6 where pH = 11.3010 [OH-¹] = [H3O+¹] = pOH = -log [OH^¹] = a. acetic b. hydrochloric g. sulfurous c. phosphoric d. perchloric h. hydrosulfuric i. OH-1 j. H30+1 k. SO4² f. sulfuric e. hydrofluoric 1. Mg+2 n. HPO3-² 0. H₂PO3-¹ p. CO3-² q. HCO3-1 r. H₂S S. HS-1 t. S-² u. 0.2500 v. 0.1000 w. 0.02500 z. 1x 10-14 x. 1.000 x 10-3 dd. 2 aa. 2.6990 ff. 2.000 x 10-3 gg. 5.000 x 10-12 mm. conjugate acid bb. 1.000 x 10-11 hh. 11.3010 oo. base CC. 11.0000 ii. 0.01250 pp. conjugate base y. 3.0000 ee. 1 jj. 0.22100 kk. 0.05656 m. Cl-1 II. acid
Chemistry: Principles and Practice
3rd Edition
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Chapter16: Reactions Between Acids And Bases
Section: Chapter Questions
Problem 16.80QE: Calculate the pH of each of the following solutions. (a) 10.0 mL of 0.300 M hydrofluoric acid plus...
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Question
![Use the "Square of Death" to determine the following for the solution in problem 6 where pH = 11.3010
[OH-¹] =
[H3O+¹] =
pOH = -log [OH-¹] =
a. acetic b. hydrochloric
g. sulfurous
f. sulfuric
c. phosphoric d. perchloric
h. hydrosulfuric i. OH-1 j. H30+1 k. SO4²
0. H₂PO3-1 p. CO3² q. HCO3-¹
e. hydrofluoric
1. Mg+2
-2
-2
n. HPO3-²
-2
r. H₂S
t. S-²
u. 0.2500
v. 0.1000
S. HS-1
y. 3.0000
w. 0.02500
z. 1x 10-14
x. 1.000 x 10-3
dd. 2
aa. 2.6990
ee. 1
ff. 2.000 x 10-³
gg. 5.000 x 10-12
mm. conjugate acid
bb. 1.000 x 10-11
hh. 11.3010
oo. base
CC. 11.0000
ii. 0.01250
pp. conjugate base
jj. 0.22100
kk. 0.05656
m. Cl-1
II. acid](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03121ea6-6cc9-419a-9866-a8a4918051de%2F142279a8-85ff-494b-a313-b83bc9cc2af0%2Fg5ovym4_processed.png&w=3840&q=75)
Transcribed Image Text:Use the "Square of Death" to determine the following for the solution in problem 6 where pH = 11.3010
[OH-¹] =
[H3O+¹] =
pOH = -log [OH-¹] =
a. acetic b. hydrochloric
g. sulfurous
f. sulfuric
c. phosphoric d. perchloric
h. hydrosulfuric i. OH-1 j. H30+1 k. SO4²
0. H₂PO3-1 p. CO3² q. HCO3-¹
e. hydrofluoric
1. Mg+2
-2
-2
n. HPO3-²
-2
r. H₂S
t. S-²
u. 0.2500
v. 0.1000
S. HS-1
y. 3.0000
w. 0.02500
z. 1x 10-14
x. 1.000 x 10-3
dd. 2
aa. 2.6990
ee. 1
ff. 2.000 x 10-³
gg. 5.000 x 10-12
mm. conjugate acid
bb. 1.000 x 10-11
hh. 11.3010
oo. base
CC. 11.0000
ii. 0.01250
pp. conjugate base
jj. 0.22100
kk. 0.05656
m. Cl-1
II. acid
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