Calculate the K₂ for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn't you could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 = H3O+1 start: equil: HX + H₂O 0.100000 M 0.099536 M 0.000464 M X-1 + 0 0.000464 M 0 0.000464 M mol of The stoichiometry tells us that if we made 0.000464 mol of H3O+1 we also made X-1 The amount of HX at equilibrium is 0.100000 M HX- M HX |][ [ x [ Ka = Ka = a. acetic h. CaCO3(s) 0. H₂PO4¹ v. H₂CO3 bb. 0.1000 hh. 2.16 x 10-6 b. X-1 i. OH-¹ P. CO3-2 W. PO43 CC. 3 c. HX d. Al+3 j. H30+1 q. HCO3-1 x. HC₂H3O2 dd. 2 ii. 1.60 x 10-5 k. SO4² r. H₂S e. CO₂ f. CaF2(aq) 1. Mg+2 S. HS-1 y. C₂H30₂-1 ee. 1 ff. 0.099536 jj. 0.2000 m. C-1 t. 5-² Z. 1x 10-14 MX1 = kk. 6.600 x 10-3 g. HF(aq) n. HPO4² u. H₂O(liq) aa. 0.000464 gg. 3.95 x 10-3
Calculate the K₂ for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn't you could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 = H3O+1 start: equil: HX + H₂O 0.100000 M 0.099536 M 0.000464 M X-1 + 0 0.000464 M 0 0.000464 M mol of The stoichiometry tells us that if we made 0.000464 mol of H3O+1 we also made X-1 The amount of HX at equilibrium is 0.100000 M HX- M HX |][ [ x [ Ka = Ka = a. acetic h. CaCO3(s) 0. H₂PO4¹ v. H₂CO3 bb. 0.1000 hh. 2.16 x 10-6 b. X-1 i. OH-¹ P. CO3-2 W. PO43 CC. 3 c. HX d. Al+3 j. H30+1 q. HCO3-1 x. HC₂H3O2 dd. 2 ii. 1.60 x 10-5 k. SO4² r. H₂S e. CO₂ f. CaF2(aq) 1. Mg+2 S. HS-1 y. C₂H30₂-1 ee. 1 ff. 0.099536 jj. 0.2000 m. C-1 t. 5-² Z. 1x 10-14 MX1 = kk. 6.600 x 10-3 g. HF(aq) n. HPO4² u. H₂O(liq) aa. 0.000464 gg. 3.95 x 10-3
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
Chapter14: Acids And Bases
Section: Chapter Questions
Problem 7RQ: Table 13-4 lists the stepwise Ka values for some polyprotic acids. What is the difference between a...
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Question
![Calculate the Ka for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn't
you could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 =
+
start:
equil:
HX + H₂O S
0.100000 M
0.099536 M
0.000464 M
X-1
0
0.000464 M
H3O+1
0
0.000464 M
mol of
The stoichiometry tells us that if we made 0.000464 mol of H3O+¹ we also made
X-1
The amount of HX at equilibrium is 0.100000 M HX-
M HX
][
[
]
D(
[
Ka
Ka
a. acetic
h. CaCO3(s)
i. OH-¹
-1
0. H₂PO4¹ p. CO3-²
v. H₂CO3
bb. 0.1000
W. PO4-³
CC. 3
hh. 2.16 x 10-6
b. X-1
c. HX
d. Al+3
j. H30+1
q. HCO3-1
x. HC₂H3O2
dd. 2
ii. 1.60 x 10-5
k. SO4²
r. H₂S
e. CO₂
ee. 1
jj. 0.2000
y. C₂H30₂-1
1. Mg+2
S. HS-1
f. CaF2(aq)
ff. 0.099536
m. Cl-¹
t. S-²
z. 1 x 10-14
MX-1 =
kk. 6.600 x 10-³
g. HF(aq)
n. HPO4²
u. H₂O(liq)
aa. 0.000464
gg. 3.95 x 10-3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03121ea6-6cc9-419a-9866-a8a4918051de%2F6d9a5fd1-7bbd-4926-836a-e706b47d1f2e%2Fwv7r0rg_processed.png&w=3840&q=75)
Transcribed Image Text:Calculate the Ka for a monoprotic acid, HX, if a 0.100000 M solution has a pH of 3.333. The data is provided. If it wasn't
you could get the concentrations by converting the pH into [H3O+¹] using [H3O+1] = 10-PH = 10-3.333 =
+
start:
equil:
HX + H₂O S
0.100000 M
0.099536 M
0.000464 M
X-1
0
0.000464 M
H3O+1
0
0.000464 M
mol of
The stoichiometry tells us that if we made 0.000464 mol of H3O+¹ we also made
X-1
The amount of HX at equilibrium is 0.100000 M HX-
M HX
][
[
]
D(
[
Ka
Ka
a. acetic
h. CaCO3(s)
i. OH-¹
-1
0. H₂PO4¹ p. CO3-²
v. H₂CO3
bb. 0.1000
W. PO4-³
CC. 3
hh. 2.16 x 10-6
b. X-1
c. HX
d. Al+3
j. H30+1
q. HCO3-1
x. HC₂H3O2
dd. 2
ii. 1.60 x 10-5
k. SO4²
r. H₂S
e. CO₂
ee. 1
jj. 0.2000
y. C₂H30₂-1
1. Mg+2
S. HS-1
f. CaF2(aq)
ff. 0.099536
m. Cl-¹
t. S-²
z. 1 x 10-14
MX-1 =
kk. 6.600 x 10-³
g. HF(aq)
n. HPO4²
u. H₂O(liq)
aa. 0.000464
gg. 3.95 x 10-3
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