Determine the pH of a solution by constructing a BCA table, constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The value of Ka for HC₂H₂O₂ is 1.8 x 10-5 Complete Parts 1-4 before submitting your answer.

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
Chapter13: Chemical Equilibrium
Section: Chapter Questions
Problem 2ALQ: The boxes shown below represent a set of initial conditions for the reaction: Draw a quantitative...
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Question
pr 6
1
Initial (M)
Change (M)
Equilibrium (M)
ere or pull up for additional resources
Q
A
N
@
2
< PREV
Based on the result of the acid-base reaction (Part 1), set up the ICE table in order to
determine the unknown concentrations of all reactants and products.
W
S
Determine the pH of a solution by constructing a BCA table, constructing an
ICE table, writing the equilibrium constant expression, and using this
information to determine the pH. The value of Ka for HC₂H₂O₂ is 1.8 x 10-5.
Complete Parts 1-4 before submitting your answer.
0.050
0.025 + x
#
3
E
D
HCzH,Oz(aq)
0
0.025
0.025 - x
$
1
4
R
F
0.20
+X
do 5
%
+
T
Question 10 of 11
2
G
...
0.10
-x
6
H₂O (1)
Y
1.0 x 10-¹
3
0.10+x
&
7
U
2.0 x 10-²
0.10 - x
* 00
8
H J
4
HO*(aq)
1
3.0 x 10-³
0.050 + x
9
K
X C V B N M
O
NEXT
+
>
RESET
4.0 x 10-²
0.050 - x
L
C,H,Oz (aq)
P
[
Transcribed Image Text:pr 6 1 Initial (M) Change (M) Equilibrium (M) ere or pull up for additional resources Q A N @ 2 < PREV Based on the result of the acid-base reaction (Part 1), set up the ICE table in order to determine the unknown concentrations of all reactants and products. W S Determine the pH of a solution by constructing a BCA table, constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The value of Ka for HC₂H₂O₂ is 1.8 x 10-5. Complete Parts 1-4 before submitting your answer. 0.050 0.025 + x # 3 E D HCzH,Oz(aq) 0 0.025 0.025 - x $ 1 4 R F 0.20 +X do 5 % + T Question 10 of 11 2 G ... 0.10 -x 6 H₂O (1) Y 1.0 x 10-¹ 3 0.10+x & 7 U 2.0 x 10-² 0.10 - x * 00 8 H J 4 HO*(aq) 1 3.0 x 10-³ 0.050 + x 9 K X C V B N M O NEXT + > RESET 4.0 x 10-² 0.050 - x L C,H,Oz (aq) P [
Thu Apr 6
Before (mol)
Change (mol)
After (mol)
Q
Tap here or pull up for additional resources
A
17
2
NEXT >
Two solutions are mixed: 20.0 mL of 0.20 M HC₂H₂O₂ and 20.0 mL of 0.10 M NaOH. Fill in the
table with the appropriate value for each involved species to determine the moles of reactant
and product after the reaction of the acid and base. You can ignore the amount of water in the
reaction.
W
S
-2.0 x 10³
Determine the pH of a solution by constructing a BCA table, constructing an
ICE table, writing the equilibrium constant expression, and using this
information to determine the pH. The value of Ka for HC₂H₂O₂ is 1.8 x 10-5.
Complete Parts 1-4 before submitting your answer.
X
3
E
D
HC,H,Oz(aq)
4.0 x 10-¹
0
4
-2.0 x 10-³
C
3.0 x 10-³
2.0 x 10-³
R
F
20.0
-3.0 x 10³
5
V
+
T
Question 10 of 11
2
G
0.20
OH (aq)
6
2.0 x 10-¹
4.0 x 10.³
-2.0 x 10-³
0
Y
B
0.10
H
3
-4.0 x 10°³
U
→
N
1.0 x 10-¹
8
J
1
4
H₂O(l)
-1.0 x 10"
9
M
K
O
+
2.0 x 10-³
O
<
RESET
H
L
CH,O;(aq)
P
0
2.0 x 10-³
2.0 x 10"¹
... 2
Transcribed Image Text:Thu Apr 6 Before (mol) Change (mol) After (mol) Q Tap here or pull up for additional resources A 17 2 NEXT > Two solutions are mixed: 20.0 mL of 0.20 M HC₂H₂O₂ and 20.0 mL of 0.10 M NaOH. Fill in the table with the appropriate value for each involved species to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of water in the reaction. W S -2.0 x 10³ Determine the pH of a solution by constructing a BCA table, constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The value of Ka for HC₂H₂O₂ is 1.8 x 10-5. Complete Parts 1-4 before submitting your answer. X 3 E D HC,H,Oz(aq) 4.0 x 10-¹ 0 4 -2.0 x 10-³ C 3.0 x 10-³ 2.0 x 10-³ R F 20.0 -3.0 x 10³ 5 V + T Question 10 of 11 2 G 0.20 OH (aq) 6 2.0 x 10-¹ 4.0 x 10.³ -2.0 x 10-³ 0 Y B 0.10 H 3 -4.0 x 10°³ U → N 1.0 x 10-¹ 8 J 1 4 H₂O(l) -1.0 x 10" 9 M K O + 2.0 x 10-³ O < RESET H L CH,O;(aq) P 0 2.0 x 10-³ 2.0 x 10"¹ ... 2
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