A solution of phenol (HC,H,O) is prepared by dissolving 0.385 g of phenol in 2.00 L of H,O. The solution has a pH = 6.29. Determine the value of Ka for phenol. 1 2 NEXT Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. HC H,O(aq) H,O(1) H,Oʻ(aq) C,H,O (aq) + + Initial (M) Change (M) Equilibrium (M)
A solution of phenol (HC,H,O) is prepared by dissolving 0.385 g of phenol in 2.00 L of H,O. The solution has a pH = 6.29. Determine the value of Ka for phenol. 1 2 NEXT Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. HC H,O(aq) H,O(1) H,Oʻ(aq) C,H,O (aq) + + Initial (M) Change (M) Equilibrium (M)
Chemistry
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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 3 of 5
A solution of phenol (HC H¸O) is prepared by dissolving 0.385 g of phenol in 2.00
L of H,O. The solution has a pH = 6.29. Determine the value of Ka for phenol.
PREV
1
2
Based on your ICE table and definition of Ka, set up the expression for Ka and then evaluate it.
Do not combine or simplify terms.
Ka =
%3D
5 RESET
[0]
[0.385]
[6.29]
[4.09 x 10*]
[2.05 x 10*]
[5.1 x 10]
[0.799]
[x]
[2x]
[4.09 x 10° + x]
[4.09 x 103 - x]
[2.05 x 10 + x]
[2.05 x 10* - x]
[5.1 x 107 + x]
[5.1 x 107 - x]
1.3 x 1010
7.8 x 10°
4.0 x 103
2.5 x 104
W
A tv A
MAR
PDF
O 28](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b05a37d-01cc-41b5-b183-5c50b16f16a3%2F13bb468e-60b1-40f9-a6dd-776251144aa6%2F5xhnnw8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 3 of 5
A solution of phenol (HC H¸O) is prepared by dissolving 0.385 g of phenol in 2.00
L of H,O. The solution has a pH = 6.29. Determine the value of Ka for phenol.
PREV
1
2
Based on your ICE table and definition of Ka, set up the expression for Ka and then evaluate it.
Do not combine or simplify terms.
Ka =
%3D
5 RESET
[0]
[0.385]
[6.29]
[4.09 x 10*]
[2.05 x 10*]
[5.1 x 10]
[0.799]
[x]
[2x]
[4.09 x 10° + x]
[4.09 x 103 - x]
[2.05 x 10 + x]
[2.05 x 10* - x]
[5.1 x 107 + x]
[5.1 x 107 - x]
1.3 x 1010
7.8 x 10°
4.0 x 103
2.5 x 104
W
A tv A
MAR
PDF
O 28

Transcribed Image Text:Question 3 of 5
A solution of phenol (HC¸H,O) is prepared by dissolving 0.385 g of phenol in 2.00
L of H,O. The solution has a pH = 6.29. Determine the value of Ka for phenol.
1
NEXT
Based on the given values, fill in the ICE table to determine concentrations of all reactants and
products.
HC H̟O(aq)
H,O(1)
H,O (aq)
CHO(aq)
+
+
Initial (M)
Change (M)
Equilibrium (M)
5 RESET
0.385
6.29
-6.29
4.09 x 103
-4.09 x 103
2.05 x 103
-2.05 x 10*
5.1 x 107
-5.1 x 107
0.799
-0.799
+x
4.09 x 10° + x
4.09 x 103 - x
2.05 x 103 + x
2.05 x 10 - x
étv 4 O
NI PDF
P
MAR
28
MacB
B
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