The pH for 0.0850 M solution of C₆H₅CH₂COOH is 2.68. Determine the value of Ka for C₆H₅CH₂COOH. Based on the given values, fill in the ICE table to determine concentrations of all reactants and products.
The pH for 0.0850 M solution of C₆H₅CH₂COOH is 2.68. Determine the value of Ka for C₆H₅CH₂COOH. Based on the given values, fill in the ICE table to determine concentrations of all reactants and products.
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Question
The pH for 0.0850 M solution of C₆H₅CH₂COOH is 2.68. Determine the value of Ka for C₆H₅CH₂COOH.
Based on the given values, fill in the ICE table to determine concentrations of all reactants and products.
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The pH for 0.0850 M solution of C6HSCH2COOH is 2.68. Determine the value of Ka
for CoHsCH2COOH.
1
2
NEXT
>
Based on the given values, fill in the ICE table to determine concentrations of all reactants and
products.
C6HSCH:COOH(aq)+
H:O(1)
H:O*(aq)
+CoHsCH2COO (aq)
Initial (M)
0.0850
Change (M)
-X
+x
+x
Equilibrium (M)
0.0829
2.1 x 10-3
2.1 x 10-3
Incorrect, 8 attempts remaining
Your Change in concentration for C6HSCH2COOH is incorrect. In this problem, you should use
the given pH to determine the concentration of H:O* or OH- at Equilibrium as appropriate.
From there, you should be able to determine the Change of each species, which is the
difference between the Equilibrium and Initial concentration. Recall that pH = -log([H:O*]) and
that [H:O*][OH] = 1 × 10-14
Your Change in concentration for H3O* is incorrect. In this problem, you should use the given
pH to determine the concentration of H:O+ or OH at Equilibrium as appropriate. From there,
you should be able to determine the Change of each species, which is the difference between
the Equilibrium and Initial concentration. Recall that pH = -log([H:O*]) and that [H:O*][OH] = 1
x 10-14
%3D
+
Your Change in concentration for C6HSCH2COO- is incorrect. In this problem, you should use
12:47 PM
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Question 1 of 29
Retry
The pH for 0.0850 M solution of C6HSCH2COOH is 2.68. Determine the value of Ka
for CoHsCH2COOH.
1
2
NEXT
>
Based on the given values, fill in the ICE table to determine concentrations of all reactants and
products.
C6HSCH:COOH(aq)+
H:O(1)
H:O*(aq)
+CoHsCH2COO (aq)
Initial (M)
0.0850
Change (M)
-X
+x
+x
Equilibrium (M)
0.0829
2.1 x 10-3
2.1 x 10-3
Incorrect, 8 attempts remaining
Your Change in concentration for C6HSCH2COOH is incorrect. In this problem, you should use
the given pH to determine the concentration of H:O* or OH- at Equilibrium as appropriate.
From there, you should be able to determine the Change of each species, which is the
difference between the Equilibrium and Initial concentration. Recall that pH = -log([H:O*]) and
that [H:O*][OH] = 1 × 10-14
Your Change in concentration for H3O* is incorrect. In this problem, you should use the given
pH to determine the concentration of H:O+ or OH at Equilibrium as appropriate. From there,
you should be able to determine the Change of each species, which is the difference between
the Equilibrium and Initial concentration. Recall that pH = -log([H:O*]) and that [H:O*][OH] = 1
x 10-14
%3D
+
Your Change in concentration for C6HSCH2COO- is incorrect. In this problem, you should use
12:47 PM
e Type here to search
10/23/2020
...
1L
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