Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Determine the pH of a solution of aspirin (acetylsalicylic acid, HC,H,O4) by constructing an
ICE table, writing the equilibrium constant expression, and using this information to
determine the pH. The Ka of aspirin is 3.3 x 10-4. Complete Parts 1-3 before submitting
your answer.
PREV
NEXT
Based on your ICE table and the definition of Ka, set up the expression for Ka in order to determine the
unknown. Each reaction participant must be represented by one tile. Do not combine terms.
[0]
0
[2.75 x 10-3+x]
[0.0153 + x]
[2.75 x 10-³]
6.17
1
[2.75 x 10-3-x]
[0.0153 -x]
Ka =
1
[3.62 x 10-³]
[3.62 x 10-3+x]
[1.53 x 10-5]
---------
[3.62 x 10-a-x] [1.53 x 10-5+x] [1.53 x 10-5-x]
2.25 x 10-3
[15.3]
2.09 x 10-3
2
3
= 3.3 × 10-4
[0.0153]
Determine the pH of a solution of aspirin (acetylsalicylic acid, HC,H,O4) by constructing am
ICE table, writing the equilibrium constant expression, and using this information to
determine the pH. The Ka of aspirin is 3.3 x 10-4. Complete Parts 1-3 before submitting
your answer.
2.68
PREV
on your ICE table and the equilibrium expression for Ka, determine the pH of the aspirin solu
pH =
[x]
[15.3 + x]
5.05 x 10-5
3
RESET
[2x]
5.30
[15.3 -x]
RESET
12.2
expand button
Transcribed Image Text:Determine the pH of a solution of aspirin (acetylsalicylic acid, HC,H,O4) by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The Ka of aspirin is 3.3 x 10-4. Complete Parts 1-3 before submitting your answer. PREV NEXT Based on your ICE table and the definition of Ka, set up the expression for Ka in order to determine the unknown. Each reaction participant must be represented by one tile. Do not combine terms. [0] 0 [2.75 x 10-3+x] [0.0153 + x] [2.75 x 10-³] 6.17 1 [2.75 x 10-3-x] [0.0153 -x] Ka = 1 [3.62 x 10-³] [3.62 x 10-3+x] [1.53 x 10-5] --------- [3.62 x 10-a-x] [1.53 x 10-5+x] [1.53 x 10-5-x] 2.25 x 10-3 [15.3] 2.09 x 10-3 2 3 = 3.3 × 10-4 [0.0153] Determine the pH of a solution of aspirin (acetylsalicylic acid, HC,H,O4) by constructing am ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The Ka of aspirin is 3.3 x 10-4. Complete Parts 1-3 before submitting your answer. 2.68 PREV on your ICE table and the equilibrium expression for Ka, determine the pH of the aspirin solu pH = [x] [15.3 + x] 5.05 x 10-5 3 RESET [2x] 5.30 [15.3 -x] RESET 12.2
Determine the pH of a solution of aspirin (acetylsalicylic acid, HC,H,O4) by constructing an
ICE table, writing the equilibrium constant expression, and using this information to
determine the pH. The Ka of aspirin is 3.3 x 10-4. Complete Parts 1-3 before submitting
your answer.
Initial (M)
Change (M)
Equilibrium (M)
-X
NEXT
>
652 mg of aspirin (HC9H₂O4) is dissolved in an aqueous solution of 237 mL aqueous solution. Fill in the ICE
table with the appropriate value for each involved species to determine concentrations of all reactants and
products.
15.3 - x
0
1
HCgH;O4(aq)
2.75 x 10-3+x
0.0153 + x
2.75 x 10-3
2.75 x 10-3-X
0.0153 - x
+
2
H₂O(1)
3.62 x 10-³
3.62 x 10-3+x
=
1.53 x 10-5
3.62 x 10-3-x
3
H3O+ (aq)
15.3
1.53 x 10-5 + x
0.0153
+
1.53 x 10-5-x
C9H₂O4 (aq)
RESET
+X
15.3 + x
expand button
Transcribed Image Text:Determine the pH of a solution of aspirin (acetylsalicylic acid, HC,H,O4) by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The Ka of aspirin is 3.3 x 10-4. Complete Parts 1-3 before submitting your answer. Initial (M) Change (M) Equilibrium (M) -X NEXT > 652 mg of aspirin (HC9H₂O4) is dissolved in an aqueous solution of 237 mL aqueous solution. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. 15.3 - x 0 1 HCgH;O4(aq) 2.75 x 10-3+x 0.0153 + x 2.75 x 10-3 2.75 x 10-3-X 0.0153 - x + 2 H₂O(1) 3.62 x 10-³ 3.62 x 10-3+x = 1.53 x 10-5 3.62 x 10-3-x 3 H3O+ (aq) 15.3 1.53 x 10-5 + x 0.0153 + 1.53 x 10-5-x C9H₂O4 (aq) RESET +X 15.3 + x
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