The pH for 0.0715 M solution of CCl,CO2H is 1.40. Determine the value of Ka for CCl,CO2H. 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

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Chapter14: Acids And Bases
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The pH for 0.0715 M solution of CCl;CO2H is 1.40.
Determine the value of Ka for CCl,CO2H.
1
2
Based on the given values, fill in the ICE table to
determine concentrations of all reactants and
products.
Cl,CO,H(a
q)
H20(1)
= H3O*(aq) +
CCl,CO; (aq
Initial (M)
Change (M)
Equilibrium
(M)
5 RESET
0.0715
1.40
-1.40
0.040
-0.040
0.032
-0.032
+x
-x
0.0715 + x
0.0715 - x
1.40 + x
1.40 - x
0.040 + x
0.0340 - x
0.032 + x
0.032 - x
Transcribed Image Text:ll Verizon LTE 12:36 PM O 29% O Question 3 of 16 Submit The pH for 0.0715 M solution of CCl;CO2H is 1.40. Determine the value of Ka for CCl,CO2H. 1 2 Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Cl,CO,H(a q) H20(1) = H3O*(aq) + CCl,CO; (aq Initial (M) Change (M) Equilibrium (M) 5 RESET 0.0715 1.40 -1.40 0.040 -0.040 0.032 -0.032 +x -x 0.0715 + x 0.0715 - x 1.40 + x 1.40 - x 0.040 + x 0.0340 - x 0.032 + x 0.032 - x
ll Verizon LTE
12:36 PM
O 29% O
Question 3 of 16
Submit
The pH for 0.0715 M solution of CCl;CO2H is 1.40.
Determine the value of Ka for CCl,CO2H.
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.
Ка
5 RESET
[0]
[0.0715]
[1.40]
[0.040]
[0.032]
[0.146]
[0.0715 + x]
[0.0715 - x]
[1.40 + x]
[1.40 - x]
[0.040 + x]
[0.040 - x]
[0.032 + x]
[0.032 - x]
[0.146 + x]
[0.146 - x]
2.0 x 101
0.050
1.3
0.80
Transcribed Image Text:ll Verizon LTE 12:36 PM O 29% O Question 3 of 16 Submit The pH for 0.0715 M solution of CCl;CO2H is 1.40. Determine the value of Ka for CCl,CO2H. 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. Ка 5 RESET [0] [0.0715] [1.40] [0.040] [0.032] [0.146] [0.0715 + x] [0.0715 - x] [1.40 + x] [1.40 - x] [0.040 + x] [0.040 - x] [0.032 + x] [0.032 - x] [0.146 + x] [0.146 - x] 2.0 x 101 0.050 1.3 0.80
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