acetic acid, has a k a of 1.7 x 10 -5 . determine the ph of a 0.10 m solution of acetic acid
DESCRIPTION
Acetic acid, has a K a of 1.7 x 10 -5 . Determine the pH of a 0.10 M solution of acetic acid. CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO -. Hint: First write out the equilibrium expression of the acid in water. - PowerPoint PPT PresentationTRANSCRIPT
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Acetic acid, has a Ka of 1.7 x 10-5. Determine the pH of a 0.10 Msolution of acetic acid.
Hint: First write out the equilibrium expression of the
acid in water.
CH3COOH + H2O ↔ H3O+ + CH3COO-
2
Now make a I.C.E. Chart and fill in the
values
CH3COOH + H2O ↔ H3O+ + CH3COO-
I.
C.
E.
0.10 0 0
-x +x +x
0.10-x +x +x
Acetic acid, has a Ka of 1.7 x 10-5. Determine the pH of a 0.10 Msolution of acetic acid.
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Acetic acid, has a Ka of 1.7 x 10-5. Determine the pH of a 0.10 Msolution of acetic acid.
Create an equilibrium expression from the
“E.” term.
CH3COOH + H2O ↔ H3O+ + CH3COO-
I.
C.
E.
0.10 0 0
-x +x +x
0.10-x +x +x
COOH][CH]COO][CHO[HK
3
33a
so: x
xx
1001071
25
..
Try dropping the -x term. If the value of x comes out toless than 5% of 0.10dropping the term isjustified.
x = 1.30 x 10-3 so dropping the term was valid.
since x = [H3O+] , pH = -log(1.30 x 10-3) = 2.88
and pH = 14 – 2.87 = 11.12
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A weak acid, HA, has a pKa of 5.82. Determine the pH of a 0.050 Msolution of HA.
Hint: First write out the equilibrium expression of the
acid in water.
HA + H2O ↔ H3O+ + A-
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A weak acid, HA, has a pKa of 5.82. Determine the pH of a 0.050 Msolution of HA.
Now make a I.C.E. Chart and fill in the
values
HA + H2O ↔ H3O+ + A-
I.
C.
E.
0.050 0 0
-x +x +x
0.050-x +x +x
6
A weak acid, HA, has a pKa of 5.82. Determine the pH of a 0.050 Msolution of HA.
Determine the Ka and create an equilibrium expression from the
“E.” term.
HA + H2O ↔ H3O+ + A-
I.
C.
E.
0.050 0 0
-x +x +x
0.050-x +x +x
Since pKa = 5.82. The value of Ka = 10-5.82 = 1.51 x 10-6
]HA[]A][OH[K 3
a
so: x050.0
x10x51.12
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Try dropping the -x term. If the value of x comes out toless than 5% of 0.050dropping the term isjustified.
x = 2.75 x 10-4 so dropping the term was valid.
since x = [H3O+] , pH = -log(2.75 x 10-4) = 3.56
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Ammonia, has a Kb of 1.8 x 10-5. Determine the pH of a 0.10 Msolution of ammonia.
Hint: First write out the equilibrium expression of the
base in water.
NH3 + H2O ↔ NH4+ + OH-
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Ammonia, has a Kb of 1.8 x 10-5. Determine the pH of a 0.10 Msolution of ammonia..
Now make a I.C.E. Chart and fill in the
values
NH3 + H2O ↔ NH4+ + OH-
I.
C.
E.
0.10 0 0
-x +x +x
0.10-x +x +x
10
Ammonia, has a Kb of 1.8 x 10-5. Determine the pH of a 0.10 Msolution of ammonia.
Create an equilibrium expression from the
“E.” term.
NH3 + H2O ↔ NH4+ + OH-
I.
C.
E.
0.10 0 0
-x +x +x
0.10-x +x +x
]B[]OH][BH[Kb
so: x10.0
x10x8.12
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Try dropping the -x term. If the value of x comes out toless than 5% of 0.10dropping the term isjustified.
x = 1.34 x 10-3 so dropping the term was valid.
since x = [OH-] , pOH = -log(1.34 x 10-3) = 2.87
and pH = 14 – 2.87 = 11.13
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A weak base, B, has a pKb of 5.82. Determine the pH of a 0.050 Msolution of B.
Hint: First write out the equilibrium expression of the
base in water.
B + H2O ↔ BH+ + OH-
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A weak base, B, has a pKb of 5.82. Determine the pH of a 0.050 Msolution of B.
Now make a I.C.E. Chart and fill in the
values
B + H2O ↔ BH+ + OH-
I.
C.
E.
0.050 0 0
-x +x +x
0.050-x +x +x
13
A weak base, B, has a pKb of 5.82. Determine the pH of a 0.050 Msolution of B.
Determine the Kb and create an equilibrium expression from the
“E.” term.
B + H2O ↔ BH+ + OH-
I.
C.
E.
0.050 0 0
-x +x +x
0.050-x +x +x
Since pKb = 5.82. The value of Kb = 10-5.82 = 1.51 x 10-6
]B[]OH][BH[Kb
so: x050.0
x10x51.12
6
Try dropping the -x term. If the value of x comes out toless than 5% of 0.050dropping the term isjustified.
x = 2.75 x 10-4 so dropping the term was valid.
since x = [OH-] , pOH = -log(2.75 x 10-4) = 3.56
and pH = 14 – 3.56 = 10.44
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Trimethlyamine, has a Kb of 6.4 x 10-5. Determine the pH of a 0.10 Msolution of trimethylamine. (CH3)3N
Hint: First write out the equilibrium expression of the
base in water.
(CH3)3N + H2O ↔ (CH3)3NH+ + OH-
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Trimethlyamine, has a Kb of 6.4 x 10-5. Determine the pH of a 0.10 Msolution of trimethylamine. (CH3)3N
Now make a I.C.E. Chart and fill in the
values
(CH3)3N + H2O ↔ (CH3)3NH+ + OH-
I.
C.
E.
0.10 0 0
-x +x +x
0.10-x +x +x
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Trimethlyamine, has a Kb of 6.4 x 10-5. Determine the pH of a 0.10 Msolution of trimethylamine. (CH3)3N
Create an equilibrium expression from the
“E.” term.
(CH3)3N + H2O ↔ (CH3)3NH+ + OH-
I.
C.
E.
0.10 0 0
-x +x +x
0.10-x +x +x
]B[]OH][BH[Kb
so: x
xx
10.0104.6
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Try dropping the -x term. If the value of x comes out toless than 5% of 0.10dropping the term isjustified.
x = 2.53 x 10-3 so dropping the term was valid.
since x = [OH-] , pOH = -log(2.53 x 10-3) = 2.60
and pH = 14 – 2.60 = 11.40
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