003050 x 100 060 dissociation chem 162 2007 exam ii

% dissociation = (0.003/0.50) x 100 = 0.60% dissociation
Chem 162-2007 exam II review session 45

ET: Given initial concentration and equilibrium concentration of weak acid, find Ka; define 1.00 x 10 -2 M as initial, and 2.77 pH as at equilibrium.
67. The pH of a 1.00 x 10 -2 M solution of cyanic acid (HOCN) is 2.77
at 25 o C. Calculate K
a
for HOCN from this result.
HOCN + H
2
O H
3
O
+
+ OCN
-
0.0100M HCN
pH 2.77 = 10
-2.77
= 1.70 x 10
-3
[H
3
O
+
]
HOCN + H
2
O H
3
O
+
+ OCN
-
Initial 0.0100
Change
Equilibrium 0.00170
HOCN + H
2
O H
3
O
+
+ OCN
-
Initial 0.0100 0 0
Change -X +X +X
Equilibrium 0.0100 - X 0.00170 +X
HOCN + H
2
O H
3
O
+
+ OCN
-
Initial 0.0100 0 0
Change -0.00170 +0.00170 +0.00170
Equilibrium 0.0083 0.00170 0.00170
K
a
= ([H
3
O
+
][OCN
-
])/[HOCN]
K
a
= ((0.00170)(0.00170))/0.0083 = 3.48 x 10 -4
Chem 162-2007 exam II review session 46

ET: Given initial concentration of and equilibrium concentration of weak base, find Kb.
91 (mod.)
Codeine (C
18
H
21
NO
3
) is a derivative of morphine, a weak base, that is used as
an analgesic, narcotic, or antitussive. It was once commonly used in cough
syrups but is now available only by prescription because of its addictive
properties. If the pH of a 1.7 x 10 -3 M solution of codeine is 9.59, calculate K
b
.
RNH
2
+ H
2
O RNH
3
+ + OH -
1.7 x 10
-3
M codeine = RNH
2
pH = 9.59
pOH = 14.00 - 9.59 = 4.41
10
-pOH
= [OH
-
]
10 -4.41 = 3.89 x 10 -5 = [OH - ]
RNH
2
+ H
2
O RNH
3
+ + OH -
Initial 1.7 x 10 -3 M 0 0
Change
Equilibrium +3.89 x 10
-5
RNH
2
+ H
2
O RNH
3
+ + OH -
Initial 1.7 x 10
-3
M 0 0
Change -X +X +X
Equilibrium 1.7x10 -3 -X +X +3.89 x 10 -5
RNH
2
+ H
2
O RNH
3
+
+ OH
-
Initial 1.7 x 10
-3
M 0 0
Change -3.89 x 10 -5 +3.89 x 10 -5 +3.89 x 10 -5
Equilibrium 1.66 x 10 -3 +3.89 x 10 -5 +3.89 x 10 -5
K
b
= ([RNH
3
+ ][OH - ])/[RNH
2
]
((+3.89 x 10 -5 ) x (+3.89x10 -5 ))/(1.66 x 10 -3 ) = 9.11 x 10 -7 = K
b
Chem 162-2007 exam II review session 47

SKIP
ET: Given equilibrium concentration and Ka, find initial concentration
24. A saturated solution of benzoic acid, C
6
H
5
COOH, has pH =
2.93. What amount of benzoic acid is dissolved* in a total of 1.00 L
of this solution? K
a
= 6.3x10 -5 for C
6
H
5
COOH
*What is the initial concentration of benzoic acid in solution?
A
. 0.023 mol
B. 0.16 mol
C. 0.30 mol
D. 0.080 mol
E. 0.016 mol
pH 2.93
[H
+
] = 10
-2.93
= 0.001175
K
a
= 6.3 x 10 -5
HB + H
2
O
←
→
B
-
+ H
3
O
+
HB(aq) +
H
2
O(l)
←
→
B
-
(aq) + H
3
O
+
(aq)
Initial Y
Change
Equilibrium +0.001175
HB(aq) +
H
2
O(l)
←
→
B
-
(aq) + H
3
O
+
(aq)
Initial Y 0 0
Change -X +X +X
Equilibrium Y - X +X +0.001175
HB(aq) +
H
2
O(l)
←
→
B - (aq) + H
3
O + (aq)
Initial Y 0 0
Change -0.001175 +0.001175 +0.001175
Equilibrium Y - 0.001175 +0.001175 +0.001175
K
a
= ([B
-
][H
3
O
+
])/[HB] = 6.3 x 10
-5
([0.001175][0.001175])/[Y - 0.001175] = 6.3 x 10 -5
Y = 2.2 x 10
-2
Chem 162-2007 exam II review session 48

POLYPROTIC ACIDS
Chem 162-2004 Exam II
Acids and Bases - Chapter 14
Solutions of acids and bases, K
a
, K
b
, pH, pOH, % dissoc.
ET: Define polyprotic acid
H
2
CO
3
is a diprotic acid with K
a1
= 4.3 x 10 -7
;
K
a2
= 5.6 x 10 -11
Calculate the pH of 0.025 M H
2
CO
3
.
A. 6.27
B . 3.98
C. 4.49
D. 5.02
E. 5.63
H
2
CO
3
+ H
2
O
←
→
HCO
3
- + H
3
O + K
a1
= 4.3 x 10 -7
HCO
3
- + H
2
O
←
→
CO
3
2- + H
3
O + K
a2
= 5.6 x 10
-11
H
2
CO
3
H
2
O
←
→
HCO
3
-
H
3
O
+
Initial 0.025 0 0
Change -X +X +X
Equilibrium 0.025-X +X +X
([HCO
3
- ][H
3
O + ])/[H
2
CO
3
] = 4.3 x 10 -7
([X][X])/[0.025-X] = 4.3 x 10 -7
X = 1.035 x 10 -4 = [H
3
O + ] = [HCO
3
- ]
Considering only this first step, pH = -log(1.035 x 10 -4 ) = 3.99
Based on the extremely small K
a2
, the HCO
3
- should dissociate very little to provide
additional [H
3
O
+
], so we can probably just use the first ICE table to get the [H
3
O
+
].