63 k a 31 x 10 8 hx h 2 x 2 initial 1002 050 change

63

(K
a
= 3.1 x 10
-8
)
HX H
2
X
2
Initial 1.00/2 = 0.50 0 0
Change -2X +X +X
Equilibrium 0.50-2X +X +X
X
2
/(0.50 - 2X)
2
= 3.1 x 10
-8
X/(0.50-2X) = 1.76 x 10
-4
X = 0.88 x 10
-4
= [H
2
] = [X
2
]
0.50 = [HX]
2002 Quiz
CHEM162-2003 5
TH
WEEK RECITATION
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
3. Consider the reaction: 2HI(g) H
2
(g) + I
2
(g)
A container originally holds 0.50 M HI. Suppose 10.0% of this
HI reacts as in the above equation, to reach equilibrium.
Calculate the value of K for this reaction.
HI H
2
I
2
Initial 0.50 0 0
Change
Equilibrium
HI H
2
I
2
64

Initial 0.50 0 0
Change -0.05 +0.025 +0.025
Equilibrium 0.45 0.025 0.025
([0.025] x [0.025])/[0.45] 2 = 3.1 x 10 -3 = K
CHEM162-2003 5 TH WEEK RECITATION
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
15.23
Calculate the pH of each of the following solutions.
a. 0.100 M propanoic acid (HC
3
H
5
O
2
, K
a
= 1.3 x 10
-5
)
(a) HP + H
2
O


H
3
O
+
+ P
-
HP H
3
O + P -
Initial 0.100 0 0
Change -X +X +X
Equilibrium 0.100 - X +X +X
X
2
/(0.1-X) = 1.3 x 10
-5
X = 1.140 x 10
-3
pH = -log (1.140 x 10 -3 ) = 2.94
Chapter 14
65

4. CHEM162-2000 HOURLY EXAM II
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
2 H
2
O(
l
)


H
3
O
+
(aq) + OH
-
(aq) K
w
= 1.0 x 10
-14
at 25  C
K
w
= 3.8 x 10
-14
at 40  C
Which statement is true concerning pure water at 40  C?
A. pH = 7.00 pOH = 7.00
B
. pH = 6.71 pOH = 6.71
C. pH = 6.71 pOH = 7.29
D. pH = 7.29 pOH = 6.71
E. pH = 7.00 pOH = 6.71
K
w
= [H + ][OH - ] = 3.8 x 10 -14
X = [H
+
] = [OH
-
]
X 2 = 3.8 x 10 -14
X = 1.95 x 10 -7 = [H + ] = [OH - ]
pH = -log[H + ] = -log(1.95 x 10 -7 ) = 6.71
pOH = -log[OH + ] = -log(1.95 x 10 -7 ) = 6.71
8. CHEM162-2000 HOURLY EXAM II
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
What is the pH of a solution prepared by dissolving 0.80 grams of NaOH (molar mass
= 40.0 g/mol) in enough water to make 3.0 liters of solution?
(0.80 g/40 g/mol
-1
)/3.0 L = 0.00667 M NaOH = 0.0067 M OH
-
pOH = -log [OH - ] = -log [0.00667] = 2.18
pH = 14.0 - 2.18 = 11.82
A. 10.89
B. 11.37
C
. 11.82
D. 12.30
E. 10.52
66

9. CHEM162-2000 HOURLY EXAM II
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
What is the pH of a 0.50 M aqueous HNO
2
solution? K
a
= 4.3 x 10
-4
for HNO
2
A. 3.37
B. 3.67
C. 4.67
D
. 1.84
E. 2.37
10. CHEM162-2000 HOURLY EXAM II
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
NaY is the sodium salt of hypothetical weak acid HY. If a 0.10 M aqueous NaY
solution has pH = 11.00, then what is the value of K
a
for HY?
Y - + H
2
O


HY + OH -
pH 11 = pOH 3
[OH
-
] = 1 x 10
-3
Y - H
2
O HY OH -
Initial 0.10 0 0
Change -1 x 10 -3 +1 x 10 -3 +1 x 10 -3
Equilibrium 0.1 - 0.001 =
0.1
1 x 10 -3 1 x 10 -3
([HY][OH - ])/[Y - ] = K
b’
([1 x 10 -3 ][1 x 10 -3 ])/[0.1] = K
b’
1 x 10
-5
= K
b’
K
w
= K
a
x K
b
K
a
= K
w
/K
b’
K
a
= (1 x 10 -14 )/(1 x 10 -5 )
K
a
= 1 x 10
-9
A. 1.0 x 10
-5
B. 1.0 x 10
-3
C. 1.0 x 10
-7
D
. 1.0 x 10
-9
E. 1.0 x 10
-11
67

12. CHEM162-2000 HOURLY EXAM II
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
Hydrazine, N
2
H
4
, is a weak base with K
b
= 1.7 x 10
-6
. What is the percent
protonation of N
2
H
4
in a 0.010 M aqueous solution of hydrazine?
A. 2.1 %
B
. 1.3 %
C. 0.54 %
D. 2.8 %
E. 0.85 %
N
2
H
4
+ H
2
O


N
2
H
5
+ + OH -
N
2
H
4
+
H
2
O


N
2
H
5
+ + OH -
Initial 0.010 0 0
Change -X +X +X
Equilibrium 0.010 - X +X +X
([X] x [X])/(0.010 - X) = 1.7 x 10 -6
((([X] x [X])/(0.010 - X) = (1.7 x 10 -6 )),X)
X = 1.30 x 10 -4
mole percent protonation = ((1.30 x 10
-4
)/0.010) x 100 = 1.3
68

14. CHEM162-2000 HOURLY EXAM II
CHAPTER 14 - ACID AND BASE EQUILIBRIA
ACID AND BASE EQUILIBRIA CALCULATIONS
It is found that 25.0 mL of 0.200 M aqueous HCl solution is completely neutralized by
20.0 mL of aqueous Ba(OH)
2
solution. What is the molarity of the Ba(OH)
2
solution?