Ions and molecules in solution from the largest

ions and molecules in solution from the largest concentration to the smallest is:
H
2
O, CH
3
COO
–
, Na
+
, CH
3
COOH, H
3
O
+
, OH
–
(b) Adapt the methods used in the answers to Question 30. The equation for the equilibrium and the
equilibrium expression are:
CH
3
COOH + H
2
O(  ) CH
3
COO
–
(aq) + H
3
O
+
(aq)
Κ
α
=
[ΧΗ
3
ΧΟΟ
-
][Η
3
Ο
+
]
[ΧΗ
3
ΧΟΟΗ]
4.95 γ ΝαΧΗ
3
ΧΟΟ
250. μΛ
1 μολ ΝαΧΗ
3
ΧΟΟ
82.0337 γ ΝαΧΗ
3
ΧΟΟ
1000 μΛ
1 Λ
1 μολ ΧΗ
3
ΧΟΟ
-
1 μολ ΝαΧΗ
3
ΧΟΟ
= 0.241 M CH
3
COO
–
As the reactants decompose, the concentrations of the products increase stoichiometrically, until they
reach equilibrium concentrations.
CH
3
COOH
CH
3
COO
–
(aq) H
3
O
+
(aq)
conc. init. (M) 0.150 0.241 0
change conc. (M) – x + x + x
eq. conc. (M) 0.150 – x 0.241 + x x
At equilibrium K
a
=
(ξ29( 0.241+ ξ 29
(0.150- ξ29
= 1.8 × 10
–5
Assume x is very small and does not affect the sum or the difference.
233

Chapter 17: Additional Aqueous Equilibria
1.8 × 10
–5 =
(ξ29( 0.241 29
(0.150 29
x = 1.1 × 10
–5
M = [H
3
O
+
]
pH = – log[H
3
O
+
] = – log(1.1 × 10
–5
) = 4.95
(c) The strong base neutralizes the acid in the solution, CH
3
COOH.
CH
3
COOH + OH
–
(aq) CH
3
COO
–
(aq) + H
2
O(  )
(conc. OH
–
) =
80. μγ ΝαΟΗ
100. μΛ
1000 μΛ
1 Λ
1 γ
1000 μγ
1 μολ ΝαΟΗ
40.00 γ ΝαΟΗ
1 μολ ΟΗ
-
1 μολ ΝαΟΗ
= 0.020 M
We take it as far to the products as possible, using the method of limiting reactants.
CH
3
COOH
OH
–
(aq) CH
3
COO
–
(aq)
conc. init. (M) 0.150 0.020 0.241
change conc. (M) – 0.020 – 0.020 + 0.020
conc. final (M) 0.130 0 0.261
The solution is still a buffer solution, containing an acid-base conjugate pair, so we can find the pH
using K
a
of the acid.
CH
3
COOH + H
2
O(  ) CH
3
COO
–
(aq) + H
3
O
+
(aq)
Κ
α
=
[ΧΗ
3
ΧΟΟ
-
][Η
3
Ο
+
]
[ΧΗ
3
ΧΟΟΗ]
As the reactants decompose, the concentrations of the products increase stoichiometrically, until they
reach equilibrium concentrations.
CH
3
COOH
CH
3
COO
–
(aq) H
3
O
+
(aq)
conc. init. (M) 0.130 0.261 0
change conc. (M) – x + x + x
eq. conc. (M) 0.130 – x 0.261 + x x
At equilibrium K
a
=
(0.261+ ξ29( ξ 29
(0.130- ξ29
= 1.8 × 10
–5
Assume x is very small and does not affect the sum or the difference.
1.8 × 10
–5 =
(0.261 29( ξ 29
(0.130 29
x = 9.0 × 10
–6
M = [H
3
O
+
]
pH = – log[H
3
O
+
] = – log(9.0 × 10
–6
) = 5.05
(d) CH
3
COOH(aq) + H
2
O(  ) CH
3
COO
–
(aq) + H
3
O
+
(aq)
77. Adapt the methods used in the answers to Questions 15 - 32.
At equilibrium, pH = 9.00, [OH
–
] = 10
pH–14.00
= 10
9.00–14.00
= 1.0 × 10
–5
Κ
β
=
[ΝΗ
4
+
][ΟΗ
-
]
[ΝΗ
3
]
1.8 × 10
–5
=
[ΝΗ
4
+
](1.0 10
-5
29
(0.93 29
[NH
4
+
] = 1.7 M ( Notice: This could also be calculated using the Henderson-Hasselbalch equation .)
234

Chapter 17: Additional Aqueous Equilibria
400. μΛ
1 Λ
1000 μΛ
1.7 μολ ΝΗ
4
+
1 Λ
1 μολ ΝΗ
4
Χλ
1 μολ ΝΗ
4
+
53.49 γ ΝΗ
4
Χλ
1 μολ ΝΗ
4
Χλ
= 36 g NH
4
Cl
78 . It is possible to adapt the methods used in the answers to Question 15 - 32; however, the Henderson-
Hasselbalch equation is easily used when we know the pK
a
of the conjugate acid.
pH = pK
a
+ log
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At equilibrium, pH = 4.0 and pK
a
= 3.79
pH – pK
a
= log
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4.0 – 3.79 = 0.2 = log
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