Ha oh k cb k w k a 1 x 10 14 1 x 10 5 1 x 10 9

HA + OH
-
K
cb
= K
w
/K
a
= (1 x 10
-14
)/(1 x 10
-5
) = 1 x 10
-9
Conjugate base + H
3
O
+
(e.g., sodium acetate + H
3
O + )
:A
-
+ H
3
O
+
→ HA + H
2
O
K = 1/K
a
= 1/(1 x 10
-5
) = 1 x 10
5
(or bring to completion; then solve either using HH or reverse + K
a
)
Weak acid + strong base
(e.g., acetic acid + sodium hydroxide)
: HA + OH
-
→ H
2
O + A
-
K = 1/K
cb
= 1/(K
w
/K
a
) = K
a
/K
w
= (1 x 10
-5
)/(1 x 10
-14
) = 1 x 10
9
(or bring to completion; then solve either using HH or reverse + K
a
)
Weak base + water
(e.g., NH
3
+ H
2
O)
: RNH
2
+ H
2
O
←
→
RNH
3
+
+ OH
-
Assume K
b
= ~1 x 10
-5
Conjugate acid + water
(e.g., sodium acetate + H
2
O)
: RNH
3
+
+ H
2
O
←
→
RNH
2
+ H
3
O
+
K
ca
= K
w
/K
b
= (1 x 10
-14
)/(1 x 10
-5
) = (1 x 10
-9
)
Conjugate acid + OH
-
(e.g., ammonium chloride + OH
-
)
: RNH
3
+
+ OH
-
→ RNH
2
+ H
2
O
K = 1/K
b
= 1/(1 x 10
-5
) = 1 x 10
5
(or bring to completion; then solve either using HH or reverse + K
b
)
Weak base + strong acid
(e.g., NH
3
+ H
3
O + )
: RNH
2
+ H
3
O
+
→ RNH
3
+
+ H
2
O
K = 1/K
ca
= 1/(K
w
/K
b
) = K
b
/K
w
= (1 x 10
-5
)/(1 x 10
-14
) = 1 x 10
9
(or bring to completion; then solve either using HH or reverse + K
b
)

Chem 162-2012 Chapter 16 (B&O) Version III 34
Acid-base Properties of Salts
ET: A salt is another name for an ionic compound, i.e., something that will dissociate into positive and negative ions in water.
111 (mod.) Are solutions of the following salts acidic, basic, or neutral?
a. NaNO
3
NaNO
3
→ Na
+
+ NO
3
-
Na
+
+ H
2
O NR
NO
3
-
+ H
2
O NR
Neutral
b. NaNO
2
NaNO
2
→ Na
+
+ NO
2
-
Na
+
+ H
2
O NR
NO
2
-
+ H
2
O HNO
2
+ OH
-
Therefore, basic solution
f. NH
4
ClO
4
NH
4
ClO
4
→ NH
4
+
+ ClO
4
-
NH
4
+
+ H
2
O H
3
O
+
+ NH
3
ClO
4
-
+ H
2
O NR
Therefore, acidic solution
NH
4
CN
NH
4
CN → NH
4
+
+ CN
-
NH
4
+
+ H
2
O H
3
O
+
+ NH
3
CN
-
+ H
2
O OH
-
+ HCN
Note that in this case neither the acidic ion nor the basic ion drops out of the reaction.
Which forms more, H
3
O
+
or OH
-
?
NH
4
+
+ H
2
O H
3
O
+
+ NH
3
K
b
= 1.8x10
-
5; Kca = K
w
/K
b
= (1x10
-14
)/(1.8x10
-5
) = 5.56 x 10
-10
CN
-
+ H
2
O OH
-
+ HCN
K
a
= 6.2x10
-10
; K
cb
= K
w
/K
a
= (1x10
-14
)/(6.2 x 10
-10
) = 1.61x10
-5
K
cb
> K
ca
. Therefore, more OH
-
is formed than H
3
O
+
. Therefore, NH
4
CN forms a basic
solution.

Chem 162-2012 Chapter 16 (B&O) Version III 35
ET: Equilibria with salts; (1) dissolve in solution; (2) get rid of spectator ions; (3) write reaction. Use Kw.
105b. Calculate the pH of 0.050 M NaCN.
K
a
for HCN = 6.2 x 10
-10
.
NaCN → Na
+
+ CN
-
Na
+
+ H
2
O → NR
CN
-
+ H
2
O → HCN + OH
-
CN
-
+ H
2
O HCN + OH
-
Initial 0.050
Change
Equilibrium
CN
-
+ H
2
O HCN + OH
-
Initial 0.050 0 0
Change -X +X +X
Equilibrium 0.050-X +X +X
K
cb
= ([HCN][OH
-
])/[CN
-
]
What is the value of K
cb
?
K
w
= K
a
x K
cb
K
cb
= (1 x (10
-14
))/(6.2 x (10
-10
))
K
cb
= 1.61 x 10
-5
(X x X)/(0.050-X) = 1.61 x 10
-5
Simplify with small K rule:
(X x X)/0.050 = 1.61 x 10
-5
X = 8.89 x 10
-4
= [OH
-
]
[H
+
] = (1 x 10
-14
)/(8.89 x 10
-4
) = 1.12 x 10
-11
pH = -log(1.12 x 10
-11
) = 10.95
Note: This is consistent with CN
-
being a base.

Chem 162-2012 Chapter 16 (B&O) Version III 36
Given M and pH; find Ka.
A 0.25M sodium hypochlorite solution (NaClO) has a pH of 10.43. Calculate
K
a
for HClO (hypochlorous acid).
NaOCl → Na
+
+ OCl
-
Na
+
= junk ion
NaOCl → Na
+
+ OCl
-
OCl
-
+ H
2
O
←
→
HOCl + OH
-
pH = 10.43
[H
+
] = 10
-pH
= 10
-10.43
= 3.72 x 10
-11
[H
+
][OH
-
] = 1x10
-14
[OH
-
] = (1x10
-14
)/[H
+
] = (1x10
-14
)/(3.72 x 10
-11
) = 2.69 x 10
-4
OCl
-
+
H
2
O
←
→
HOCl + OH
-
Initial 0.25
Change
Equilibrium 2.69 x 10
-4
OCl
-
+
H
2
O
←
→
HOCl + OH
-
Initial 0.25 0 0
Change -X +X +X
Equilibrium 0.25-X +X 2.69 x 10
-4
0 + X = 2.69 x 10
-4
; X = 2.69 x 10
-4
OCl
-
+
H
2
O
←
→
HOCl + OH
-
Initial 0.25 0 0
Change -2.69 x 10
-4
2.69 x 10
-4
2.69 x 10
-4
Equilibrium 0.25-2.69 x 10
-4
= 0.25
2.69 x 10
-4
2.69 x 10
-4