H 2 so 4 h 2 o h 3 o hso 4 initial 20 0 0 20

H
2
SO
4
+ H
2
O → H
3
O
+
HSO
4
-
Initial 2.0 0 0
Change -2.0 +2.0 +2.0
Equilibrium 0 2.0 2.0
2
nd
K
a
: HSO
4
-
+ H
2
O H
3
O
+
+ SO
4
2-
HSO
4
-
H
2
O H
3
O
+
SO
4
2-
New initial 2.0 2.0 0
Change -X +X +X
Equilibrium 2.0 - X 2.0 + X +X
([H
3
O
+
][SO
4
2-
])/[HSO
4
-
] = K
a2
= 1.2 x 10
-2
([2.0 + X] x [X])/[2.0 – X] = 1.2 x 10
-2
Small K rule: ([2.0] x [X])/[2.0] = 1.2 x 10
-2
Note that B&O recommends using small and large K rules at <1 x 10
-2
and >1x10
2
, respectively.
X = 0.012
H
3
O
+
= 2.0 + 0.012 = 2.012
pH = -log[2.012] = -0.304

Diprotic Bases
The carbonate ion CO
3
2-
, is a diprotic base; i.e., it can accept two protons to make
H
2
CO
3
. In water, it accepts these protons in two steps, analogous to H
2
CO
3
donating its protons in two steps.
CO
3
2-
+ H
2
O ⇄ HCO
3
-
+ OH
-
K
b1
= 2.1 x 10
-4
HCO
3
-
+ H
2
O ⇄ H
2
CO
3
+ OH
-
K
b2
= 2.4 x 10
-8

ET: Consider adding equations for SA+H
2
O, SA + OH-, SB + H
2
O, SB + H
3
O
+
.
ET: First discuss equations, then reverse equations invert K, then K
a
x K
cb
, or K
ca
x K
b
= 10
-14
Write the equil. equations and identify K’s* for the following reactions:
Weak acid + water
(e.g., acetic acid + H
2
O)
: HA + H
2
O
←
→
H
3
O
+
+ A
-
Assume K
a
= K
wa
= ~1 x 10
-5
Conjugate base + H
3
O
+
(e.g., sodium acetate + H
3
O + )
:A
-
+ H
3
O
+
→ HA + H
2
O
K = K
cbsa
= 1/K
a
= 1/(1 x 10
-5
) = 1 x 10
5
Conjugate base + water
(e.g., sodium acetate + H
2
O)
: A
-
+ H
2
O
←
→
HA + OH
-
K
cb
= K
w
/K
a
= (1 x 10
-14
)/(1 x 10
-5
) = 1 x 10
-9
Weak acid + strong base
(e.g., acetic acid + sodium hydroxide)
: HA + OH
-
→ H
2
O + A
-
K = K
wasb
= 1/K
cb
= 1/(K
w
/K
a
) = K
a
/K
w
= (1 x 10
-5
)/(1 x 10
-14
) = 1 x 10
9
Weak base + water
(e.g., NH
3
+ H
2
O)
: RNH
2
+ H
2
O
←
→
RNH
3
+
+ OH
-
Assume K
b
= K
wb
= ~1 x 10
-5
Conjugate acid + OH
-
(e.g., ammonium chloride + OH
-
)
: RNH
3
+
+ OH
-
→ RNH
2
+ H
2
O
K = K
casb
= 1/K
b
= 1/(1 x 10
-5
) = 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
)
Weak base + strong acid
(e.g., NH
3
+ H
3
O + )
: RNH
2
+ H
3
O
+
→ RNH
3
+
+ H
2
O
K = K
wbsa
= 1/K
ca
= 1/(K
w
/K
b
) = K
b
/K
w
= (1 x 10
-5
)/(1 x 10
-14
) = 1 x 10
9
*K
a
= K
acid
= K
wa
= K
weakacid
; K
cb
= K
conjugatebase
; K
cbsa
= K
conjugatebasestrongacid
; K
wasb
= K
weakacidstrongbase
;
K
b
= K
base
= K
weakbase
; K
ca
=
Kconjugate acid
; K
b
= K
casb
= K
conjugateacidstrongbase
; K
wbsa
= K
weakbasestrongacid

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
Therefore, neutral solution
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 in NH
4
CN reaction, H
3
O
+
or OH
-
?
NH
4
+
+ H
2
O H
3
O
+
+ NH
3
K
ca
= K
w
/K
b
= (1x10
-14
)/(1.8x10
-5
) = 5.56 x 10
-10
CN
-
+ H
2
O OH
-
+ HCN
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.

ET: Equilibria with salts; (1) dissolve in solution; (2) get rid of spectator ions; (3) write reaction. Use Kw.