If the pH of an aqueous solution of 0.15 M HF is 2, what is the percent ionization of HF?
Answers
The percent "ionization" is the same as the percent dissociation. Since you know the pH is
2
, and you can find the pKa in your book or online (
≈
3.15
), and since this acid is a weak acid when put into water, it is a buffer problem.
HF
⇌
H
+
+
F
−
We can use the Henderson-Hasselbalch equation.
p
H
=
p
K
a
+
log
[
A
−
]
[
HA
]
where
A
−
is the conjugate base (
F
−
) and
HA
is the weak acid (
HF
).
2
=
3.15
+
log
[
A
−
]
[
HA
]
−
1.15
=
log
[
A
−
]
[
HA
]
10
−
1.15
=
[
A
−
]
[
HA
]
But we know that as
HF
dissociates into
F
−
, the same number of
mol
s of
F
−
is created as the number of
mol
s of
HF
dissociated, because there is no other source that gives
F
−
.
Additionally, both species are within the same solution, so their concentrations are determined by the same total volume. When we compare concentrations, we simply get:
[
A
−
]
[
HA
]
=
mol A
−
L soln
mol HA
L soln
=
mol A
−
mol HA