Explain how the pH value of 0.1M potassium hydroxide solution compares with that of 0.1M aqueous ammonia
Answers
Answer:Ka = 1.8 x 10-5
a)pH = 8.92
b)pH = 4.74
c)pH = 11.55
d)pH = 4.46
e)pH = 4.73
Explanation:For starters, I think that the question is either missing the option
4.75
or one of the options given to you was mistyped.
The idea here is that adding a very small amount of strong base to the buffer will increase the
pH
of the buffer ever so slightly.
Even without doing any calculations, you can say that adding
0.001
moles of potassium hydroxide would be enough to increase the
pH
of the buffer by
0.01
, from
4.74
, what you have at first, to
4.75
.
As you know, the
pH
of a weak acid - conjugate base buffer can be calculated using the Henderson - Hasselbalch equation.
pH
=
p
K
a
+
log
(
[
conjugate base
]
[
weak acid
]
)
Now, when you have equal concentrations of the weak acid and of the conjugate base, you get
log
(
[
conjugate base
]
[
weak acid
]
)
=
log
(
1
)
=
0
and
pH
=
p
K
a
So before adding the potassium hydroxide, your buffer has
[
CH
3
COOH
]
=
[
CH
3
COO
−
]
=
0.1 M
and
pH
=
−
log
(
1.8
⋅
10
−
5
)
pH
=
4.74
Now, adding the strong base will cause the concentration of the weak acid to decrease and the concentration of the conjugate base to increase.
So after you add the potassium hydroxide, you have
[
CH
3
COOH
]
<
[
CH
3
COO
−
]
This means that
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
>
0
and so
pH
=
4.74
+
(
something > 0
)
which gets you
pH
>
4.74
Since the number of moles of potassium hydroxide is very small compared to the initial concentration of the weak acid and of the conjugate base (assuming a
1-L
solution), you can say that an option like
4.75
would be an ideal fit here.
pH
>
4.74
→
but very close to 4.74
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
You can follow the same logic and say that if you start with the same buffer and add
0.001
moles of hydrochloric acid, a strong acid, the
pH
would be
4.73
.
This time, the strong acid would consume some of the conjugate base and produce weak acid, so
(
[
CH
3
COOH
]
>
[
CH
3
COO
−
]
This means that
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
<
0
and so
pH
=
4.74
+
(
something < 0
)
which gets you
pH
<
4.74
In this case,
4.73
would be an ideal fit because the number of moles of hydrochloric acid is very small compared to the initial concentration of the weak acid and of the conjugate base (assuming a
1-L
solution).
pH
<
4.74
→
but very close to 4.74
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Now, back to the problem at hand. If you assume that you're working with a
1-L
solution and that the addition of the strong base does not change the volume of the buffer, you can use the molarity and the number of moles interchangeably to say that after you add
0.001
moles of potassium hydroxide, you will have
{
(
0.1
−
0.001
)
moles CH
3
COOH
(
0.1
+
0.001
)
moles CH
3
COO
−
And so
pH
=
p
K
a
+
log
(
(
0.1
+
0.001
)
moles
(
0.1
−
0.001
)
moles
)
pH
=
4.7487
≈
4.75
As predicted, we have
pH
=
4.75
>
4.74
→
but very close to 4.74