Calculate the dissociation constant acid of an acid 0.1 solution of which as pH=5 ?
Answers
Answer:
acid here so that the ionization equilibrium can be written like this
HA
(
a
q
)
+
H
2
O
(
l
)
⇌
A
−
(
a
q
)
+
H
3
O
+
(
a
q
)
Now, you know that the solution has
pH
=
5
As you know, the
pH
of the solution is defined as
pH
=
−
log
(
[
H
3
O
+
]
)
−−−−−−−−−−−−−−−−−−−−
You can rearrange this equation to find the equilibrium concentration of hydronium cations.
log
(
[
H
3
O
+
]
)
=
−
pH
This is equivalent to
10
log
(
[
H
3
O
+
]
)
=
10
−
pH
which gets you
[
H
3
O
+
]
=
10
−
pH
In your case, you will have
[
H
3
O
+
]
=
10
−
5.0
=
1.0
⋅
10
−
5
M
Now, notice that every mole of
HA
that ionizes produces
1
mole of
A
−
, the conjugate base of the acid, and
1
mole of hydronium cations.
This means that, at equilibrium, the solution has
[
A
−
]
=
[
H
3
O
+
]
→
produced in a
1
:
1
mole ratio
In your case, you have
[
A
−
]
=
1.0
⋅
10
−
5
M
The initial concentration of the acid will decrease because some of the molecules ionize to produce
A
−
and
H
3
O
+
.
So, in order for the ionization to produce
[
H
3
O
+
]
, the initial concentration of the acid must decrease by
[
H
3
O
+
]
.
This means that, at equilibrium, the concentration of the weak acid will be equal to
[
HA
]
=
[
HA
]
initial
−
[
H
3
O
+
]
In your case, you will have
[
HA
]
=
0.01 M
−
1.0
⋅
10
−
5
.
M
[
HA
]
=
0.00999 M
By definition, the acid dissociation constant,
K
a
, will be equal to
K
a
=
[
A
−
]
⋅
[
H
3
O
+
]
[
HA
]
Plug in your values to find
K
a
=
1.0
⋅
10
−
5
M
⋅
1.0
⋅
10
−
5
.
M
0.00999
M
K
a
=
1.001
⋅
10
−
8
M
Rounded to one significant figure and expresses without added units, the answer will be
K
a
=
1
⋅
10
−
8
−−−−−−−−−−−−
Answer:
Your answer is in the attachment.
Hope it helps
Mark me as brainlist please
Please thanks my answer
