What is the pH and pOH of a 3.0 M solution of NH3, where K for NH3 = 1.76 x 10-5?
Answers
Answer:
As you know, ammonia acts as a weak base in aqueous solution, so right from the start, you should expect the
pH
of the solution to be
>
7
.
NH
3
(
a
q
)
+
H
2
O
(
l
)
⇌
NH
+
4
(
a
q
)
+
OH
−
(
a
q
)
The ratio that exists between the equilibrium concentrations of the ammonium cations and of the hydroxide anions and the equilibrium concentration of ammonia is given by the base dissociation constant,
K
b
.
K
b
=
[
NH
+
4
]
⋅
[
OH
−
]
[
NH
3
]
Now, ammonia will only partially ionize to produce ammonium cations and hydroxide anions. If you take
x
M
to be the concentration of ammonia that ionizes, you can say that, at equilibrium, the solution will contain
[
NH
+
4
]
=
[
OH
−
]
=
x
M
This happens because every mole of ammonia that ionizes produces
1
mole of ammonium cations and
1
mole of hydroxide anions.
So if
x
M
ionizes, you can expect the solution to contain
x
M
of the two ions.
The solution will also contain
[
NH
3
]
=
(
1.5
−
x
)
M
When
x
M
ionizes, the initial concentration of ammonia will decrease by
x
M
.
This means that the expression of the base dissociation constant will now take the form
K
b
=
x
⋅
x
1.5
−
x
which is equal to
1.80
⋅
10
−
5
=
x
2
1.5
−
x
Notice that the value of the base dissociation constant is significantly smaller than the initial concentration of the base. This tells you that you can use the approximation
1.5
−
x
≈
1.5
because the concentration of ammonia that ionizes will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will lie to the left.
You now have
1.80
⋅
10
−
5
=
x
2
1.5
Rearrange and solve for
x
to get
x
=
√
1.5
⋅
1.80
⋅
10
−
5
=
0.005196
Since
x
M
represents the equilibrium concentration of the hydroxide anions, you can say that
[
OH
−
]
=
0.005196 M
Now, an aqueous solution at
25
∘
C
has
pH + pOH = 14
Since
pOH
=
−
log
(
[
OH
−
]
)
you can say that the
pH
of the solution is given by
pH
=
14
+
log
(
[
OH
−
]
)
Plug in your value to find
pH
=
14
+
log
(
0.005196
)
=
11.72
−−−
Explanation:
please make me as braniest