Solubility of Cd(OH) 2 in aqueous solution of 0.01 M NaOH will be (K sp (Cd(OH) 2 ) = 2.5 × 10 –14
Answers
Explanation:
Explanation:
Your strategy here will be to calculate the concentrations of the cadmium(II) cations,
Cd
2
+
, and of the hydroxide anions,
OH
−
, and figure out if these concentrations are high enough to ensure that a precipitate is formed.
So, you know that the pH of the solution is equal to
9
. As you know, an aqueous solution at room temperature has
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
+
POH
=
14
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−
This means that the pOH of the solution, which can be used to find the concentration of hydroxide anions, will be equal to
pOH
=
14
−
9
=
5
This means that you have
[
OH
−
]
=
10
−
pOH
=
10
−
5
M
Cadmium chloride,
CdCl
2
, is a soluble ionic compound that dissociates completely in aqueous solution to form cadium(II) cations and chloride anions
CdCl
2(aq]
→
Cd
2
+
(aq]
∣
+
2
Cl
−
(aq]
Notice that one mole of cadmium chloride produces one mole of cadmium cations in solution. This means that you have
[
Cd
2
+
]
=
[
CdCl
2
]
=
10
−
2
M
The partial dissociation of cadmium hydroxide in aqueous solution looks like this
Cd
(
OH
)
2(s]
⇌
Cd
2
+
(aq]
+
2
OH
−
(aq]
By definition, the solubility product constant,
K
s
p
, for cadmium hydroxide will be equal to
K
s
p
=
[
Cd
2
+
]
⋅
[
OH
−
]
2
Now, in order to determine if a precipitate if formed or not, you need to calculate the reaction quotient,
Q
s
p
.
The difference between
K
s
p
and
Q
s
p
lies in the fact that the former is calculated using equilibrium concentrations, while the latter is calculated using the concentrations of the ions at a given time.
In other words,
Q
s
p
can be used to determine if a precipitate will form. In order for a precipitate to form, you need to have
Q
s
p
>
K
s
p
At this point, a precipitate will form until you reach saturation, i.e until
Q
s
p
=
K
s
p
.
If
Q
s
p
<
K
s
p
, the solution is unsaturated and a precipitate will not form.
Plug in the concentrations of the two ions to get
Q
s
p
=
10
−
2
M
⋅
(
10
−
5
)
2
M
2
=
10
−
12
M
3
Since you have
10
−
12
M
3
>
2.5
⋅
10
−
14
M
3
cadmium hydroxide will indeed precipitate out of solution.