What's the solubility (in grams per liter) of LaF3 in pure water?
Answers
what I got.
Explanation:
I'll show you how to solve part (a) and leave part (b) to you as practice.
Lanthanum trifluoride,
LaF
3
, is considered Insoluble in water, so right from the start you know that when the salt is dissolved in water, an equilibrium is established between the undissolved salt and the dissolved ions.
LaF
3
(
s
)
⇌
La
3
+
(
a
q
)
+
3
F
−
(
a
q
)
The position of the equilibrium is given to you by the value of the solubility product constant,
K
s
p
.
More specifically, the smaller the
K
s
p
, the more the equilibrium will lie to the left, i.e. the less of the salt actually dissolves in solution.
Now, you want to know how much lanthanum trifluoride can be dissolved in a solution that is
0.049 M
in potassium fluoride,
KF
, which, unlike lanthanum trifluoride, is soluble in water.
In other words, potassium fluoride dissociates completely in aqueous solution to form potassium cations,
K
+
, and fluoride anions,
F
−
KF
(
a
q
)
→
K
+
(
a
q
)
+
F
−
(
a
q
)
Since every mole of potassium fluoride that dissociates form
1
mole of fluoride anions, you will have
[
F
−
]
=
1
×
[
KF
]
=
0.049 M
The presence of the fluoride anions will affect the position of the equilibrium in accordance with Le Chatelier's Principle.
More specifically, the equilibrium will shift to the left in an attempt to decrease the extra concentration of dissolved fluoride anions. As a result, the solubility of the salt will decrease
→
this is known as the common-ion effect.
So, if you take
s
to be the molar solubility of lanthanum trifluoride in your solution, you can use an ICE table to write
LaF
3
(
s
)
⇌
La
3
+
(
a
q
)
+
3
F
−
(
a
q
)
I
a
a
a
−
a
a
a
a
a
a
a
a
a
a
a
a
0
a
a
a
a
a
a
a
a
a
0.049
C
a
a
a
−
a
a
a
a
a
a
a
a
a
a
(
+
s
)
a
a
a
a
a
a
(
+
3
s
)
E
a
a
a
−
a
a
a
a
a
a
a
a
a
a
a
a
s
a
a
a
a
a
a
a
0.049
+
3
s
By definition, the solubility product constant is equal to
K
s
p
=
[
La
3
+
]
⋅
[
F
−
]
3
In this case, you have
K
s
p
=
s
⋅
(
0.049
+
3
s
)
3
(*)
This will be equivalent to
K
s
p
=
s
⋅
[
0.049
3
+
(
3
s
)
3
+
3
⋅
0.049
2
⋅
(
3
s
)
+
3
⋅
0.049
⋅
(
3
s
)
2
]
K
s
p
=
s
⋅
(
0.00011765
+
27
s
3
+
0.02161
s
+
1.323
s
2
)
K
s
p
=
27
s
4
+
1.323
s
3
+
0.02161
s
2
+
0.00011765
s
This can be rearranged to give
27
s
4
+
1.323
s
3
+
0.02161
s
2
+
0.00011765
s
−
K
s
p
=
0
Since the problem doesn't provide you with a value for the
K
s
p
of the salt, I'll pick one
K
s
p
=
2.0
⋅
10
−
19
mark as brainlest plzzz