Calculate the solubility of cd3(po4)2 in pure water
Answers
Explanation:
We address the solubility equilibrium...
C
a
3
(
P
O
4
)
2
(
s
)
⇌
3
C
a
2
+
+
2
P
O
3
−
4
...
...now this expression is a little unrealistic...and I will expand on this later. However, for the nonce, we will assume this reactivity.
And now...
K
sp
=
1.08
×
10
−
23
≡
[
C
a
2
+
]
3
[
P
O
3
−
4
]
2
And if we let
S
=
solubility of calcium phosphate
...then clearly
[
C
a
2
+
]
=
3
S
, and
[
P
O
3
−
4
]
=
2
S
...and if we substitute these values back into the solubility expression....
K
sp
=
1.08
×
10
−
23
≡
[
C
a
2
+
]
3
[
P
O
3
−
4
]
2
=
(
3
S
)
3
(
2
S
)
2
=
108
S
5
...
And so
S
=
5
√
K
sp
108
=
5
√
1.08
×
10
−
23
108
=
1.0
×
10
−
5
⋅
m
o
l
⋅
L
−
1
(if I have pressed on my calculator buttons properly).
And this gives a gram solubility of
1
×
10
−
5
⋅
m
o
l
⋅
L
−
1
×
310.3
⋅
g
⋅
m
o
l
−
1
=
3.1
⋅
m
g
⋅
L
−
1
...i.e. approx.
3 ppm
...
And note that in aqueous solution, the phosphate anion would probably be present as
H
P
O
2
−
4
...obviously, there would be a competing equilibrium, but this requires a bit more work...
