Calculate the molar solubility of agcl in 1 l solution which contains 10 g of cacl2
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Answer:
Explanation:
K
sp
A
g
C
l
(
s
)
≡
1.77
×
10
−
10
....and this equilibrium constant (which WILL be given you) addresses the following equilibrium...
A
g
C
l
(
s
)
⇌
A
g
+
+
C
l
−
..
And so
K
sp
A
g
C
l
(
s
)
=
[
A
g
+
]
[
C
l
−
]
≡
1.77
×
10
−
10
Now it is a fact that
[
A
g
+
]
=
S
solubility of AgCl
for this problem....but we artificially raised
[
C
l
−
]
, i.e. we have
salted out
the stuff....
And so
S
×
(
0.216
⋅
m
o
l
⋅
L
−
1
+
S
)
=
1.77
×
10
−
10
And if we make the reasonable approx. that
0.108
⋅
m
o
l
⋅
L
−
1
>>
S
, then....
S
1
=
1.77
×
10
−
10
0.216
⋅
m
o
l
⋅
L
−
1
=
8.19
×
10
−
10
⋅
m
o
l
⋅
L
−
1
...the value is indeed small compared to
0.216
⋅
m
o
l
⋅
L
−
1
...so our approx. was justified
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