The units of Ksp for the following reaction are
PbCl2 -> Pb^2 +2CL^-1 :
Answers
Explanation:
We assess a solubility equilibrium....we get Ksp=1.64×10−5
hope it helped you
Explanation:
We assess a solubility equilibrium....we get
K
sp
=
1.64
×
10
−
5
Explanation:
We assess a solubility equilibrium...i.e....
P
b
C
l
2
(
s
)
H
2
O
⇌
P
b
2
+
+
2
C
l
−
And as normal, we write out the equilibrium expression....
K
sp
=
[
P
b
2
+
]
[
C
l
−
]
2
[
P
b
C
l
2
(
s
)
]
=
[
P
b
2
+
]
[
C
l
−
]
2
=
?
?
But we have been given the equilibrium concentrations....
K
sp
=
[
P
b
2
+
]
[
C
l
−
]
2
=
1.6
×
10
−
2
×
(
3.2
×
10
−
2
)
2
=
1.64
×
10
−
5
...
And what is the solubility in terms of grams per litre....?
Well, clearly
S
the solubility of lead chloride
=
[
P
b
2
+
]
..
...and clearly....
[
P
b
2
+
]
=
S
, and also clearly,
[
C
l
−
]
=
2
S
...does you agree..?
And so
K
sp
=
S
(
2
S
)
2
=
4
S
3
...
And so
S
=
3
√
K
sp
4
=
3
√
1.64
×
10
−
5
4
=
0.0176
⋅
m
o
l
⋅
L
−
1
...
...the which is a solubility of....
0.0176
⋅
m
o
l
⋅
L
−
1
×
278.10
⋅
g
⋅
m
o
l
−
1
=
4.90
⋅
g
⋅
L
−
1