The solubility of lead chromate : s= 5.29 x 10⁻⁷ mol/L
Further explanation
Given
Ksp for PbCl,=2.8x10*-13 mole/dmº
PbCrO4
Required
The solubility
Solution
P
b
C
r
O
₄
(
s
)
⇌
P
b²⁺(
a
q)
+C
r
O
₄
⁻²(
a
q
)
s s s
s= solubility
Ksp P
b
C
r
O
₄ = s²
K
s
p =
[
P
b
²⁺]
[C
r
O
₄
⁻²
]=
2.8x10⁻¹³
s = √2.8x10⁻¹³
s= 5.29 x 10⁻⁷ mol/L
