The solubility of PbCl2 is:

Question

The solubility of PbCl2 is: (A)  √Ksp  (B)  (Ksp)1/3  (C)  (Ksp/4)1/3  (D)  (8Ksp)1/2

Tardigrade Admin · Accepted Answer

Let the solubility of PbCl 2=x mol / L undersetx mol / LPbCl 2 leftharpoons undersetxPb 2++ underset2x2 Cl - Ks p=[ Pb 2+][ Cl -]2 or =(x) ×(2 x)2 or =4 x3 ∴ x=√[3](Ks p/4) =((Ks p/4))1 / 3

Q. The solubility of $P b C l_{2}$ is:

$K_{s p}$

$(K_{s p})^{1/3}$

$(K_{s p /4})^{1/3}$

$(8 K_{s p})^{1/2}$

Let the solubility of $P b C l_{2} = x m o l / L$
$x m o l / L P b C l_{2} ⇌ x P b^{2 +} + 2 x 2 C l^{-}$
$K_{s p} = [P b^{2 +}] [C l^{-}]^{2}$
or $= (x) \times (2 x)^{2}$
or $= 4 x^{3}$
$∴ x = 3 \frac{K _{s p}}{4}$
$= (\frac{K _{s p}}{4})^{1/3}$

Q. The solubility of PbCl2​ is:

Ksp​​

(Ksp​)1/3

(Ksp/4​)1/3

(8Ksp​)1/2