Solubility of lead chloride is K. Then the number of moles of Cl- ionsin 1000 ml of saturated lead chloride solution will be

Question

Solubility of lead chloride is K. Then the number of moles of Cl- ionsin 1000 ml of saturated lead chloride solution will be (A)  ( (K/4) )1/3  (B)  (2K)1/3  (C)  K1/2  (D)  (K/2)

Tardigrade Admin · Accepted Answer

: For lead chloride (PbCl2) Ksp=[Pb2+][Cl-]2 Ksp=x.(2x)2 Ksp=x.4x2 x=( (K/4) )1/3

Q. Solubility of lead chloride is K. Then the number of moles of $C l^{-}$ ionsin 1000 ml of saturated lead chloride solution will be

$(\frac{K}{4})^{1/3}$

$(2 K)^{1/3}$

$K^{1/2}$

$\frac{K}{2}$

: For lead chloride $(P b C l_{2})$ $K_{s p} = [P b^{2 +}] [C l^{-}]^{2}$ $K_{s p} = x . (2 x)^{2}$ $K_{s p} = x .4 x^{2}$ $x = (\frac{K}{4})^{1/3}$

Q. Solubility of lead chloride is K. Then the number of moles of Cl− ionsin 1000 ml of saturated lead chloride solution will be

(4K​)1/3

(2K)1/3

K1/2

2K​