Solubility product of PbCl2 at 298 K is 1 × 10-6. At this temperature, solubility of PbCl2 (in mol/L) is

Question

Solubility product of PbCl2 at 298 K is 1 × 10-6. At this temperature, solubility of PbCl2 (in mol/L) is (A) (1 × 10-6 )1/2 (B) (1 × 10-6 )1/3 (C) (0.25 × 10-6 )1/3 (D) (2.5 × 10-6 )1/2

Tardigrade Admin · Accepted Answer

PbCl2 leftharpoons Pb2+ + 2Cl- Ksp = [Pb2+ ][Cl-]2 = (S)(2S)2 = 4S3 1 × 10-6 = 4S3 S = ( frac1 × 10-64 )1/3 = (0.25 × 10-6 )1/3

Q. Solubility product of $P b C l_{2}$ at $298 K$ is $1 \times 1 0^{- 6}$ . At this temperature, solubility of $P b C l_{2} (in m o l / L)$ is

$(1 \times 1 0^{- 6})^{1/2}$

$(1 \times 1 0^{- 6})^{1/3}$

$(0.25 \times 1 0^{- 6})^{1/3}$

$(2.5 \times 1 0^{- 6})^{1/2}$

$P b C l_{2} ⇌ P b^{2 +} + 2 C l^{-}$
$K_{s p} = [P b^{2 +}] [C l^{-}]^{2} = (S) (2 S)^{2} = 4 S^{3}$
$1 \times 1 0^{- 6} = 4 S^{3}$
$S = (f r a c 1 \times 1 0^{- 6} 4)^{1/3} = (0.25 \times 1 0^{- 6})^{1/3}$

Q. Solubility product of PbCl2​ at 298K is 1×10−6. At this temperature, solubility of PbCl2​(inmol/L) is

(1×10−6)1/2

(1×10−6)1/3

(0.25×10−6)1/3

(2.5×10−6)1/2