The solubility of PbI2 at 25°C is 0.7 g L-1. The solubility product of PbI2 at this temperature is (molar mass of PbI2R = 461.2 g mol-1)

Question

The solubility of PbI2 at 25°C is 0.7 g L-1. The solubility product of PbI2 at this temperature is (molar mass of PbI2R = 461.2 g mol-1) (A) 1.40 × 10-9 (B) 0.14 × 10-9 (C) 140 × 10-9 (D) 14.0 × 10-9

Tardigrade Admin · Accepted Answer

PbI2 arrow underset textSPb++ + underset text2S 2I- Ksp = s × (2s)2 = 4s3 = 4 × ((0.7/461.2))3 = 14.0 × 10-9

Q. The solubility of $P b I_{2}$ at $25° C$ is $0.7 g L^{- 1}$ . The solubility product of $P b I_{2}$ at this temperature is (molar mass of $P b I_{2} R = 461.2 g m o l^{- 1}$ )

$1.40 \times 1 0^{- 9}$

$0.14 \times 1 0^{- 9}$

$140 \times 1 0^{- 9}$

$14.0 \times 1 0^{- 9}$

$P b I_{2} \to S P b^{++} + 2S 2 I^{-}$
$K_{s p} = s \times (2 s)^{2} = 4 s^{3}$
$= 4 \times (\frac{0.7}{461.2})^{3} = 14.0 \times 1 0^{- 9}$

Q. The solubility of PbI2​ at 25°C is 0.7gL−1. The solubility product of PbI2​ at this temperature is (molar mass of PbI2​R=461.2gmol−1)

1.40×10−9

0.14×10−9

140×10−9

14.0×10−9