The solubility product of PbS at 298 K is 4.2 × 10-28,The concentrati

1.	The solubility product of PbS at 298 K is 4.2 × 10-28,The concentration of Pb++ ion is 0.001 M. Calculate S2- ion concentration at which PbS just gets precipitated.
Answer» Given : Solubility product of PbS = K_sp = 4.2 × 10^-28 Concentration of Pb⁺⁺ = [Pb⁺⁺] = 0.001 M Concentration of S^-- = [S^--] = ? For PbS, PbS_(s) ⇌ Pb⁺⁺+ S^– ∴ K_sp = [Pb⁺⁺] × [S^--] ∴ [S^–] = \(\frac{K_{sp}}{[Pb^{++}]}\) = \(\frac{4.2\times 10^{-28}}{0.001}\) = 4.2 × 10^-25 M To precipitate Pb⁺⁺ as PbS, ionic product must be greater than 4.2 × 10^-28. Hence, [S^--] > 4.2 × 10^-25 M. ∴ Concentration of S^-- required > 4.2 × 10^-25 M.

The solubility product of PbS at 298 K is 4.2 × 10-28,The concentration of Pb++ ion is 0.001 M. Calculate S2- ion concentration at which PbS just gets precipitated.

Answer»

Given :

Solubility product of PbS = K_sp = 4.2 × 10^-28

Concentration of Pb⁺⁺ = [Pb⁺⁺] = 0.001 M

Concentration of S^-- = [S^--] = ?

For PbS,

PbS_(s) ⇌ Pb⁺⁺+ S^–

∴ K_sp = [Pb⁺⁺] × [S^--]

∴ [S^–] = \(\frac{K_{sp}}{[Pb^{++}]}\)

= \(\frac{4.2\times 10^{-28}}{0.001}\)

= 4.2 × 10^-25 M

To precipitate Pb⁺⁺ as PbS, ionic product must be greater than 4.2 × 10^-28.

Hence,

[S^--] > 4.2 × 10^-25 M.

∴ Concentration of S^-- required > 4.2 × 10^-25 M.