Question

Consider the solution containing 0.181 M lead ions and 0.174

Mmercury (II) ions. the Ksp for lead sulfide
is3.4*10-28 and that for mercury (II) sulfide
is4.0*10-53. calculate the concentration in (M)
ofsulfide ions required to reduce the concentration of mercury
(II)ions to 1.0*10-6 M?

Answer 1

[ S2- ] = 4.0 E-47 M

Step-by-step explanation:

• PbS(s) → Pb2+ + S2-
• HgS(s) → Hg2+ + S2-

∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]

∴ [Pb2+] = 0.181 M

∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]

∴ [Hg2+] = 0.174 M

∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:

∴ [ Hg2+ ] = 1.0 E-6 M

⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M