Question

A solution contains 0.10 M Pb2+ and 0.10 M Cu2.. Which cation will precipitate first when a solution of NazS is slowly added to the mixture? Refer to the information sheet for solubility constants. P A) Pb2+ B) Cu2+ C) impossible to tell D) both cations

Answer 1

b) Cu2+

Step-by-step explanation:

• information sheet for solubility constants:

Ksp PbS = 3.4 E-28

Ksp CuS = 6.0 E-37

• PbS ↔ Pb2+ + S2-

∴ Ksp = 3.4 E-28 = [ Pb2+ ] * [ S2- ]

∴ [ Pb2+ ] = 0.10 M

⇒ [ S2- ] = 3.4 E-28 / 0.10 = 3.4 E-27 M

• CuS ↔ Cu2+ + S2-

∴ Ksp = 6.0 E-37 = [ Cu2+ ] * [ S2- ]

∴ [ Cu2+ ] = 0.10 M

⇒ [ S2- ] = 6.0 E-37 / 0.10 = 6.0 E-36 M

we have:

(1) [ S2- ] PbS >> [ S2- ] CuS

(2) Ksp PbS >> Ksp CuS

from (1) and (2) it can determined, that separation can be carried out and also the cation that precipitates first is the Cu2+