Question

Lead thiocyanate, Pb(SCN)₂, has a
`K_(sp)` value of 2.00×10⁻⁵. 1. Calculate the molar solubility of lead thiocyanate in pure water. The molar solubility is the maximum amount of lead thiocyanate the solution can hold. Express your answer with the appropriate units. 1.71×10⁻² M Common-Ion Effect Consider the dissolution of AB(s) : AB(s)⇌A (aq) + B⁻(aq) Le Châtelier's principle tells us that an increase in either [A ] or [B⁻] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A or B⁻ ions. 2. Calculate the molar solubility of lead thiocyanate in 0.800 M KSCN.

Answer 1

1) S Pb(SCN)2 = 1.71 E-2 M...molar solubility in pure water

2) S Pb(SCN)2 = 3.125 E-5 M......molar solubility in 0.800 M KSCN

Step-by-step explanation:

• Pb(SCN)2(s) ↔ Pb2+(aq) + 2SCN-(aq)

S S 2S

∴ Ksp = [Pb2+].[SCN-]² = (S).(2S)² = 4S³ = 2.00 E-5

⇒ S³ = 5.0 E-6

⇒ S = ∛5.0 E-6 = 0.017099 M = 1.71 E-2 M

2) Common-Ion Effect:

• KSCN → K+ + SCN-

0.800M 0.800M 0.800M

• Pb(SCN)2 ↔ Pb2+ + 2SCN-

S S 2S+0.800

⇒ KspPb(SCN)2 = (S).(2S+0.800)² = 2.00 E-5

If we compared the concentration ( 0.800M ) with the value of Ksp ( 2.00 E-5 ) where 0.800>>Ksp; therefore, we can neglect the solubility as adding:

⇒ 2.00 E-5 = (S).(0.800)²

⇒ 2.00 E-5 = 0.64.S

⇒ S = 3.125 E-5 M

checking the assumption:

%S = (3.125 E-5 / 0.800)×100 = 3.906 E-3 %......we can make the assumption