Answer:
a) 1.71 × 10⁻³ M
b) 8.00 × 10⁻⁵ M
Explanation:
In order to calculate the solubility (S) of Pb(SCN)₂ we will use an ICE chart. We identify 3 stages (Initial, Change, Equilibrium) and complete each row with the concentration or change in the concentration.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product (Ksp) is:
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (2S)² = 4S³
S = 1.71 × 10⁻³ M
b) Calculate the molar solubility of lead thiocyanate in 0.500 M KSCN.
KSCN is a strong electrolyte that dissociates to give 0.500 M K⁺ and 0.500M SCN⁻.
Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)
I 0 0.500
C +S +2S
E S 0.500 + 2S
Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (0.500 + 2S)²
In the term (0.500 + 2S)², 2S is negligible.
Ksp = 2.00 × 10⁻⁵ = S . (0.500)²
S = 8.00 × 10⁻⁵ M
