question has 4 parts related to solubility and dissociation

Question

Calculate the solubility of PbCl₂ in 0.75 M NaOH. The K_f of Pb(OH)₃^- is 8E13 and the K_sp of PbCl₂ is 1.7E-5. 

a) Write the chemical equation and the ion product expression for the dissolution of PbCl₂

Calculate the solubility of PbCl₂ in 0.75 M NaOH. The K_f of Pb(OH)₃^- is 8E13 and the K_sp of PbCl₂ is 1.7E-5. 

b) Write the chemical equation and the equilibrium expression for the formation of Pb(OH)₃^-

Calculate the solubility of PbCl₂ in 0.75 M NaOH. The K_f of Pb(OH)₃^- is 8E13 and the K_sp of PbCl₂ is 1.7E-5. 

c) Write the overall chemical reaction and calculate its associated equilibrium constant K_overall

Calculate the solubility of PbCl₂ in 0.75 M NaOH. The K_f of Pb(OH)₃^- is 8E13 and the K_sp of PbCl₂ is 1.7E-5. 

d) Calculate the solubility of PbCl₂ in 0.75 M NaOH

Answer 1

a) PbCl₂(s) ==> Pb²⁺(aq) + 2Cl^-(aq) ... Ksp = 1.7x10^-5

b) Pb²⁺(aq) + 3OH-(aq) ==> Pb(OH)₃^-(s) ... Kf = 8x10¹³

—————————————————————————————

c) PbCl2(s) + 3OH-(aq) ==> Pb(OH)3(s) + 2Cl- ... K_overall = Ksp*Kf = 1.7x10^-5 * 8x10¹³ = 1.36x10⁹

.................0.75 M...................................0.....Initial

................-3x.........................................+2x..Change

...............0.75-3x....................................2x...Equilibrium

1.36x10⁹ = [Cl-]² / [OH-]³ = 2x² / (0.75-3x)³

Solve for x and that should be the molar solubility of PbCl₂ under the conditions given.