Solubility question

Question

Need help with this question with working out if possible :

 2. The Ksp for lead(II) chloride is 1.6 x 10-5 . Calculate the molar solubility of lead(II) chloride  

a) in pure water (3 marks) 

 b) in 0.10 mol/L calcium chloride (5 marks)  

Answer 1

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq)

a) Solubility in pure water:

Ksp = [Pb²⁺][Cl^-]²

Let [Pb²⁺] = x and the [Cl^-] = 2x

1.6x10^-5 = (x)(2x)² = 4x³

x³ = 4.0x10^-6

x = 1.59x10^-2 M = molar solubility of PbCl₂ in pure water

b) Solubility in 0.10 M CaCl₂ - Common Ion Effect

The common ion is Cl^- and the [Cl^-] will be taken to be 0.20 M since CaCl₂ ==> Ca²⁺ + 2Cl^-

Ksp = [Pb²⁺][Cl-]²

1.6x10^-5 = [Pb²⁺][0.2]²

[Pb²⁺] = 4x10^-4 M = molar solubility of PbCl₂ in 0.1 M CaCl₂