Question

1. Estimate the solubility of PbCl₂ in g/100mL at 25C. Use the Ksp value 1.7x10^-5.

Answer 1

At equilibrium:

PbCl2 <----> Pb2+ + 2 Cl-

   s 2s

Ksp = [Pb2+][Cl-]^2

1.7*10^-5=(s)*(2s)^2

1.7*10^-5= 4(s)^3

s = 1.62*10^-2 M

Molar mass of PbCl2,

MM = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

Molar mass of PbCl2= 278.1 g/mol

s = 1.62*10^-2 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.62*10^-2 mol/L * 278.1 g/mol

s = 4.505 g/L

We need find mass in 100 mL solution

volume = 100 mL = 0.100 L

So,

mass = s*volume

= 4.505 g/L * 0.100 L

= 0.4505 g

Answer: 0.45 g