Question

A saturated solution of leadi) sulfate can be prepared by diluting 0.0183 g of PbS0, (MM- 303.26 g/mo0 to 480 mL. What is the K of lead) sulfate? Answer Check

Answer 1

Molar mass of PbSO4 = 303.26 g/mol

mass(PbSO4)= 0.0183 g

use:

number of mol of PbSO4,

n = mass of PbSO4/molar mass of PbSO4

=(1.83*10^-2 g)/(303.26 g/mol)

= 6.034*10^-5 mol

volume , V = 4.8*10^2 mL

= 0.48 L

use:

Molarity,

M = number of mol / volume in L

= 6.034*10^-5/0.48

= 1.257*10^-4 M

This is solubility of PbSO4

At equilibrium:

PbSO4 <----> Pb2+ + SO42-

   s s

Ksp = [Pb2+][SO42-]

Ksp = (s)*(s)

Ksp = 1(s)^2

Ksp = 1(1.257*10^-4)^2

Ksp = 1.58*10^-8

Answer: 1.58*10^-8