Suppose you add 0.2469 g of PbCl2(s) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of PbX2+(aq) is 0.0159 M and the concentration of C l − ( a q ) ClX−(aq) is 0.0318 M. What is the value of the equilibrium constant, Ksp, for the dissolution of PbCl2?

6.431 x 10∧-5 {mol/l)³

Explanation:

The Ksp (PbCl₂) of dissolve PbCl₂ is 6.431×10⁻⁵ (mol/L)³

Explanation:

                      PbCl₂ (s)   ⇒   Pb²⁺ (aq) + 2Cl⁻ (aq)

=           a                      -                 -

=    a - s                 s                2s

T=time , a=initial condition , s=solubility on concentration

Con. [Pb²⁺(aq)] = 0.0159 M

Con. [Cl⁻(aq)] = 0.0318 M

Ksp = [Pb²⁺] [Cl⁻]²

      = (S) (2S)²

      = (0.0159 M) (2 × 0.0318)²

Ksp (PbCl₂) = 6.431×10⁻⁵ M³

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.