Question

Consider the dissolution equation of lead(II) chloride.

PbCl2 (s) ⇆ Pb2+ (aq) + 2Cl− (aq)
Suppose you add 0.2307 g of PbCl2 (s) to 50.0 mL of water. In the resulting saturated solution, you find that the concentration of Pb2+ (aq) is 0.0159 M and the concentration of Cl− (aq) is 0.0318 M.
What is the value of the equilibrium constant, Ksp, for the dissolution of PbCl2?

Answer 1

The solubility product of lead(II) chloride is
`1.61* 10^(-5)`.

Step-by-step explanation:

Concentration of lead (II) ions =
`[Pb^(2+)]=0.0159 M`

Concentration of chloride ion =
`[Cl^-]=0.0318 M`

`PbCl_2(s)\rightleftharpoons Pb^(2+)(aq)+2Cl^-(aq)`

The expression of a solubility product will be given as:

`K_(sp)=[Pb^(2+)][Cl^-]^2`

`=0.0159 M* (0.0318 M)^2=1.61* 10^(-5)`

The solubility product of lead(II) chloride is
`1.61* 10^(-5)`.