Question

Enter your answer in the provided box. Lead(II) nitrate is added slowly to a solution that is 0.0400 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol / L) required to initiate the precipitation of PbCl2.

(Ksp for PbCl2 is 2.40 × 10−4.)

Answer 1

the concentration of Pb2+ ions is
`[Pb^(2+)] } = 0.15 mol/L`

Step-by-step explanation:

From the question we are told that

The concentration of
`Cl^(-)` is
`C_s = 0.0400 M`

The reaction is

`Pb^(2+) + 2Cl^- ----> PbCl_2`

The solubility constant for this reaction is mathematically represented as

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

Substituting values

`2.4 0 * 10^(-4) = 0.0400^2 * [Pb^(2+)]`

`[Pb^(2+)] } = (2.4 0 * 10^(-4) )/(0.0400^2 )`

`[Pb^(2+)] } = 0.15 mol/L`