Question

Lead(II) nitrate is added slowly to a solution that is 0.0800 M in CT ions. Calculate the concentration of Pb2+ ions (in mol/L) required to initiate the precipitation of PbCl2- (Ksp for PbCl2 is 2.40 x 10-4.) Type here to search

Answer 1

Answer : The concentration of
`Pb^(2+)` ion is 0.0375 M.

Explanation :

The balanced equilibrium reaction will be:

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

The expression for solubility constant for this reaction will be,

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

Now put all the given values in this expression, we get:

`2.40* 10^(-4)=[Pb^(2+)]* (0.0800)^2`

`[Pb^(2+)]=0.0375M`

Therefore, the concentration of
`Pb^(2+)` ion is 0.0375 M.