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:

$2.40\times 10^{-4}=[Pb^{2+}]\times (0.0800)^2$

$[Pb^{2+}]=0.0375M$

Therefore, the concentration of $Pb^{2+}$ ion is 0.0375 M.