Question

Lead(II) nitrate is added slowly to a solution that is 0.0800 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

Answer:

$[Pb^{2+}]=3.9 \times 10^{-2}M$

this is the concentration required to initiate precipitation

Explanation:

$PbCl_2$  ⇄ $Pb^{2+}+2Cl^-$

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

$Ip=[Pb^{2}][2Cl^-]^2=Ksp$

$Ksp=2.4\times 10^{-4}$

lets solubility=S

$[Pb^{2+}] = S$

$[Cl^-]=2S$

$Ksp=[Pb^{2+}]\times [Cl^-]^2$

$Ksp=S \times (2S)^2$

$Ksp=4S^3$

$S=\sqrt[3]{\frac{Ksp}{4} }$

$S=3.9\times 10^{-2}$

$[Pb^{2+}]=3.9 \times 10^{-2}M$ this is the concentration required to initiate precipitation