Calculate the equilibrium molarity of aqueous Pb2+

Question

For the aqueous [PbCI₄]^2- complex K_f =2.5 x 10¹⁵ at 25 °C.

Suppose equal volumes of 0.0074M Pb(NO₃)₂ solution and 0.36M KCI solution are mixed. Calculate the equilibrium molarity of aqueous Pb²⁺

Round your answer to 2 significant digits.

____M

Answer 1

Pb²⁺(aq) + 4Cl^-(aq) ==> [PbCl₄]^2-(aq) .. balanced equation for complex formation .. Kf = 2.5x10¹⁵

Since equal volumes of the 2 reactants are used, the final concentrations will be 1/2 of the original concentrations. Thus...

[Pb2+] = 1/2 x 0.0074 M = 3.7x10^-3 M

[Cl-] = 1/2 x 0.36 M = 0.18 M

Pb²⁺(aq) + 4Cl^-(aq) <==> [PbCl₄]^2-(aq)

3.7x10^-3........0.18..................0............Initial

-3.7x10^-3....-3.7x10^-3.......+3.7x10^-3....Change

0..............0.18-3.7x10^-3...3.7x10^-3....Equilibrium

New ICE table for reverse:

Pb²⁺(aq) + 4Cl^-(aq) <==> [PbCl₄]^2-(aq)

0.........0.18-3.7x10^-3............3.7x10^-3.........Initial

+x..............+x......................-x....................Change

x.......0.18 - 3.7x10^-3+x..........3.7x10^-3-x.......Equilibrium

Kf = [PbCl₄]^2- / [Pb²⁺][Cl^-]⁴

2.5x10¹⁵ = 3.7x10^-3-x / (x)(0.18 - 3.7x10^-3+x)⁴

Because Kf is so large, we can simplify as follows:

2.5x10¹⁵ = 3.7x10^-3 / (x)(0.18)⁴

2.5x10¹⁵ = 3.7x10^-3 / 1.05x10^-3 x

x = [Pb²⁺] = 1.41x10^-15 M

(be sure to check all of the math)