Question

Calculate the concentration of free mercury (II) ion, [Hg²⁺], in an aqueous solution that is initially 0.16 M Hg(NO₃)₂ and 3.2 M NaCl given that the K_f of HgCl₄^2- is 1.7*10¹⁶.

The answer is 2.2*10^-19 M, how?

Answer 1

The equation for the complexation reaction is

        Hg⁺²               + 4 Cl^-         -------> [HgCl₄]^-2

      0.16M                  3.2M                       0       initial concentrations

   0.16-x                   3.2-4x                       x       after reaction

Thus Kf = x/[(0.16-x)(3.2-4x)⁴ which is equal to 1.7 x 10¹⁶

Soving for x we get