In: Chemistry
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 Kf of HgCl42- is 1.7*1016.
The answer is 2.2*10-19 M, how?
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