Question

In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate the concentration of free Zn2+ ion when 1.57×10-2 mol Zn(NO3)2(s) is added to 1.00 L of solution in which [OH- ] is held constant (buffered at pH 12.20). For Zn(OH)42-, Kf = 4.6×1017. [Zn2+]=___M please explain

Answer 1

As pH = 12.20,

pOH = 14 - 12.20 = 1.8

[OH^-] = 10^-1.8 = 0.0158

[Zn²⁺] added = 1.57 x 10^-2 mol in 1 liter = 0.0157 M

As K_f is very high, most of Zn²⁺ will be present as complex.

K_f = [Zn(OH)₄^2-]/[Zn²⁺][OH^-]⁴

4.6 x 10¹⁷ = (0.0157)/[Zn²⁺](0.0158)⁴

[Zn²⁺] = 5.4 x 10^-13

So, free [Zn²⁺] ions = 5.4 x 10^-13 M