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In the presence of excess OH-, the Zn2+(aq) ion forms a hydroxide complex ion, Zn(OH)42-. Calculate...

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

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

Kf = [Zn(OH)42-]/[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

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