Question

In the presence of excess OH^-, the Zn²⁺(aq) ion forms a hydroxide complex ion, Zn(OH)₄^2-. Calculate the concentration of free Zn²⁺ ion when 1.46×10^-2 mol Zn(CH₃COO)₂(s) is added to 1.00 L of solution in which [OH^- ] is held constant (buffered at pH 12.80). For Zn(OH)₄^2-, K_f = 4.6×10¹⁷.

Answer 1

Zn+2 + 4OH- --> Zn(OH)-2

Kf = [Zn(OH)-2] /([Zn+2][OH-]^4)

initially

[Zn+2] = 0.0146 ( since it is added to liter)

[OH-] = 10^-pOH = 10^-(14-pH) = 10^-(14-12.80) = 0.063095

[Zn(OH)-2] = 0

in euqilibrium

[Zn+2] = 0.0146 - x

[OH-] = 0.063095 -4x --> constnat so = 0.063095

[Zn(OH)-2] = +x

substitute in Kf

Kf = [Zn(OH)-2] /([Zn+2][OH-]^4)

4.6*10^17 = (x)/(0.0146 - x)(0.063095)^4

(4.6*10^17)((0.063095)^4) = x /(0.0146 - x)

(7.29*10^12)*((0.0146 - x)) = x

0.0146 - x = x/(7.29*10^12)

x = 0.0146

then

[Zn(OH)-2] = +x = 0.0146 M approx