Question

Find the pH of a solution prepared from 1.0 L of a 0.20 M solution of Ba(OH)₂ and excess Zn(OH)₂(s). The Ksp of Zn(OH)₂ is 3×10⁻¹⁵ and the Kf of Zn(OH)₄²⁻ is 2×10¹⁵.

Answer 1

The concentration of Ba(OH)2 = 0.2 M

Ksp of Ba(OH)2 = 0.0005

          Ba(OH)₂ --> Ba⁺² + 2OH^-

so concentration of OH- = 0.4 M

Zn(OH)₂ --> Zn⁺² + 2OH^-   Ksp = 3X10^-`15

Zn⁺² + 4OH-   --> Zn(OH)₄^-2   Kf = 2X10^15

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Zn(OH)2 + 2OH- --> Zn(OH)₄^-2    K = Ksp X Kf = 3X2X10^-15 X 10^14 = 6

Now Zn(OH)2 is in excess so K = [Zn(OH)₄^-2] / [OH-]²

Let us calculate the molar solubility of Zn(OH)₄^-2 in presence of 0.2 M Ba(OH)2

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

initial             0.2 0

Change    -x x

Equilibrium    0.2-x x         

K = 6 = x / (0.2-x)²        

6 = x / 0.04 + x^2 -0.4x

0.24 + 6x^2 - 2.4x = x

6x^2 -3.4x + 0.24 = 0

on solving for x

x = 0.0826

so concentration of OH- = 0.2 - 0.0826 = 0.1174 M

pOH = -log[OH-] = 0.930

pH = 14-0.93 = 13.07