What is the molar solubility of zinc hydroxide at pH 12.34? For Zn(OH)₂,K_sp = 2.1 × 10^-16; for Zn(OH)₄^2-,K_f = 2.8 × 10¹⁵.

Question

Quizplus · Accepted Answer

The equilibrium reaction for the dissolution of zinc hydroxide is:

Zn(OH)2(s) ⇌ Zn2+(aq) + 2OH-(aq)

The solubility product constant (Ksp) expression for this reaction is:

Ksp = [Zn2+][OH-]^2

At pH 12.34, the concentration of hydroxide ions can be calculated from the expression for the dissociation constant of water (Kw):

Kw = [H+][OH-] = 1.0 × 10^-14

pH + pOH = 14, so pOH = 14 - 12.34 = 1.66

[OH-] = 10^-pOH = 10^-1.66 = 2.17 × 10^-2 M

Substituting this value and the given Ksp into the Ksp expression:

2.1 × 10^-16 = [Zn2+](2.17 × 10^-2)^2

[Zn2+] = 2.8 × 10^-4 M

This is the molar solubility of zinc hydroxide at pH 12.34, so the best choice is C.

What Is the Molar Solubility of Zinc Hydroxide at PH