   Chapter 20, Problem 89CWP

Chapter
Section
Textbook Problem

# Calculate the solubility of Mg(OH)2 (Ksp = 8.9 × 10−12) in an aqueous solution buffered at pH = 9.42.

Interpretation Introduction

Interpretation: The solubility of Mg(OH)2 at the given pH of the given solubility product constant is to be calculated.

Concept introduction: The concentration of hydroxide ion is calculated as,

[OH]=antilog[(14pH)]

The solubility product constant is calculated by the formula,

Ksp=[Mg2+][OH]2

To determine: The solubility of Mg(OH)2 for the given pH .

Explanation

Given

The pH of the solution is 9.42 .

The solubility product constant, Ksp , is 8.9×1012 .

The concentration of hydroxide ion is calculated as,

[OH]=antilog[(14pH)]

Substitute the value of pH in the above equation,

[OH]=antilog[(149.42)]=antilog(4.58)=2.63×105M

The dissolution equilibrium of Mg(OH)2 is,

Mg(OH)2(s)Mg2+(aq)+2OH(aq)

The solubility of Mg(OH)2 is assumed to be x .

The concentration of the ions formed in the above reaction is shown in the table below,

Mg(OH)2(s)Mg2+(aq)+2OH(aq)Ininial(M):02.63×105Change(M):+x+2xEquilibrium(M):x2

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