   Chapter 21, Problem 40PS

Chapter
Section
Textbook Problem

Ca(OH)2 has a Ksp of 5.5 × 10−5, whereas Ksp for Mg(OH)2 is 5.6 × 10−12. Calculate the equilibrium constant for the reaction Ca(OH) 2 +  Mg 2+ ( aq )   ⇄  Ca 2+ (aq) + Mg(OH) 2 ( s ) Explain why this reaction can be used in the commercial isolation of magnesium from seawater.

Interpretation Introduction

Interpretation:

To calculate the equilibrium constant for the given reaction and explain why the given reaction is used in commercial isolation of magnesium from sea water.

Concept introduction:

The equilibrium constant is a quantity that relates the amount of product and reactant form at equilibrium at a given temperature. It is expressed as a ratio of molar concentrations of product over the reactants.

The equilibrium constant for any reaction is the reciprocal of the solubility product of that reaction. The solubility product is expressed by the symbol Ksp and is defined as the product of concentrations of the ions formed in the reaction.

Explanation

The equilibrium constant for the given reaction is calculated below.

Given:

The given reaction is,

Ca(OH)2(s)+Mg2+(aq)Ca2+(aq)+Mg(OH)2(s)

The Ksp for Ca(OH)2 is 5.5×105.

The Ksp for Mg(OH)2 is 5.6×1012.

For Ca(OH)2,

Ca(OH)2(s)Ca2+(aq)+2OH(aq)

The solubility product is written as,

Ksp=[Ca2+][OH]2

The concentration of calcium ion and hydroxide ion is expressed by s.

Substitute the value of Ksp.

5.5×105=(s)(2s)25.5×105=4s3s=0.0239

Thus,

[Ca2+]=0

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