   Chapter 16, Problem 91AE

Chapter
Section
Textbook Problem

# The solubility rules outlined in Chapter 4 say that BaCOH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble hydroxides. Calculate the pH of a saturated solution of each of these marginally soluble hydroxides.

Interpretation Introduction

Interpretation:

The pH of a saturated solution of the given hydroxides is to be calculated.

Concept introduction:

The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The pH of a solution is calculated by the formula, pH=log[H+]

The sum, pH+pOH=14

Explanation

To determine: The pH of a saturated solution of the given hydroxides.

pH of a saturated solution of Ba(OH)2

Concentration of hydroxide ion of Ba(OH)2

The Ksp of Ba(OH)2 is 5.0×103 .

The dissociation reaction of Ba(OH)2 is,

Ba(OH)2Ba2++2OH

The solubility of Ba2+ is assumed to be smol/L .

The solubility product of dissociation of Ba(OH)2 is calculated by the formula,

Ksp=[Ba2+][OH]2

The solubility of Ba(OH)2 can be calculated from the concentration of its ions at equilibrium.

It is assumed that smol/L of solid is dissolved to reach the equilibrium. The meaning of 1:2 stoichiometry of salt is,

smol/LBa(OH)2smol/LBa2++2smol/LOH

Formula

The solubility product of Ba(OH)2 is calculated as,

Ksp=[Ba2+][OH]2=(s)(2s)2=4s3

Where,

• Ksp is solubility product.
• [Ba2+] is concentration of Ba2+ .
• [OH] is concentration of OH .
• s is the solubility.

Substitute the given value of Ksp in the above expression.

4s3=5.0×103s=0.11mol/L

The [OH] =2s=2(0.11mol/L)=0.22mol/L_

pH Calculation:

The [OH] is 0.22mol/L .

The pOH is calculated by the formula,

pOH=log[OH]

Substitute the value of [OH] in the above expression.

pOH=log[0.22]=0.66

The pH for the solution is =14pOH=140.66=13.34_

The pH of a saturated solution of Ba(OH)2 is 13.34_ .

pH of a saturated solution of Sr(OH)2

Concentration of hydroxide ion of Sr(OH)2

The Ksp of Sr(OH)2 is 3.2×104 .

The dissociation reaction of Sr(OH)2 is,

Sr(OH)2Sr2++2OH

The solubility of Sr2+ is assumed to be smol/L .

The solubility product of dissociation of Sr(OH)2 is calculated by the formula,

Ksp=[Sr2+][OH]2

The solubility of Ba(OH)2 can be calculated from the concentration of its ions at equilibrium.

It is assumed that smol/L of solid is dissolved to reach the equilibrium. The meaning of 1:2 stoichiometry of salt is,

smol/LSr(OH)2smol/LSr2++2smol/LOH

Formula

The solubility product of Sr(OH)2 is calculated as,

Ksp=[Sr2+][OH]2=(s)(2s)2=4s3

Where,

• Ksp is solubility product

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