   # Calculate the molar solubility of Al(OH) 3 , K sp = 2 × 10 −32 . ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 33E
Textbook Problem
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## Calculate the molar solubility of Al(OH)3, Ksp = 2 × 10−32.

Interpretation Introduction

Interpretation: The solubility product of Al(OH)3 is given. The molar solubility of Al(OH)3 is to be calculated.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

### Explanation of Solution

Explanation

To determine: The solubility of Al(OH)3 from the given Ksp value.

The solubility of Al(OH)3 is 5.14×109mol/L_ .

Given

Solubility product of Al(OH)3 is 2×1032 .

Since, solid Al(OH)3 is placed in contact with water. Therefore, compound present before the reaction is Al(OH)3 and H2O . The dissociation reaction of Al(OH)3 is,

Al(OH)3(s)Al3+(aq)+3OH(aq)

Since, Al3+ does not dissolved initially, hence,

[Al3+]initial=0

It is assumed that initial concentration of OH is 1×107 .

The solubility of Al(OH)3 can be calculated from the concentration of ions at equilibrium.

It is assumed that smol/L of solid is dissolved to reach the equilibrium. The meaning of 1:3 stoichiometry of salt is,

smol/LAl(OH)3smol/LAl3++3smol/LOH

Make the ICE table for the dissociation reaction of Al(OH)3

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