 # Calculate the pH of a 0.072 M aqueous solution of aluminum chloride, AlCl 3 . The acid ionization of hydrated aluminum ion is Al ( H 2 O ) 6 3 + ( a q ) + H 2 O ( l ) ⇌ Al ( H 2 O ) 5 OH 2 + ( a q ) + H 3 O + ( a q ) and K 4 is 1.4 × 10 −3 . ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343 ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section
Chapter 16, Problem 16.101QP
Textbook Problem

## Calculate the pH of a 0.072 M aqueous solution of aluminum chloride, AlCl3. The acid ionization of hydrated aluminum ion is Al ( H 2 O ) 6 3 + ( a q ) + H 2 O ( l ) ⇌ Al ( H 2 O ) 5 OH 2 + ( a q ) + H 3 O + ( a q ) and K4 is 1.4 × 10−3.

Expert Solution
Interpretation Introduction

Interpretation:

The pH of a 0.072 M aqueous solution of Aluminium chloride has to be calculated

Concept Introdution:

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion [H3O+] concentration.

pH=-log[H3O+]

### Explanation of Solution

To Calculate: The pH of a 0.072 M aqueous solution of Aluminium chloride

Given data:

The concentration of aqueous Aluminium chloride solution = 0.072 M

The acid ionization of hydrated aluminium ion is:

Al(H2O)63+(aq) + H2O(l)  Al(H2O)5OH2+(aq) + H3O+(aq)

The Ka value is 1.4×105

pH calculation:

Assemble Al3+ is Al(H2O)63+

Now, set up an equilibrium table for the hydrolysis reaction of Al(H2O)63+

 Al(H2O)63+(aq) + H2O(l) ⇌ Al(H2O)5OH2+(aq) + H3O+(aq) Initial (M) 0.072 −x 0.072-x 0.00 0.00 Change (M) +x +x Equilibrium (M) x x

Substitute equilibrium concentrations into the equilibrium-constant equation

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