Explanation A l C l 3 produces A l 3 + and C l − ions in the solution. A l C l 3 ( s ) → A l 3 + ( a q ) + 3 C l − ( a q ) The C l − ions do not affect the p H of a water solution. A l 3 + ion react with water to form hydrated aqueous aluminum ion. A l 3 + ( a q ) + 6 H 2 O ( a q ) → [ A l ( H 2 O ) 6 ] 3 + K a value of [ A l ( H 2 O ) 6 ] 3 + is 1.0 × 10 − 5 . The equation for the equilibrium can be written as given below. [ A l ( H 2 O ) 6 ] 3 + ( a q ) + H 2 O ( l ) ⇌ A l ( H 2 O ) 5 ( O H ) 2 + ( a q ) + H 3 O + ( a q ) The acid ionization constant for the above equation can be written as given below. K a = [ A l ( H 2 O ) 5 ( O H ) 2 + ] [ H 3 O + ] [ A l ( H 2 O ) 6 3 + ] A table can be set up as given below. A l ( H 2 O ) 6 3 + ( a q ) A l ( H 2 O ) 5 ( O H ) 2 + ( a q ) H 3 O + ( a q ) I n i t i a l c o n c

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

5th Edition

ISBN: 9781285460420

Author: John W. Moore; Conrad L. Stanitski

Publisher: Cengage Learning US

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Chapter 14, Problem 123QRT

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Interpretation:

The pH of 0.10 M AlCl3 has to be calculated.

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Chapter 14, Problem 122QRT

Chapter 14, Problem 124QRT

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Chapter 14 Solutions

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

Ch. 14.1 - Identify each molecule or ion as a Brnsted-Lowry...Ch. 14.1 - Using Le Chatelier’s Principle Use Le Chatelier’s...Ch. 14.1 - Prob. 14.3E Ch. 14.1 - Complete the table. Ch. 14.1 - Prob. 14.4E Ch. 14.1 - Prob. 14.5E Ch. 14.2 - Prob. 14.6CE Ch. 14.2 - Prob. 14.7E Ch. 14.3 - Prob. 14.2PSP Ch. 14.3 - Prob. 14.3PSP

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