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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

When 5 M ammonia is added to a solution containing Cu(OH)2(s), the precipitate will eventually dissolve in solution. Why? If 5 M HNO3 is then added, the Cu(OH)2 precipitate re-forms. Why? In general, what effect does the ability of a cation to form a complex ion have on the solubility of salts containing that cation?

Interpretation Introduction

Interpretation: The reason of non-formation of precipitate when 5M ammonia is added to the solution of Cu(OH)2 but there occurs formation of precipitate of Cu(OH)2 when 5M HNO3 is added and the effect of ability of a cation to form a complex ion over the solubility of complex ion is to be stated.

Concept introduction: The formation of precipitate occurs when there occurs the reaction between the cations and anions present in the solution. It is determined by using the rules of solubility of ionic compounds.

Explanation

Explanation

To determine: The reason of non-formation of precipitate when 5M ammonia is added to the solution of Cu(OH)2 but there occurs formation of precipitate of Cu(OH)2 when 5M HNO3 is added and the effect of ability of a cation to form a complex ion over the solubility of complex ion.

The reaction between Cu2+ ions Ammonia forms the complex as Cu(NH3)42+ . Therefore, the addition of 5M ammonia to the solution of Cu(OH)2 , leads to the dissolution of precipitate.

The solubility of Cu(OH)2 is given as,

Cu(OH)2Cu2++2OH

Here, the formation of Cu2+ ions occurs. These formed ions combine with Ammonia and forms the complex as Cu(NH3)42+ . As there occurs the consumption of Cu2+ ions, it leads to the dissolution of more Cu(OH)2 according to the Le Chatelier’s principle and this will occur the point till the ionic product is lower than the solubility product. Once the ionic product becomes greater than the solubility product, precipitation starts

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