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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 93 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 93

93: Suppose you have a solution that might contain any or all of the following cations: Ni2+, Ag+ Sr2+, and Mn2+. Addition of HCI solution causes a precipitate to form. After filtering off the precipitate, H2S04 solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of NaOH is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to identify the precipitates formed during each of the above step reactions. Step1: Precipitate is the insoluble product formed during a reaction. When we add HCl to a solution 2+ + 2+ 2+ containing Ni , Ag , Sr , or Mn , the chloride salt of the cation will be formed. From the guidelines of solubility, all chlorides, bromides, and iodides are soluble except those of silver, mercury, and lead (e.g., AgCl, Hg Cl , and PbCl ). Thus, the precipitate 2 2 2 formed in this step is AgCl and the ions present in it are Ag and Cl . + - Step2: + After filtering off the precipitate, Ag is removed from the solution. So, on addition of H SO 2 4 2+ 2+ 2+ to the solution containing Ni , Sr , or Mn , the sulfate salt...

Step 2 of 3

Chapter 4, Problem 93 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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