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Mercury(I) ions (Hg22+) can be removed from solution by

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

Solution for problem 81E 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 81E

Mercury(I) ions (Hg22+) can be removed from solution by precipitation with Cl- . Suppose that a solution contains aqueous Hg2(NO3)2. Write complete ionic and net ionic equations to show the reaction of aqueous Hg2(NO3)2 with aqueous sodium chloride to form solid Hg2Cl2 and aqueous sodium nitrate.

Step-by-Step Solution:
Step 1 of 3

Solution: Step1: a) In order to calculate the volume of 0.115 M HClO , we will use the 4utralization equation, i.e., M V1 M1V , 2e2 M and V are1 e molar1y and volume of the acid and M , V are the molarity and volume of the base. 2 2 Substituting the values of molarity and volume of the acid and base in the above equation, we get, 0.115 M X V = 0.0875 M X 50 mL 1 0.115 M X V = 4.31 (M X mL) V1 4.375 / 0.115 [(M X mL)/ mL] V1 38.04 mL Thus, required volume of HClO is 38.04 mL4 Step2: b) Mass of Mg(OH) = 2.87 g 2 Molar mass of Mg(OH) = 58.32 g/mol 2 Therefore, number of moles of Mg(OH) = 2.87 / 58.322 = 0.049 mol The reaction between HCl and Mg(OH) takes place in 2e following manner: 2HCl(aq) + Mg(OH) -----------> MgCl (aq) + 2H O 2 2 2 In the above equation, it is seen that, 1 mole of Mg(OH) requires 2 mole2of HCl for complete neutralization. Therefore, 0.049 moles of Mg(OH) will require (2 x 0.049 = 0.098) moles of HCl for 2 complete neutralization. Now, using the expression for molarity, we can calculate the volume of HCl required. Volume of HCl in litres = Number of moles of HCl / Molarity = 0.098 mol / 0.128 M = 0.7656 L = 0.7656 X 1000 mL = 765.6 mL Thus, the required volume of HCl is 765.6 mL Step3: c) Given mass of KCl = 785 mg = 785 / 1000 g = 0.785 g Molar mass of KCl = 74.54 g/ mol Therefore, number...

Step 2 of 3

Chapter 4, Problem 81E is Solved
Step 3 of 3

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

This full solution covers the following key subjects: solution, needed, neutralize, Volume, koh. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. Since the solution to 81E from 4 chapter was answered, more than 371 students have viewed the full step-by-step answer. The answer to “Mercury(I) ions (Hg22+) can be removed from solution by precipitation with Cl- . Suppose that a solution contains aqueous Hg2(NO3)2. Write complete ionic and net ionic equations to show the reaction of aqueous Hg2(NO3)2 with aqueous sodium chloride to form solid Hg2Cl2 and aqueous sodium nitrate.” is broken down into a number of easy to follow steps, and 46 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. The full step-by-step solution to problem: 81E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM.

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