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Solved: The arsenic in a 1.22-g sample of a pesticide was

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

Solution for problem 113 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 113

The arsenic in a 1.22-g sample of a pesticide was converted to As043- by suitable chemical treatment. It was then titrated using Ag+ to form Ag3As04 as a precipitate, (a) What is the oxidation state of As in As043-? (b) Name Ag3As04 by analogy to the corresponding compound containing phosphorus in place of arsenic, (c) If it took 25.0 ml of 0.102 M Ag+ to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Step-by-Step Solution:
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Solution: Step1: 3- a) Calculation of oxidation state of As in AsO : 4 Since, Arsenate ion is a polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, x + 4(-2) = -3 [x being the oxidation number of As] x - 8 = -3 x = -3 + 8 x = +5 Thus, the oxidation number of As in AsO 4-is +4. Step2: b) The name of Ag AsO 3 silv4 arsenate. Step3: 25.0 Given, volume of silver ion solution used = 25.0 mL = 1000 L = 0.025 L Molarity of silver ion solution = 0.102 M Therefore, number of moles of silver ion = volume of solution in litres x molarity = 0.025 L x 0.102 M = 0.00255 mol Step4: Now, the reaction between silver ion and arsenate ion takes place in the following...

Step 2 of 3

Chapter 4, Problem 113 is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Solved: The arsenic in a 1.22-g sample of a pesticide was

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