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: (a) A strontium hydroxide solution is prepared by

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

Solution for problem 103AE 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 103AE

103AE: (a) A strontium hydroxide solution is prepared by dissolving 12.50 g of Sr(OH)2 in water to make 50.00 ml of solution. What is the molarity of this solution? (b) Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions, (c) If 23.9 ml of the strontium hydroxide solution was needed to neutralize a 37.5 ml aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?

Step-by-Step Solution:
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Solution: Step1: a) Given, mass of strontium hydroxide used = 12.50 g Molar mass of strontium hydroxide = 121.63 g/mol So, number of moles of strontium hydroxide = given mass / molar mass = 12.50 / 121.63 = 0.103 mol Volume of the solution = 50 mL = 50/1000 L = 0.050 L Therefore, molarity of the strontium hydroxide solution = number of moles / volume of solution in litres = 0.103 mol / 0.050 L = 2.06 M Thus, molarity of the strontium hydroxide solution is 2.06 M. Step2: b) Strontium hydroxide reacts with nitric acid to give strontium nitrate and water. The balanced molecular equation for the above reaction is: Sr(OH) 2q) + 2HNO (a3 --------> Sr(NO ) 3 2+ 2H O(2 -----(1) Step3: 23.9 c) Given, volume of the strontium hydroxide solution = 23.9 mL = 1000L = 0.0239 L Molarity...

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Chapter 4, Problem 103AE is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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