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A solution of NaOH was standardized by titration of a known quantity of the primary

Quantitative Chemical Analysis | 8th Edition | ISBN: 9781429218153 | Authors: Daniel C. Harris ISBN: 9781429218153 475

Solution for problem 1-E Chapter 1

Quantitative Chemical Analysis | 8th Edition

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Quantitative Chemical Analysis | 8th Edition | ISBN: 9781429218153 | Authors: Daniel C. Harris

Quantitative Chemical Analysis | 8th Edition

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Problem 1-E

A solution of NaOH was standardized by titration of a known quantity of the primary standard, potassium hydrogen phthalate (page 223): Potassium hydrogen phthalate (FM 204.221) C8H5O4K NaOH C8H4O4NaK H2O The NaOH was then used to find the concentration of an unknown solution of H2SO4:(a) Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of NaOH solution to reach the end point detected by phenolphthalein indicator. Find the concentration of NaOH (mol NaOH/kg solution). (b) A 10.00-mL aliquot of H2SO4 solution required 57.911 g of NaOH solution to reach the phenolphthalein end point. Find the molarity of H2SO4.

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Chapter 50: How are Equilibrium Concentrations Determined 2/20/16 50.1 How are Equilibrium Concentrations Determined  How can the Direction of a Reaction be Predicted o Reaction quotient­ Q, same as the equilibrium constant but concentrations in Q expression not necessarily at equilibrium o Q doesn’t have a fixed value o System at equilibrium Q=K but otherwise Q doesn’t equal K o Only forward reaction can occur o Forward reaction rate has to be greater than reverse reaction rate o Numerator will increase denominator will decrease o Q will become greater than zero but starts as a small number but less than K o Q less than K­ rate of forward reaction is greater than the

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Textbook: Quantitative Chemical Analysis
Edition: 8
Author: Daniel C. Harris
ISBN: 9781429218153

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A solution of NaOH was standardized by titration of a known quantity of the primary