A 0.8250 g of soda ash sample was dissolved in distilled H O and diluted to 250-mL using a volumetric flask. A 50.00 mL aliquot portion of the dilute soda ash solution needed 4.50 mL of 0.1200 M HCl to reach the phenolphthalein end point and 10.00 mL to reach the methyl orange end point. From the volumes of HCl needed to reach the phenolphthalein and methyl orange endpoints, how many millimoles of NaHCO are there in the aliquot portion that was titrated?
A 0.8250 g of soda ash sample was dissolved in distilled H O and diluted to 250-mL using a volumetric flask. A 50.00 mL aliquot portion of the dilute soda ash solution needed 4.50 mL of 0.1200 M HCl to reach the phenolphthalein end point and 10.00 mL to reach the methyl orange end point. From the volumes of HCl needed to reach the phenolphthalein and methyl orange endpoints, how many millimoles of NaHCO are there in the aliquot portion that was titrated?
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter16: Reactions Between Acids And Bases
Section: Chapter Questions
Problem 16.102QE
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A 0.8250 g of soda ash sample was dissolved in distilled H O and diluted to 250-mL using a volumetric flask. A 50.00 mL aliquot portion of the dilute soda ash solution needed 4.50 mL of 0.1200 M HCl to reach the phenolphthalein end point and 10.00 mL to reach the methyl orange end point.
From the volumes of HCl needed to reach the phenolphthalein and methyl orange endpoints, how many millimoles of NaHCO are there in the aliquot portion that was titrated?
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