-0.500 g sample consisting of NazCO3, NaHCO3, and inert materials was dissolved in sufficient vater to give 250.0 mL of solution. A 25.00 mL aliquot of this solution was boiled with 50.00 ml of 0.01255 M HC. Titration of the excess HCl to a phenolphthalein endpoint required 2.34 mL of 0.01063 M NaOH. The carbonate species in the second 25.00 mL aliquot was precipitated with an excess of BaCl, and treated with 25.00 mL of the NaOH. Titration of the excess base required 7.63 ML of the standard acid. Calculate % Na2CO3 and NaHCO3 in the sample.

Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.40QAP
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A 0.500 g sample consisting of NażCO3, NaHCO3, and inert materials was dissolved in sufficient
water to give 250.0 mL of solution. A 25.00 mL aliquot of this solution was boiled with 50.00 ml
of 0.01255 M HCI. Titration of the excess HCI to a phenolphthalein endpoint required 2.34 mL of
0.01063 M NaOH. The carbonate species in the second 25.00 mL aliquot was precipitated with an
excess of BaC, and treated with 25.00 mL of the NaOH. Titration of the excess base reguired 7.63
mL of the standard acid. Calculate % NazCO3 and NaHCO3 in the sample.
Transcribed Image Text:A 0.500 g sample consisting of NażCO3, NaHCO3, and inert materials was dissolved in sufficient water to give 250.0 mL of solution. A 25.00 mL aliquot of this solution was boiled with 50.00 ml of 0.01255 M HCI. Titration of the excess HCI to a phenolphthalein endpoint required 2.34 mL of 0.01063 M NaOH. The carbonate species in the second 25.00 mL aliquot was precipitated with an excess of BaC, and treated with 25.00 mL of the NaOH. Titration of the excess base reguired 7.63 mL of the standard acid. Calculate % NazCO3 and NaHCO3 in the sample.
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