A second 2500-ml aliquot was then treated with an excess of


a 0.5000-g sample containing NaHCO3, Na2CO3, and H2O was dissolved and diluted to 250.0 mL. A 25.00-mL aliquot was then boiled with 50.00 mL of 0.01255 M HCl. After cooling, the excess acid in the solution required 2.34 mL of 0.01063 M NaOH when titrated to a phenolphthalein end point. A second 25.00-mL aliquot was then treated with an excess of BaCl2 and 25.00 mL of the base; precipitation of all the carbonate resulted, and 7.63 mL of the HCl was required to titrate the excess base. Calculate the composition of the mixture.

Request for Solution File

Ask an Expert for Answer!!
Chemistry: A second 2500-ml aliquot was then treated with an excess of
Reference No:- TGS0615150

Expected delivery within 24 Hours