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How to solve for the % of NaHCO3 and Na2CO3?
A 0.5000 g sample containing NaHCO3 (MW: 84.01 g/mol), Na2CO3(105.99 g/mol), and H2O
(18.02g/mol) was dissolved and diluted to 250.0 ml aliquot was then boiled with 50.00ml of0.01255 M HCl. After cooling, the excess acid in the solution required 2.34 ml of 0.01063 M NaOH when titrated up to the phenolphthalein end point. A second 25.00 ml aliquot was then titrated with excess BaCl2 and 25.00 ml of the base; precipitation of all carbonates resulted, and 7.63 ml of the HCl was required to titrate the excess base.
0.01255 M HCl. After cooling, the excess acid in the solution required 2.34 ml of 0.01063 M
NaOH when titrated up to the phenolphthalein end point. A second 25.00 ml aliquot was then
titrated with excess BaCl2 and 25.00 ml of the base; precipitation of all carbonates resulted, and
7.63 ml of the HCl was required to titrate the excess base.
- Attachment 1
- Attachment 2
