Calculate weight percentages of Na_2CO_3 and NaHCO_3?

Question

A sample weighing 0.3375 g and containing Na_2CO_3 (106.0 g/mol), NaHCO_3 (84.01 g/mol), and inert material is dissolved and titrated with 0.1014 M HCl. The first end point volume (phenolphthalein end point) is 11.60 mL and the second end point volume (bromocresol green end point) is 41.93 mL. Calculate the weight percentages of Na_2CO_3 and NaHCO_3.

Answer 1

initially the 11.60 ml reacts with the Na2CO3:

1 (CO3)-2 &1 HCl --> (HCO3)-1

0.0116 Litres @ 0.1014 mol/litre = 0.001176 moles of HCL

0.001176 mol HCL =  0.001176 mol Na2CO3

0.001176 mol Na2CO3 @ 105.99 g/mol = 0.1247g Na2CO3

0.1247g Na2CO3 / 0.3375 g mix = 36.94% Na2CO3

your first answer: 36.94% Na2CO3

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another 11.60 ml of HCl converts that (HCO3)- onto H2CO3

that makes 23.20 ml out of the total 41.93ml for the (CO3)-2 , & that leaves 18.70 ml to react with the original NaHCO3

0.01870 L @ 0.1014 mol/litre = 0.001896 moles of HCl

0.001896 mol HCL = 0.001896 mol NaHCO3

0.001896 mol NaHCO3 @ 84.007 g/mol = 0.1593g NaHCO3

0.1593NaHCO3 / 0.3375 g mix = 47.20% NaHCO3

your next answer: 47.20% NaHCO3