Question

A .523g sample of a mix of Na2co3 and Nahco3, is titrated with .1 HCl requiring 17 ml to reach the phenolphthalein end point and a total of 43.8 ml to reach the methyl orange end point. What is the percent of each of na2co3 and nahco3 in the mixture?

Answer 1

Here is the way this works.
At the beginning of the titration you have a mixture of CO3^2- + HCO3^-.
When you titrated with the first 17 mL you don't touch the HCO3^-; you titrate all of CO3^- halfway. That is
CO3^2- + H^+ ==> HCO3^-. With that information you can calculate the percent Na2CO3.
mL x M x milliequivalent weight = 17 x 0.1 x 0.106 = ? grams Na2CO3.
Then %Na2CO3 = (g Na2CO3/mass sample)*100 = ?

The next part of the titration titrates what's left. What's left is all of the CO3^- (which is now HCO3^-) + the HCO3^- there from the start. It takes 43.8 mL to titrate from beginning to end of which 34 mL (17 mL x 2) = the volume needed to titrate the carbonate. (It take 17 mL to titrate the carbonate half way so it takes 34 mL to titrate it all the way). So 43.8-34 = 9.8 mL to titrate the HCO3^- that was in the sample initially.
grams HCO3^- = 9.8 x 0.1 x 0.084 = g.
%NaHCO3 = (mass NaHCO3/mass sample)*100 = ?

Answer 2

thank you DrBob222