1. Consider a 1.00g unknown sample that contains 62.0% sodium carbonate and 38.0% sodium bicarbonate, NaHCO3, by mass. When this sample is dissolved in water, how many mL of 0.200M HCl will be required to convert all of the carbonate to bicarbonate? How many mL will be required to convert the resulting bicarbonate solution to carbonic acid, H2CO3(aq)?

2. Based on your answer to question 1, what feature or features of your titration curve will tell you whether you have a single base or a mixture of bases?

I have done titrations before but I don't understand what to do with the percentages and what they really mean. As well as answering the question, if possible, explain what the percentages mean. Thanks!

1 answer

The percentages just tell you how much Na2CO3 and NaHCO3 you start with.
1 x 0.62 =0.62g Na2CO3.
1 x 0.38 = 0.38g NaHCO3.

I will estimate from here but you confirm and use better accuracy. It makes the fractions come out better.

mols Na2CO3 = 0.62/106 = 0.00585
mols NaHCO3 = 0.38/84 = 0.00452
So we place 0.00585 mols Na2CO3 and 0.00452 mols NaHCO3 in a flask and titrate with 0.2M HCl.

Na2CO3 + HCl ==> NaHCO3 + NaCl. Note that the NaHCO3 initially there isn't titrated at all.
It will take 0.00585 mols HCl to change all of the CO3^2- to HCO3^-. M = mols/L so
L HCl = mols/M = 0.00585/0.2 = 0.0292L or 29.2 mL to titrate all of the carbonate to bicarbonate.

For part 2 of #1, we now have 0.00585 mols HCO3^- from the CO3^- titration (halfway) + 0.00452 mols HCO3^- there initially for a total of 0.0104
NaHCO3 + HCl ==> NaCl + H2O + CO2
0.0104 mol NaHCO3 will require 0.0104 mols 0.2M HCl. How many L is that?
M = mols/L or L = mols/M = 0.0104/0.2 = 0.05185 = 51.85 mL of 0.2M HCl for the second titration. So total = 29.2 from first + 51.85 for the second = about 81 mL. What message do you get from this? If the mixture were not a mixture (all Na2CO3) how much would it have taken? If it took 29.2 for the first H^+ to be added ti must take double that or 58.2 total. BUT the total is 81 so something must have been mixed with the Na2CO3 to start. These numbers will come out more exact if you use better numbers as you go through the calculation.