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Acid-base titration calculations

jamiescriven

Any help with question 2 would be hugely appreciated! Thank you.

Reply 1

jamiescriven

Original post by JimJam456

So your two reactants are HCl and NaHCO3.

(33.2X0.1)/1000= 3.32x10-3 mols of HCl.
The molar ratio is 1:1 so there are also 3.32x10-3 mols of NaHCO3
This is for 25cm3 however so multiply by 10 which gives 0.332 mols of NaHCO3 in 250cm3.
This needs to be converted into grams:
The Mr for NaHCO3 is 84.
84X0.332= 2.7888g of NaHCO3 in the tablet.
(2.7888/mass of tablet in total)X100= 58.6% which is the % by mass of NaHCO3 in the tablet. :smile:

NaHCO3 is sodium bicarbonate right? But in the question it says the second reactant is hydrated sodium carbonate, so wouldn't that be NaCO3.xH2O?

Reply 2

B_9710

Original post by jamiescriven

NaHCO3 is sodium bicarbonate right? But in the question it says the second reactant is hydrated sodium carbonate, so wouldn't that be NaCO3.xH2O?

It would but you're only interested in finding the mass of NaHCO₃ and then comparing it to the actual mass of the tablet, so the actual structure of the hydrated bicarbonate is irrelevant.