How to calculate the weightpercent Na2CO3 in impure base sample???
I did my calculations but itsomehow doesnt make sense
First we find the moles of HCl
=0.1M HCl x volume of HCl used to titrate
      =0.1M x 10.5mL HCl
      =1.05 mol HCl
The ratio between Na2CO3 and Hcl is 1:2
Moles Na2CO3
= moles of HCl/2
=1.05mol HCl/2
=0.525mol Na2CO3
Mass Na2CO3
= moles of Na2CO3 x 105.9906 g/mol
      =0.525 mol Na2CO3 x105.9906 g/mol
      =55.65 g
%Â Â Â = mass of Na2CO3 x100/mass of sample fromthe impure solid
      =55.65g x 100/ 0.1328g =41905.1205 ???
What did I do wrong??? (do I need to include the fact that thatthe impure base was first mixed with 25 mL of distilled water???The HCl used is 0.1M for titrate to its endpoint)
Impure base | 1 |
Mass (g) | 0.1328 |
Trial 1 volume of HCl used in titration (mL) | 10.5 |
Trial 2 /2nd titration(after boiling) : volume of HCl used intitration (mL) | 3 |
