[Giovanni]

Hello, I am new to this forum and would just like some help with a direct titration i have recently conducted. I used NaOH to neutralize tartaric acid to prove it was diprotic. I had NaOH in the burette and 20ml of acid in a conical flask. i made standard solutions of both the base and acid. I used NaOH pellets to make a 200ml standard solution of NaOH. here are my calculations:

Before titration:
NaOH : weight=0.796g, relative molecular mass= 39.98, mol=0.019, concentration= 0.0995 mol/L.

C4H6O6: 20ml of concentration 0.5mol/L,  made a 200ml standard solution, so x10 dilution new v=200ml new concentration= 0.05mol/L.

After titration:

i got an average titre of 21.86ml. Now i have done my calculations but i am confused now because it does not lead to the mol ratio 2 NaOH:1 C4H6O6 because  C4H6O6 needs to give off two protons to NaOH

reaction i worked out to be:  H2C4H4O6+ 2NaOH → Na2C4H4O6+2H2O but i have to show through calculations

so i was wondering have i made any mistakes and what do my final calculations have to be

thank you, Giovanni

[Arkcon]

Well, there's plenty of things for us to work on.  But let's see your calculation: 21.86 ml of your NaOH solution  is how many moles?  To neutralize how many moles of acetic acid?  What is the mole ratio that you get?  1.5, 1.79, what?

[Giovanni]

ok just did it again now with a fresh mind:

naoh: v=21.86, c=0.0995 therefore n=0.00217

acid: c=0.05 v=0.02 n=0.001

so ratio is roughly 2:1, i hope this is correct and thank you for yourhelp