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Giovanni

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Titration Calculation

« on: February 26, 2015, 08:08:14 pm » 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 NaOHreaction i worked out to be: H2C4H4O6+ 2NaOH → Na2C4H4O6+2H2O but i have to show through calculationsso i was wondering have i made any mistakes and what do my final calculations have to bethank you, Giovanni Logged

Arkcon

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Re: Titration Calculation

« Reply #1 on: February 26, 2015, 08:57:51 pm » 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? Logged Hey, I'm not judging. I just like to shoot straight. I'm a man of science.

Giovanni

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Re: Titration Calculation

« Reply #2 on: February 26, 2015, 09:05:35 pm » ok just did it again now with a fresh mind:naoh: v=21.86, c=0.0995 therefore n=0.00217acid: c=0.05 v=0.02 n=0.001so ratio is roughly 2:1, i hope this is correct and thank you for yourhelp Logged

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