Calculations for titration.

Titration of Tartaric acid calculations help!?

• Please help on my titration lab, I've worked on it for a while and it's due later today!! The following list is what I have to do: Moles of NaOH which reacted: Moles of tartaric acid which reacted with OH^-(mol) Mass of tartaric acid present in weighed sample of unknown Percent tartaric acid by mass in unknown What I have for my data are: Mass of unknown acid(tartaric acid): 1.0112g NaOH solution: 27mL. Please help! I'll greatly appreciate it! Thank you!!

• Answer:

  OK here is the balanced equation C4H6O6 + 2NaOH -> Na2C4H4O6 + 2H2O moles of NaOH at the equivalence point = .12 x 27 / 1000 = 0.00324 moles so moles of the acid at the eq point = 0.00324 /2 = 0.00162 moles Molar mass of C4H6O6 is 150.1 g/mol so mass of acid in the titration = 0.24316 g % acid in the sample = 0.24316 / 1.0112 = 24.0 % there is a vote button for you to give me the points thanks ..