Question

what is the molecular weight (G/mol) of a diprotic acid 0.43 sample that needed 35.46 ml of 0.100 M of NaOH solution to reach a phenolphthalein point?

moles of NaOH = .100 x 35.46 / 1000 = 0.003546

2NaOH + H2A = Na2A + 2H2O so moles of H2A = 0.001773

these are in .43 g so mass of 1 mole = .43 /0.001773

= 242.53 g / mole .. molar mass of the acid

Related Questions

A 3.455 g sample of vinegar is titrated. It requires 16.83 mL of 0.2145 M NaOH to get to the phenolp... A 10 mL sample of tartaric acid (a diprotic acid) is titrated with 20 mL of 1.0 M NaOH. What is the... SOMEBODY PLEASE HELP ME! This is my last question for the chemistry and thank you. :D A BS Medic... A BS Medical Technology student was asked to determine the molecular weight of a diprotic acid using...