Titration/Molarity Help - Urgent (1 Viewer)

[Serdain]

The question is from Conquering Chem 2, question 19 chapter 5.

"The standard Sodium Hydroxide Solution from Excercise 17 (0.165 Mol/L) was used to determine the concentration of acetic acid in a particular brand of vinegar. 50 mL of the vinegar (by pipette) was accurately diluted to 250 mL (Volumetric flask), and then 25 mL aliquots of this diluted solution were titrated with the sodium hydroxide solution. The results of successive titrations were 21.95, 21.60, 21.55 and 21.60 mL. Calculate the molarity and %(w/v) of acetic acid in the original vinegar."

I've tried this a dozen times and I keep getting 0.07128 Mol/L when the answers say 0.712 mol/L. I haven't tried the w/v thing yet because I can't get the right answer. My mid-term exams are tomorrow so any help asap would be greatly appreciated, thanks :worried:

 

[Affinity]

you need one mole of caustic soda for one mole of vinegar, you can roughly see that the book's answer's right since:
you need approximately the same amount of sodium hydroxide solution to neutralise 25 ml of diluted acid... so the diluted acid is about 0.165 mol to get the concentration for the original, one multiplies that by 5.

calculations:
average amount of sodium hydroxide needed:
21.675
so concentration of acid should be 0.165*21.675/25 =0.143055
and to get the original multiply by 5. giving .715 mol

the book's solution probably considers 21.95 to be an outlier and therefore excluded...... which IMHO is.... ^&^&**^%$%^

 

[Serdain]

Ah that was it. The times by 5 and the outlier thing. Thanks so much, saved my life.