Vinegar analysis help!!!

[delco714]

Lemon juice has a pH of about 2.5. Assuming that the acidity of lemon juice is due solely to citric acid, that citric acid is a monoprotic acid, and that the density of lemon juice is 1.0 g/mL, then the citric acid concentration calculates to 0.5% by mass. Estimate the volume of 0.0100 M NaOH required to neutralize a 3.71-g sample of lemon juice. The molar mass of citric acid is 190.12 g/mol. ANSWER IN ml of 0.0100 M NaOH.

 

i did:

 

3.71g X 0.005 = 0.0186 g citric acid

0.0186g / (190.12g/mole) = 0.0000977 mole

0.0000977 mole X 39.997 g/mole NaOH = 0.00391g NaOH

V = 0.00391 / 0.0100 M = .39008 L NaOH

= 390.08 mL NaOH

 

says it's wrong..

Edited October 26, 2008 by delco714

[John Cuthber]

I think you have multiplied by the molar mass of NaOH twice.

.0000977 moles of acid takes the same number of moles of NaOH and you can calculate the volume of that without needing the molar mass of NaOH.

0.0000997mol/0.0100 mol/litre =x mol

[delco714]

but... MOLE / (MOLE/L) = x Liters...which is 0.00997 L.. which does = 9.97 mL... did you mean Liters and mistakenly put mole?

[John Cuthber]

Oops! that last unit should be litres.

10 ml is a perfectly reasonable answer for a titration.