Lemon juice has actually a pH of around 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 thickness 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 compelled 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.

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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

claims it"s wrong..

Edited October 26, 2008 by delco714

### John Cuthber

Posted October 26, 2008

### John Cuthber

Chemistry Expert Resident Experts 3871
Posted October 26, 2008

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 have the right to calculate the volume of that without needing the molar mass of NaOH.

0.0000997mol/0.0100 mol/litre =x mol

### delco714

Posted October 26, 2008

### delco714

Quark Members 10
Author
Posted October 26, 2008

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

### John Cuthber

Posted October 26, 2008

### John Cuthber

Chemisattempt Expert Resident Experts 3871
Posted October 26, 2008

Oops! that last unit must be litres.

10 ml is a perfectly reasonable answer for a titration.

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